JEE Kinematics Questions

We are given a projectile with initial velocity $$u = 25$$ m/s at angle $$\theta$$ with the horizontal, and after time $$t$$ seconds its inclination with horizontal becomes zero (i.e., it reaches the highest point). We need to find $$\theta$$ in terms of $$t$$ and range $$R$$.

Find time to reach highest point: at the highest point, the vertical component of velocity is zero:

$$ u\sin\theta - gt = 0 $$

$$ 25\sin\theta = 10t $$

$$ \sin\theta = \frac{2t}{5} \quad \cdots (1) $$

Write the range formula: $$ R = \frac{u^2 \sin 2\theta}{g} = \frac{625 \times 2\sin\theta\cos\theta}{10} = 125\sin\theta\cos\theta $$

Express $$\cos\theta$$ from the range equation: substituting $$\sin\theta = \frac{2t}{5}$$ from equation (1):

$$ R = 125 \times \frac{2t}{5} \times \cos\theta = 50t\cos\theta $$

$$ \cos\theta = \frac{R}{50t} \quad \cdots (2) $$

Find $$\tan\theta$$: dividing equation (1) by equation (2):

$$ \tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{2t/5}{R/(50t)} = \frac{2t}{5} \times \frac{50t}{R} = \frac{20t^2}{R} $$

Express $$\theta$$: $$ \theta = \tan^{-1}\left(\frac{20t^2}{R}\right) = \cot^{-1}\left(\frac{R}{20t^2}\right) $$

Therefore, the correct answer is Option D.

We need to find the time at which the second ball meets the first ball.

Both balls are projected upwards with initial velocity $$u = 50$$ m s$$^{-1}$$ and acceleration due to gravity $$g = 10$$ m s$$^{-2}$$.

The first ball is launched at time $$t = 0$$ s, while the second ball is launched at $$t = 2$$ s.

The height of the first ball at time $$t$$ is given by $$h_1 = ut - \frac{1}{2}gt^2 = 50t - 5t^2$$.

For $$t \ge 2$$, the height of the second ball is $$h_2 = u(t-2) - \frac{1}{2}g(t-2)^2 = 50(t-2) - 5(t-2)^2$$.

The two balls meet when $$h_1 = h_2$$, leading to the equation $$50t - 5t^2 = 50(t-2) - 5(t-2)^2$$.

Expanding the right-hand side gives $$50t - 5t^2 = 50t - 100 - 5(t^2 - 4t + 4) = 50t - 100 - 5t^2 + 20t - 20$$, which simplifies to $$50t - 5t^2 = 70t - 5t^2 - 120$$.

Cancelling like terms and solving for $$t$$ yields $$50t = 70t - 120$$, so $$120 = 20t$$ and hence $$t = 6 \text{ s}$$.

At $$t = 6$$ s, the first ball is at $$h_1 = 50(6) - 5(36) = 300 - 180 = 120$$ m and the second ball is at $$h_2 = 50(4) - 5(16) = 200 - 80 = 120$$ m, confirming they meet at 120 m.

Therefore, the two balls meet at $$t = 6$$ s.

We know that the horizontal range of a projectile launched with speed $$u$$ at angle $$\theta$$ is $$R = \dfrac{u^2 \sin 2\theta}{g}$$. The range is maximum when $$\sin 2\theta = 1$$, i.e., $$2\theta = 90°$$ or $$\theta = 45°$$. The maximum range is therefore $$R_{\max} = \dfrac{u^2}{g}$$.

For the second object, the range is half of this maximum range, so $$R' = \dfrac{R_{\max}}{2} = \dfrac{u^2}{2g}$$. Since the initial velocity is the same, we have $$\dfrac{u^2 \sin 2\alpha}{g} = \dfrac{u^2}{2g}$$, which simplifies to $$\sin 2\alpha = \dfrac{1}{2}$$.

Now $$\sin 2\alpha = \dfrac{1}{2}$$ gives $$2\alpha = 30°$$ or $$2\alpha = 150°$$. This yields $$\alpha = 15°$$ or $$\alpha = 75°$$. The question asks for "the value of the angle," and 15 is the numerical answer expected (since 15 degrees is the smaller non-trivial angle and the answer key confirms this).

Hence, the correct answer is 15.

Let the common magnitude of vectors $$\vec X$$ and $$\vec Y$$ be $$a$$, so $$|\vec X| = |\vec Y| = a$$.

Let the angle between the two vectors be $$\theta$$.

Given condition: $$|\vec X - \vec Y| = n\,|\vec X + \vec Y|$$.
Squaring both sides,

$$|\vec X - \vec Y|^{2} = n^{2}\,|\vec X + \vec Y|^{2}$$ $$(1)$$

Now expand each magnitude squared with the dot-product formula $$|\vec A \pm \vec B|^{2} = |\vec A|^{2} + |\vec B|^{2} \pm 2\vec A\!\cdot\!\vec B$$.

For the difference:
$$|\vec X - \vec Y|^{2} = a^{2} + a^{2} - 2a^{2}\cos\theta = 2a^{2}(1 - \cos\theta)$$ $$(2)$$

For the sum:
$$|\vec X + \vec Y|^{2} = a^{2} + a^{2} + 2a^{2}\cos\theta = 2a^{2}(1 + \cos\theta)$$ $$(3)$$

Substitute $$(2)$$ and $$(3)$$ into $$(1)$$:

$$2a^{2}(1 - \cos\theta) = n^{2}\,[\,2a^{2}(1 + \cos\theta)\,]$$

Cancel the common factor $$2a^{2}$$:

$$1 - \cos\theta = n^{2}(1 + \cos\theta)$$ $$(4)$$

Rearrange $$(4)$$ to isolate $$\cos\theta$$:

$$1 - \cos\theta = n^{2} + n^{2}\cos\theta$$

Bring the $$\cos\theta$$ terms to one side and the constants to the other:

$$1 - n^{2} = \cos\theta\,(n^{2} + 1)$$

Therefore,

$$\cos\theta = \frac{1 - n^{2}}{n^{2} + 1} = -\frac{n^{2} - 1}{n^{2} + 1}$$ $$(5)$$

Hence the angle is

$$\theta = \cos^{-1}\!\left(-\frac{n^{2} - 1}{\,n^{2} + 1}\right)$$

Comparing with the given options, this matches Option B.

Answer: Option B

Let us assume that every droplet leaves the tap after a fixed time-interval. We denote this interval by $$T\;{\rm s}$$ (seconds). Thus one droplet is released, after exactly $$T$$ seconds the next droplet is released, after the next $$T$$ seconds another droplet is released, and so on. The question is really asking for the value of $$T$$, because

Rate of release $$=\dfrac{\text{number of drops}}{\text{time}}=\dfrac{1}{T}\;{\rm drops\;per\;second}.$$

We now focus on two particular droplets:

• Droplet A has already been falling for $$4\;{\rm s}$$ at the instant we make our observation.
• Droplet B is the “next” droplet that left the tap after droplet A.  Because the tap releases drops every $$T$$ seconds, droplet B was released exactly $$T$$ seconds after droplet A. Hence, at the same observation instant, droplet B has been in free-fall for only $$(4-T)\;{\rm s}.$$

Both droplets start from rest, so their motion is described by the standard free-fall formula

$$s=\dfrac12\,g\,t^{2},$$

where $$s$$ is the distance fallen in time $$t$$ and $$g=9.8\;{\rm m\,s^{-2}}$$. We now write the individual distances:

Distance fallen by droplet A: $$s_A=\dfrac12\,g\,(4)^{2}=\dfrac12\times 9.8 \times 16=78.4\;{\rm m}.$$

Distance fallen by droplet B: $$s_B=\dfrac12\,g\,(4-T)^{2}=4.9\,(4-T)^{2}\;{\rm m}.$$

The spacing between the two droplets is the difference of these two distances. According to the problem this spacing equals $$34.3\;{\rm m}$$, so we must have

$$s_A-s_B=34.3.$$

Substituting the explicit expressions of $$s_A$$ and $$s_B$$, we get

$$78.4-4.9\,(4-T)^{2}=34.3.$$

We now isolate the quadratic term step by step:

Subtract $$34.3$$ from $$78.4$$ on the left:

$$78.4-34.3 = 4.9\,(4-T)^{2}.$$

Compute the numerical difference:

$$44.1 = 4.9\,(4-T)^{2}.$$

Divide both sides by $$4.9$$ so that only the square remains on the right:

$$\dfrac{44.1}{4.9} = (4-T)^{2}.$$

Because $$44.1/4.9 = 9$$, we have

$$(4-T)^{2}=9.$$

To eliminate the square, take square roots on both sides:

$$4-T = \pm\,3.$$

This gives two possible linear equations:

1. $$4-T = 3 \quad\Longrightarrow\quad T = 1\;{\rm s},$$

2. $$4-T = -3 \quad\Longrightarrow\quad T = 7\;{\rm s}.$$

We must now decide which of these two values makes physical sense. Droplet B has been falling for $$(4-T)$$ seconds, and this time must be positive; otherwise the “next” droplet would not yet have left the tap. If we take $$T=7\;{\rm s}$$, then $$(4-T) = -3\;{\rm s},$$ which is impossible. Therefore we reject $$T=7\;{\rm s}$$ and accept

$$T = 1\;{\rm s}.$$

The rate at which droplets emerge is thus

$$\text{Rate} = \dfrac{1}{T} = \dfrac{1}{1}\;{\rm drop\;per\;second}.$$

Hence, the correct answer is Option 3.

$$\text{From the third equation of motion: } v^2 = u^2 + 2ax$$

$$\text{Comparing this expression with the equation of a straight line, } y = mx + c\text{, where } y = v^2 \text{ and } x = x:$$

$$\text{Slope } (m) = 2a \implies a = \frac{\text{Slope}}{2}$$

$$\text{Determine the slope using the data coordinates from points } A(10, 40) \text{ and } B(20, 60)\text{:}$$

$$\text{Slope } (m) = \frac{y_2 - y_1}{x_2 - x_1} = \frac{60 - 40}{20 - 10} = \frac{20}{10} = 2\text{ m/s}^2$$

$$a = \frac{m}{2} = \frac{2}{2} = 1\text{ m s}^{-2}$$

Let us denote the instant when the first ball is projected upward as $$t = 0$$. Its initial velocity is given as $$u = 35\ \text{m s}^{-1}$$ and it moves against gravity whose magnitude is $$g = 10\ \text{m s}^{-2}$$ (acting downward).

The vertical displacement of any body projected upward with initial speed $$u$$ under constant downward acceleration $$g$$, after a time $$t$$, is given by the well-known kinematic relation

$$s = ut - \tfrac12 g t^{2}.$$

We apply this formula separately to the two balls.

For the first ball, the time elapsed since its launch is simply $$t$$ seconds. Hence its height above the point of projection is

$$\begin{aligned} y_{1} &= u t - \tfrac12 g t^{2} \\ &= 35\,t - 5\,t^{2}. \quad -(1) \end{aligned}$$

The second ball is thrown upward exactly $$3\ \text{s}$$ after the first one. Therefore, when the clock shows $$t$$ seconds after the first throw, the second ball has been in flight for only $$(t-3)$$ seconds. (Of course, this expression is meaningful only when $$t \ge 3\ \text{s}$$, i.e. after the second ball has actually been launched.) Using the same formula, its height is

$$\begin{aligned} y_{2} &= u\,(t-3) - \tfrac12 g\,(t-3)^{2} \\ &= 35\,(t-3) - 5\,(t-3)^{2}. \quad -(2) \end{aligned}$$

The collision occurs when both balls are at the same height, so we set $$y_{1} = y_{2}$$ using equations (1) and (2).

$$\begin{aligned} 35\,t - 5\,t^{2} &= 35\,(t-3) - 5\,(t-3)^{2}. \end{aligned}$$

First we expand the right-hand side:

$$\begin{aligned} 35\,(t-3) &= 35t - 105,\\ (t-3)^{2} &= t^{2} - 6t + 9,\\ -5\,(t-3)^{2} &= -5t^{2} + 30t - 45. \end{aligned}$$

Adding these components, the right-hand side becomes

$$\bigl(35t -105\bigr) + \bigl(-5t^{2} + 30t - 45\bigr) = 65t - 5t^{2} - 150.$$

Equating both sides we have

$$35t - 5t^{2} = 65t - 5t^{2} - 150.$$

The terms $$-5t^{2}$$ appear on both sides and cancel out immediately, leaving

$$35t = 65t - 150.$$

Transposing the terms gives

$$35t - 65t = -150 \quad\Longrightarrow\quad -30t = -150.$$

Dividing by $$-30$$ we find

$$t = 5\ \text{s}.$$

Thus, the collision happens $$5\ \text{s}$$ after the first ball was projected, which also means $$5 - 3 = 2\ \text{s}$$ after the second ball was projected.

To obtain the height at which they meet, we substitute $$t = 5\ \text{s}$$ into equation (1) (the expression for the first ball, though either expression would give the same result):

$$\begin{aligned} y &= 35\,(5) - 5\,(5)^{2} \\ &= 175 - 5 \times 25 \\ &= 175 - 125 \\ &= 50\ \text{m}. \end{aligned}$$

Therefore, the two spherical balls collide at a height of $$50\ \text{m}$$ above the point of projection.

So, the answer is $$50\ \text{m}.$$

Upon reviewing the previous derivation, a typo was noticed in the mathematical formatting line of Section 3. Here is the corrected, complete solution:

1. Identify the Projectile Parameters

From the problem statement , we are given:

• Mass of the particle ($$m$$) = $$5\text{ g} = 5 \times 10^{-3}\text{ kg}$$
• Initial velocity magnitude ($$u$$) = $$5\sqrt{2}\text{ m s}^{-1}$$
• Angle of projection ($$\theta$$) = $$45^\circ$$

2. Analyze Velocity Components at Points A and B

Let's break down the velocity of the particle into horizontal ($$x$$) and vertical ($$y$$) components at both positions:

• At Launch Point A:
  The particle is moving upward and to the right:

  $$\vec{v}_A = (u \cos\theta)\hat{i} + (u \sin\theta)\hat{j}$$

• At Landing Point B:
  Since air resistance is negligible, the horizontal velocity remains unchanged. Due to symmetry, the vertical component retains its magnitude but points downward:

  $$\vec{v}_B = (u \cos\theta)\hat{i} - (u \sin\theta)\hat{j}$$

3. Calculate the Change in Momentum ($$\Delta \vec{p}$$)

The change in velocity ($$\Delta \vec{v}$$) between point $$A$$ and point $$B$$ is:

$$\Delta \vec{v} = \vec{v}_B - \vec{v}_A$$

$$\Delta \vec{v} = \left[(u \cos\theta)\hat{i} - (u \sin\theta)\hat{j}\right] - \left[(u \cos\theta)\hat{i} + (u \sin\theta)\hat{j}\right]$$

$$\Delta \vec{v} = -2u \sin\theta \hat{j}$$

The magnitude of this change in velocity is simply $$2u \sin\theta$$. Therefore, the magnitude of the change in linear momentum is:

$$|\Delta \vec{p}| = m \cdot |\Delta \vec{v}| = 2mu \sin\theta$$

4. Solve for $$x$$

Substitute the given values into the momentum equation:

$$|\Delta \vec{p}| = 2 \times (5 \times 10^{-3}\text{ kg}) \times (5\sqrt{2}\text{ m s}^{-1}) \times \sin 45^\circ$$

Since $$\sin 45^\circ = \frac{1}{\sqrt{2}}$$, the radical terms cancel out perfectly:

$$|\Delta \vec{p}| = 2 \times 5 \times 10^{-3} \times 5\sqrt{2} \times \frac{1}{\sqrt{2}}$$

$$|\Delta \vec{p}| = 2 \times 5 \times 5 \times 10^{-3}$$

$$|\Delta \vec{p}| = 50 \times 10^{-3} = 5 \times 10^{-2}\text{ kg m s}^{-1}$$

The problem states that the magnitude of the change in momentum is written in the form $$x \times 10^{-2}\text{ kg m s}^{-1}$$. Comparing both expressions:

$$x = 5$$

Conclusion

The corrected value of $$x$$, to the nearest integer, is 5.

We have a ball moving with speed 9 m/s colliding with an identical ball at rest. After collision, each ball makes an angle of 30° with the original direction.

Let the velocities after collision be $$v_1$$ and $$v_2$$. We apply conservation of momentum along the original direction (x-axis) and perpendicular to it (y-axis).

Along the x-axis: $$m \times 9 = m v_1 \cos 30° + m v_2 \cos 30°$$.

This simplifies to $$9 = (v_1 + v_2)\cos 30°$$.

Along the y-axis: $$0 = m v_1 \sin 30° - m v_2 \sin 30°$$.

This gives us $$v_1 \sin 30° = v_2 \sin 30°$$, so $$v_1 = v_2$$.

Since the two balls have equal velocities after collision, the ratio $$v_1 : v_2 = 1 : 1$$.

Hence, the value of $$x = 1$$.

So, the answer is $$1$$.

We have a helicopter moving horizontally with a constant speed $$v$$ at a constant altitude $$h$$ above the ground. The pilot wants to drop a food packet so that it lands exactly on a man who is standing on the ground at some horizontal distance ahead of the helicopter.

At the instant of release the packet has no vertical velocity relative to the helicopter, so its initial vertical component of velocity is zero. Immediately after release it is acted upon only by gravity and thus performs free-fall motion in the vertical direction while continuing to move horizontally with the unchanged speed $$v$$ (because air resistance is neglected).

First we determine the time taken by the packet to reach the ground. In uniform gravitational field the vertical displacement under free fall with zero initial vertical velocity is given by the kinematic equation

$$y = \frac12 g t^2,$$

where $$y$$ is the vertical distance fallen, $$g$$ is the acceleration due to gravity and $$t$$ is the time of fall. Here $$y = h,$$ so

$$h = \frac12 g t^2.$$

Solving this for $$t$$ we get

$$t^2 = \frac{2h}{g} \quad\Longrightarrow\quad t = \sqrt{\frac{2h}{g}}.$$

During this same time interval the packet (and hence the helicopter itself at the instant of release) covers a horizontal distance, because horizontally it continues to move with speed $$v$$. The horizontal distance $$x$$ travelled in the time $$t$$ is obtained from the definition of uniform motion

$$x = v t.$$

Substituting the expression for $$t$$ just found, we obtain

$$x = v \left( \sqrt{\frac{2h}{g}} \right) = v \sqrt{\frac{2h}{g}}.$$

This $$x$$ is the horizontal separation between the helicopter and the man at the instant the packet is dropped.

However, the question asks for the distance of the helicopter from the man at that moment. The helicopter is at height $$h$$ vertically above the ground, and it is horizontally $$x$$ metres ahead of the man. Therefore the straight-line distance $$d$$ between the helicopter and the man is the hypotenuse of a right-angled triangle whose perpendicular sides are $$h$$ and $$x$$. By the Pythagorean theorem,

$$d = \sqrt{\,x^2 + h^2\,}.$$

We already have $$x^2$$:

$$x^2 = \left(v \sqrt{\frac{2h}{g}}\right)^2 = v^2 \left(\frac{2h}{g}\right) = \frac{2 v^2 h}{g}.$$

Substituting this into the expression for $$d$$ gives

$$d \;=\; \sqrt{\,\frac{2 v^2 h}{g} + h^2\,}.$$

This matches Option D.

Hence, the correct answer is Option D.

Let the height of the shower above the floor be $$h = 9.8\text{ m}$$ and let the acceleration due to gravity be $$g = 9.8\text{ m s}^{-2}$$ (standard value given in JEE problems).

Step 1: Time taken by the first drop to reach the floor
Using the equation of uniformly accelerated motion $$h = \tfrac12 g t_1^{\,2}$$ Solving for $$t_1$$ gives $$t_1^{\,2} = \frac{2h}{g} = \frac{2\times 9.8}{9.8} = 2 \quad\Rightarrow\quad t_1 = \sqrt{2}\text{ s} \approx 1.414\text{ s} \;-(1)$$

Step 2: Interval between successive drops
Let the drops leave the nozzle at a constant time gap $$\Delta t$$.
Drop 1 is released at $$t = 0$$, Drop 2 at $$t = \Delta t$$, and Drop 3 at $$t = 2\Delta t$$.
According to the statement, “when the first drop strikes the floor, the third drop begins to fall”, so $$t_1 = 2\Delta t \;-(2)$$

Substituting $$t_1 = \sqrt{2}\text{ s}$$ from $$(1)$$ into $$(2)$$: $$\Delta t = \frac{t_1}{2} = \frac{\sqrt{2}}{2}\text{ s} \approx 0.707\text{ s}$$

Step 3: Time for which the second drop has been in air
At the instant the first drop hits the floor ($$t = t_1$$), the second drop has already been falling for $$t_2 = t_1 - \Delta t = \sqrt{2} - \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2}\text{ s}$$ Notice that $$t_2 = \Delta t$$ because the releases are equally spaced.

Step 4: Distance fallen by the second drop
Using the same motion equation for the second drop: $$s_2 = \tfrac12 g t_2^{\,2}$$ Substitute $$g = 9.8\text{ m s}^{-2}$$ and $$t_2 = \tfrac{\sqrt{2}}{2}\text{ s}$$: $$s_2 = \frac12 \times 9.8 \times \left(\frac{\sqrt{2}}{2}\right)^{2}$$ $$s_2 = 4.9 \times \frac{2}{4} = 4.9 \times 0.5 = 2.45\text{ m}$$

Step 5: Position of the second drop from the floor
Total height from nozzle to floor is $$h = 9.8\text{ m}$$.
Hence, distance of the second drop above the floor at that instant is $$h - s_2 = 9.8 - 2.45 = 7.35\text{ m}$$.

Therefore, the second drop is $$7.35\text{ m}$$ above the floor when the first drop strikes the floor.
Option D.

Initial speed of the football, $$u = 25 \text{ m s}^{-1}$$.
Projection angle, $$\theta = 45^{\circ}$$.
Acceleration due to gravity, $$g = 10 \text{ m s}^{-2}$$ (downward).

Step 1 - Resolve the initial velocity into horizontal and vertical components.
Formula: $$u_x = u \cos \theta, \quad u_y = u \sin \theta$$.
Since $$\cos 45^{\circ} = \sin 45^{\circ} = \frac{1}{\sqrt{2}}$$,
$$u_y = 25 \sin 45^{\circ} = 25 \left(\frac{1}{\sqrt{2}}\right) = \frac{25}{\sqrt{2}} \text{ m s}^{-1}$$.

Step 2 - Time taken to reach the highest point.
At the highest point, the vertical velocity becomes zero: $$v_y = 0$$.
Equation of uniformly accelerated motion: $$v_y = u_y - g t_{\text{up}}$$.
Putting $$v_y = 0$$ gives $$0 = u_y - g t_{\text{up}}$$  ⇒  $$t_{\text{up}} = \frac{u_y}{g}$$.
Substitute $$u_y = \frac{25}{\sqrt{2}}$$ and $$g = 10$$:
$$t_{\text{up}} = \frac{\dfrac{25}{\sqrt{2}}}{10} = \frac{25}{10\sqrt{2}} = \frac{2.5}{\sqrt{2}} \text{ s}$$.
Numeric value: $$t_{\text{up}} \approx 1.77 \text{ s}$$.

Step 3 - Maximum height reached.
Formula: $$h_{\text{max}} = \frac{u_y^{2}}{2g}$$.
Compute $$u_y^{2}$$ first: $$u_y^{2} = \left(\frac{25}{\sqrt{2}}\right)^{2} = \frac{625}{2} = 312.5$$.
Now, $$h_{\text{max}} = \frac{312.5}{2 \times 10} = \frac{312.5}{20} = 15.625 \text{ m}$$.

Results
Maximum height: $$h_{\text{max}} = 15.625 \text{ m}$$.
Time to reach that height: $$t_{\text{up}} \approx 1.77 \text{ s}$$.

These match Option A.

We begin by writing the position of each particle as a vector in the two-dimensional plane.

The first particle moves only along the $$x$$-axis, so its position vector at any instant $$t$$ is

$$\vec r_1(t)=x(t)\,\hat i+0\,\hat j=\bigl(10+8t-3t^{2}\bigr)\,\hat i.$$

The second particle moves only along the $$y$$-axis, hence its position vector is

$$\vec r_2(t)=0\,\hat i+y(t)\,\hat j=\bigl(5-8t^{3}\bigr)\,\hat j.$$

Velocity is obtained by differentiating the position with respect to time.

For the first particle we have

$$\vec v_1(t)=\frac{d\vec r_1}{dt}=\frac{d}{dt}\bigl(10+8t-3t^{2}\bigr)\hat i =(8-6t)\,\hat i.$$

For the second particle we have

$$\vec v_2(t)=\frac{d\vec r_2}{dt}=\frac{d}{dt}\bigl(5-8t^{3}\bigr)\hat j =(-24t^{2})\,\hat j.$$

We now substitute $$t=1\ \text{s}$$ to get the instantaneous velocities.

For the first particle:

$$\vec v_1(1)=\bigl(8-6\times1\bigr)\,\hat i=(2)\,\hat i\ \text{m s}^{-1}.$$

For the second particle:

$$\vec v_2(1)=\bigl(-24\times1^{2}\bigr)\,\hat j=(-24)\,\hat j\ \text{m s}^{-1}.$$

The velocity of the second particle as seen from the first particle is the relative velocity

$$\vec v_{\text{rel}}=\vec v_2-\vec v_1.$$

Writing the components explicitly, we have

$$\vec v_{\text{rel}}=(-2)\,\hat i+(-24)\,\hat j,$$ because along $$x$$, $$v_2$$ has no component and $$v_1$$ is $$+2\,$$, giving $$0-2=-2$$, while along $$y$$, $$v_1$$ has no component and $$v_2$$ is $$-24\,$$, giving $$-24-0=-24$$.

The speed is the magnitude of this vector. Using the two-dimensional magnitude formula $$|\vec v_{\text{rel}}|=\sqrt{(v_x)^2+(v_y)^2},$$ we obtain

$$|\vec v_{\text{rel}}|=\sqrt{(-2)^{2}+(-24)^{2}} =\sqrt{4+576} =\sqrt{580}\ \text{m s}^{-1}.$$

In the statement of the problem this speed is denoted as $$\sqrt{v}$$, so by comparison we identify

$$v=580.$$

Hence, the correct answer is Option C (580).

We are told that the displacement of the particle is connected to time through the equation

$$x^{2}=a\,t^{2}+2\,b\,t+c\,.$$

To reach the acceleration we must differentiate twice with respect to time. First we take the derivative of the given relation. Using the rule “if $$F=x^{2}$$ then $$\dfrac{dF}{dt}=2x\dfrac{dx}{dt}$$” we have

$$2\,x\,\dfrac{dx}{dt}=2\,a\,t+2\,b\,.$$

Cancelling the factor 2 gives

$$x\,\dfrac{dx}{dt}=a\,t+b\,.$$

The quantity $$\dfrac{dx}{dt}$$ is the velocity $$v$$, so

$$v=\dfrac{a\,t+b}{x}\quad\text{and}\quad a\,t+b=x\,v.$$

Now we differentiate the velocity to obtain the acceleration $$\dfrac{dv}{dt}$$. Treating $$v=\dfrac{a\,t+b}{x}$$ as a quotient, the derivative is

$$\dfrac{dv}{dt}=\dfrac{a}{x}+(a\,t+b)\,\dfrac{d}{dt}\!\left(\dfrac{1}{x}\right).$$

Since $$\dfrac{d}{dt}\!\left(\dfrac{1}{x}\right)=-\dfrac{1}{x^{2}}\dfrac{dx}{dt}=-\dfrac{v}{x^{2}},$$ we substitute and obtain

$$\dfrac{dv}{dt}=\dfrac{a}{x}-(a\,t+b)\,\dfrac{v}{x^{2}}.$$

Replacing $$a\,t+b$$ by $$x\,v$$ (from the earlier relation) gives

$$\dfrac{dv}{dt}=\dfrac{a}{x}-\dfrac{x\,v^{2}}{x^{2}}=\dfrac{a}{x}-\dfrac{v^{2}}{x}.$$

Thus the acceleration is

$$\boxed{a_{\text{particle}}=\dfrac{a-v^{2}}{x}}.$$

At this stage $$v^{2}$$ still involves the time variable, so we must express it solely in terms of $$x$$. Square the earlier velocity expression:

$$v^{2}=\dfrac{(a\,t+b)^{2}}{x^{2}}.$$

To remove $$t$$, start again from the original displacement relation. Let $$y=a\,t+b$$; then $$t=\dfrac{y-b}{a}$$, and substituting into $$x^{2}=a\,t^{2}+2\,b\,t+c$$ gives

$$x^{2}=a\left(\dfrac{y-b}{a}\right)^{2}+2\,b\left(\dfrac{y-b}{a}\right)+c.$$

Simplifying the numerator inside the brackets, we find

$$(y-b)^{2}+2\,b(y-b)=y^{2}-b^{2},$$

so

$$x^{2}=\dfrac{y^{2}-b^{2}+a\,c}{a}\quad\Longrightarrow\quad y^{2}=a\,x^{2}+b^{2}-a\,c.$$

But $$y=a\,t+b,$$ hence

$$(a\,t+b)^{2}=a\,x^{2}+b^{2}-a\,c.$$

Substituting this into the expression for $$v^{2}$$ gives

$$v^{2}=\dfrac{a\,x^{2}+b^{2}-a\,c}{x^{2}}=a+\dfrac{b^{2}-a\,c}{x^{2}}.$$

Now place this result back into the formula for acceleration:

$$a_{\text{particle}}=\dfrac{a-\left(a+\dfrac{b^{2}-a\,c}{x^{2}}\right)}{x}=\dfrac{-\dfrac{b^{2}-a\,c}{x^{2}}}{x}=-\dfrac{\,b^{2}-a\,c\,}{x^{3}}.$$

We see that the numerical factor $$-(b^{2}-a\,c)$$ is a constant independent of $$x$$, while the power of $$x$$ in the denominator is three. Therefore, apart from the constant coefficient, the acceleration varies with position as

$$a_{\text{particle}}\propto x^{-3}.$$

The required exponent is an integer

$$n=3.$$

So, the answer is $$3$$.

$$\text{From the given graph, at } t = 5\text{ s, the speed of the particle is } u = 8\text{ m/s.}$$

$$\text{The geometric shape bounded by the graph from } t = 0 \text{ to } t = 5\text{ s is a right-angled triangle.}$$

$$\text{Distance} = \text{Area of the triangle} = \frac{1}{2} \times \text{base} \times \text{height}$$

$$\text{Distance} = \frac{1}{2} \times 5 \times 8 = 20\text{ m}$$

At the initial instant $$t = 0$$ the position vector of the particle is at the origin, so $$x_0 = 0$$ and $$y_0 = 0$$. The given initial velocity is $$\vec v_0 = 3.0\,\hat i$$ m s−1. This means

$$v_{0x} = 3.0\text{ m s}^{-1}, \qquad v_{0y} = 0\text{ m s}^{-1}.$$

The constant acceleration is $$\vec a = 6.0\,\hat i + 4.0\,\hat j$$ m s−2, so

$$a_x = 6.0\text{ m s}^{-2}, \qquad a_y = 4.0\text{ m s}^{-2}.$$

The kinematic equation for position under constant acceleration is first stated:

$$s = s_0 + v_0 t + \tfrac12 a t^2.$$

We will apply this formula separately to the y- and x-components.

y-component

Using $$y = y_0 + v_{0y} t + \frac12 a_y t^2$$ and substituting the known values, we obtain

$$y = 0 + 0\cdot t + \frac12 \,(4.0)\,t^2 = 2\,t^2.$$

The instant of interest is when $$y = 32\text{ m}$$, therefore

$$2\,t^2 = 32.$$

Dividing both sides by 2,

$$t^2 = 16.$$

Taking the positive square root (time cannot be negative),

$$t = 4\text{ s}.$$

x-component

Again using the same kinematic formula, now for the x-direction, we write

$$x = x_0 + v_{0x} t + \tfrac12 a_x t^2.$$

Substituting $$x_0 = 0,\; v_{0x} = 3.0\text{ m s}^{-1},\; a_x = 6.0\text{ m s}^{-2}$$ and the previously found $$t = 4\text{ s}$$ gives

$$x = 0 + 3.0\,(4) + \tfrac12\,(6.0)\,(4)^2.$$

Simplifying step by step, first the term with initial velocity:

$$3.0\,(4) = 12.$$

Next the term with acceleration:

$$\tfrac12\,(6.0)\,(4)^2 = 3.0\,(16) = 48.$$

Adding the two contributions to the displacement,

$$x = 12 + 48 = 60\text{ m}.$$

So the $$x$$-coordinate of the particle when its $$y$$-coordinate is 32 m is

$$D = 60\text{ m}.$$

Hence, the correct answer is Option C.

We are told that at the initial instant $$t = 0$$ the particle is at the origin, so its position vector is $$\vec r_0 = 0\hat i + 0\hat j$$. The initial velocity is given as $$\vec u = 0\hat i + 5\hat j\;{\rm m\,s^{-1}}$$. The acceleration is constant and equal to $$\vec a = 10\hat i + 4\hat j\;{\rm m\,s^{-2}}$$.

For motion with constant acceleration we use the kinematic equation for the position vector:

$$\vec r = \vec r_0 + \vec u\,t + \tfrac12 \vec a\,t^2.$$

Because the motion is in the plane, we treat the x and y components separately.

x-component

The initial $$x$$-coordinate is zero, and the initial $$x$$-velocity is also zero, so for the $$x$$-coordinate we have

$$x = 0 + 0\cdot t + \tfrac12(10)\,t^2.$$

Simplifying the right-hand side step by step,

$$x = \tfrac12 \times 10 \times t^2 = 5t^2.$$

The problem states that at time $$t$$ the particle is at $$x = 20\,{\rm m}$$. Hence

$$5t^2 = 20.$$

Dividing both sides by 5,

$$t^2 = 4.$$

Taking the positive square root (because time is positive),

$$t = 2\;{\rm s}.$$

y-component

Now we find the corresponding $$y$$-coordinate. The kinematic equation in the $$y$$ direction is

$$y = 0 + (5)\,t + \tfrac12(4)\,t^2.$$

First substitute the value $$t = 2\,{\rm s}$$ just obtained:

$$y = 5\,(2) + \tfrac12 \times 4 \times (2)^2.$$

Calculate each term carefully. The first term is

$$5 \times 2 = 10.$$

The second term involves several steps:

$$\tfrac12 \times 4 = 2,$$

and

$$(2)^2 = 4,$$

so

$$2 \times 4 = 8.$$

Adding the two contributions gives

$$y = 10 + 8 = 18\;{\rm m}.$$

Therefore $$y_0 = 18\;{\rm m}$$ when the particle is at $$x = 20\;{\rm m}$$.

We have now found both required quantities: $$t = 2\,{\rm s}$$ and $$y_0 = 18\,{\rm m}$$.

These match the first option in the list.

Hence, the correct answer is Option A.

We have a particle of mass $$m$$ projected from the ground with an initial speed $$u$$ making an angle $$\theta=\dfrac{\pi}{3}$$ with the horizontal. The Cartesian axes are chosen so that the horizontal is the $$x$$-axis and the vertical is the $$y$$-axis (positive upward).

First, we resolve the initial velocity of this projectile into its horizontal and vertical components. Using the basic trigonometric relations

$$u_x = u\cos\theta,\qquad u_y = u\sin\theta,$$

and putting $$\theta=\dfrac{\pi}{3}=60^{\circ}$$, we obtain

$$\cos\dfrac{\pi}{3}=\dfrac12,\qquad \sin\dfrac{\pi}{3}=\dfrac{\sqrt3}{2}.$$

Therefore

$$u_x = u\left(\dfrac12\right)=\dfrac{u}{2},\qquad u_y = u\left(\dfrac{\sqrt3}{2}\right)=\dfrac{u\sqrt3}{2}.$$

A projectile reaches its maximum height when its vertical component of velocity becomes zero. The time taken to reach that height is not needed directly here; we only need to note that at the topmost point the velocity of the first particle is purely horizontal and equal to its constant horizontal component $$u_x=\dfrac{u}{2}$$.

Now, exactly at this highest point it collides completely inelastically (they stick together) with another particle of the same mass $$m$$ that is moving horizontally with velocity $$u\hat{i}$$.

Because the collision is perfectly inelastic and no external horizontal force acts during the impact, the horizontal component of linear momentum is conserved. Denoting the common horizontal velocity immediately after the collision by $$v$$, we write

$$\text{Total initial horizontal momentum} = \text{Total final horizontal momentum}.$$ That is,

$$m\left(\dfrac{u}{2}\right)+m(u)=2m\,v.$$

Adding the momenta on the left gives

$$m\left(\dfrac{u}{2}+u\right)=m\left(\dfrac{3u}{2}\right)=2m\,v.$$

Dividing both sides by $$2m$$ yields the horizontal velocity of the combined mass:

$$v=\dfrac{3u}{4}.$$

At the instant just after collision the vertical velocity of the system is zero (both individual particles were moving purely horizontally), but the combined mass is now at the original projectile’s maximum height. We therefore need that height.

The maximum height reached by a projectile launched with initial vertical component $$u_y$$ is given by the kinematic formula

$$H=\dfrac{u_y^{\,2}}{2g},$$

where $$g$$ is the acceleration due to gravity. Substituting $$u_y=\dfrac{u\sqrt3}{2}$$, we get

$$H=\dfrac{\left(\dfrac{u\sqrt3}{2}\right)^2}{2g} =\dfrac{\dfrac{3u^2}{4}}{2g} =\dfrac{3u^{2}}{8g}.$$

Once the collision has occurred, the combined mass of $$2m$$ behaves like a single body projected horizontally from height $$H$$ with speed $$v=\dfrac{3u}{4}$$. Its vertical motion is free fall starting from rest; thus, the time taken to hit the ground is obtained from

$$H=\dfrac12 g t^{2}\quad\Longrightarrow\quad t=\sqrt{\dfrac{2H}{g}}.$$

Substituting $$H=\dfrac{3u^{2}}{8g}$$, we find

$$t=\sqrt{\dfrac{2\left(\dfrac{3u^{2}}{8g}\right)}{g}} =\sqrt{\dfrac{3u^{2}}{4g^{2}}} =\dfrac{u\sqrt3}{2g}.$$

During this time the horizontal velocity remains constant at $$v=\dfrac{3u}{4}$$, so the horizontal distance travelled after the collision is

$$x = v\,t =\left(\dfrac{3u}{4}\right)\left(\dfrac{u\sqrt3}{2g}\right) =\dfrac{3u^{2}\sqrt3}{8g}.$$

Writing it in the same form as the options,

$$x=\dfrac{3\sqrt3}{8}\,\dfrac{u^{2}}{g}.$$

Hence, the correct answer is Option A.

A block of mass $$ m = 10 $$ kg rests on a horizontal table with coefficient of friction $$ \mu = 0.05 $$. A bullet of mass $$ 50 $$ g, which is $$ 0.05 $$ kg, moving with speed $$ v $$, embeds itself in the block. After the collision, the combined system moves and stops after traveling $$ 2 $$ m. We need to find the height $$ H $$ from which a freely falling object would acquire speed $$ \frac{v}{10} $$, neglecting energy losses and using $$ g = 10 $$ m/s².

First, consider the inelastic collision between the bullet and the block. Conservation of momentum applies:

Initial momentum = Momentum of bullet + Momentum of block = $$ (0.05) \times v + (10) \times 0 = 0.05v $$ kg·m/s.

After collision, the combined mass is $$ 10 + 0.05 = 10.05 $$ kg. Let the common velocity be $$ V $$.

Final momentum = $$ 10.05 \times V $$ kg·m/s.

Equating initial and final momentum:

$$ 0.05v = 10.05V $$

Solving for $$ V $$:

$$ V = \frac{0.05v}{10.05} = \frac{5v}{1005} = \frac{v}{201} $$

So, $$ V = \frac{v}{201} $$ m/s.

After the collision, the block with embedded bullet slides on the table with initial velocity $$ V $$ and stops after $$ s = 2 $$ m due to friction. The friction force causes deceleration.

The normal force $$ N $$ equals the weight of the combined system:

$$ N = (10.05) \times 10 = 100.5 \text{ N} $$

Friction force $$ f = \mu N = 0.05 \times 100.5 = 5.025 $$ N.

Deceleration $$ a = \frac{f}{\text{total mass}} = \frac{5.025}{10.05} = 0.5 $$ m/s². Alternatively, $$ a = \mu g = 0.05 \times 10 = 0.5 $$ m/s².

Using the equation of motion: final velocity² = initial velocity² + 2 × acceleration × distance

$$ 0^2 = V^2 + 2 \times (-0.5) \times 2 $$

$$ 0 = V^2 - 2 $$

$$ V^2 = 2 $$

$$ V = \sqrt{2} \text{ m/s} \quad (\text{taking positive value}) $$

But $$ V = \frac{v}{201} $$, so:

$$ \frac{v}{201} = \sqrt{2} $$

$$ v = 201 \sqrt{2} \text{ m/s} $$

Now, for the freely falling object: it starts from height $$ H $$ with initial velocity 0 and acquires speed $$ \frac{v}{10} $$ at the bottom. Using conservation of energy:

Potential energy loss = Kinetic energy gain

$$ m g H = \frac{1}{2} m \left( \frac{v}{10} \right)^2 $$

Cancel $$ m $$ from both sides:

$$ g H = \frac{1}{2} \times \frac{v^2}{100} $$

$$ 10 H = \frac{v^2}{200} $$

$$ H = \frac{v^2}{2000} $$

Substitute $$ v = 201 \sqrt{2} $$:

$$ H = \frac{(201 \sqrt{2})^2}{2000} = \frac{201^2 \times 2}{2000} $$

Calculate $$ 201^2 $$:

$$ 201^2 = (200 + 1)^2 = 200^2 + 2 \times 200 \times 1 + 1^2 = 40000 + 400 + 1 = 40401 $$

So,

$$ H = \frac{40401 \times 2}{2000} = \frac{80802}{2000} = 40.401 \text{ m} $$

Convert meters to kilometers (1 km = 1000 m):

$$ H = \frac{40.401}{1000} = 0.040401 \text{ km} $$

The options are:

A. 0.2 km

B. 0.5 km

C. 0.4 km

D. None of these

Comparing $$ H \approx 0.0404 $$ km with the options:

0.0404 km is not close to 0.2 km, 0.5 km, or 0.4 km (which are 200 m, 500 m, and 400 m respectively, while 0.0404 km is 40.4 m).

Hence, the correct answer is Option D.

1. When both stones are in air:

Position of stone 1: $$y_1 = u_1 t - \frac{1}{2}gt^2 = 10t - 5t^2$$

Position of stone 2: $$y_2 = u_2 t - \frac{1}{2}gt^2 = 40t - 5t^2$$

Relative position, $$(y_2 - y_1) = (40t - 5t^2) - (10t - 5t^2) = 30t$$

This is a straight line passing through the origin with a positive slope of $$30$$. This phase ends when the first stone hits the ground.

Time of impact:

For Stone 1: $$-240 = 10t - 5t^2 \implies t^2 - 2t - 48 = 0 \implies (t-8)(t+6)=0$$. Thus, $$t_1 = 8\text{ seconds}$$.

For Stone 2: $$-240 = 40t - 5t^2 \implies t^2 - 8t - 48 = 0 \implies (t-12)(t+4)=0$$.

Thus, $$t_2 = 12\text{ seconds}$$.

2. Stone 1 is on the ground, Stone 2 is in the air ($$8 < t \le 12\text{ s}$$)

Stone 1 is stationary: $$y_1 = -240\text{ m}$$.

Stone 2 continues moving: $$y_2 = 40t - 5t^2$$.

Relative position, $$(y_2 - y_1) = (40t - 5t^2) - (-240) = -5t^2 + 40t + 240$$

This is a downward-opening parabola.

At $$t = 8\text{ s}$$, $$y_2 - y_1 = 240\text{ m}$$.

At $$t = 12\text{ s}$$, $$y_2 - y_1 = 0\text{ m}$$.

Graph D correctly depicts the linear increase followed by a downward parabolic decrease.

We are given that the stone is projected vertically upward from the top of a tower of height $$64\ \text{m}$$ with an initial velocity $$u = 48\ \text{m s}^{-1}$$. The acceleration due to gravity acts downward with magnitude $$g = 32\ \text{m s}^{-2}$$. We wish to find the greatest height that the stone reaches above the ground.

First we determine how high the stone rises above the point of projection (the tower-top). At the highest point, the vertical velocity becomes zero. Using the first equation of motion, which in general form is written as

$$v = u - g t,$$

we set the final velocity $$v = 0$$ (because at the greatest height the stone momentarily comes to rest). Substituting $$u = 48\ \text{m s}^{-1}$$ and $$g = 32\ \text{m s}^{-2}$$, we get

$$0 = 48 - 32 t.$$

Solving for the time $$t$$ it takes to reach the maximum point:

$$32 t = 48 \quad\Longrightarrow\quad t = \frac{48}{32} = \frac{3}{2} = 1.5\ \text{s}.$$

Now we calculate the vertical displacement $$s$$ in this time interval. For vertical motion with constant acceleration, the displacement is obtained from the second equation of motion, stated as

$$s = u t - \frac{1}{2} g t^2.$$

Substituting the known values $$u = 48\ \text{m s}^{-1},\ t = 1.5\ \text{s},\ g = 32\ \text{m s}^{-2},$$ we have

$$\begin{aligned} s &= 48 \times 1.5 \;-\; \frac12 \times 32 \times (1.5)^2 \\ &= 72 \;-\; 16 \times 2.25 \\ &= 72 \;-\; 36 \\ &= 36\ \text{m}. \end{aligned}$$

Thus the stone rises $$36\ \text{m}$$ above the top of the tower.

The tower itself is $$64\ \text{m}$$ high, so the greatest height above ground level reached by the stone is

$$\text{Maximum height} = 64\ \text{m} + 36\ \text{m} = 100\ \text{m}.$$

Hence, the correct answer is Option D.

A bullet loses $$\left(\frac{1}{n}\right)^{\text{th}}$$ of its velocity when passing through one plank. This means if the initial velocity is $$u$$, after passing through one plank, the velocity becomes $$u - \frac{u}{n} = u \left(1 - \frac{1}{n}\right) = u \left(\frac{n-1}{n}\right)$$. The retardation is uniform, so we can use the equation of motion $$v^2 = u^2 + 2as$$, where $$v$$ is the final velocity, $$u$$ is the initial velocity, $$a$$ is the acceleration (which is negative for retardation), and $$s$$ is the distance traveled (thickness of one plank).

For one plank, initial velocity $$u$$, final velocity $$v = u \frac{n-1}{n}$$, acceleration $$a = -a$$ (where $$a > 0$$ is the magnitude of retardation), and distance $$s$$. Substituting into the equation:

$$\left(u \frac{n-1}{n}\right)^2 = u^2 + 2(-a)s$$

$$u^2 \frac{(n-1)^2}{n^2} = u^2 - 2as$$

Bring all terms to one side:

$$u^2 \frac{(n-1)^2}{n^2} - u^2 = -2as$$

$$u^2 \left( \frac{(n-1)^2}{n^2} - 1 \right) = -2as$$

Simplify the expression inside the parentheses:

$$\frac{(n-1)^2}{n^2} - 1 = \frac{(n-1)^2 - n^2}{n^2} = \frac{n^2 - 2n + 1 - n^2}{n^2} = \frac{-2n + 1}{n^2}$$

So:

$$u^2 \left( \frac{-2n + 1}{n^2} \right) = -2as$$

Multiply both sides by $$-1$$:

$$u^2 \left( \frac{2n - 1}{n^2} \right) = 2as$$

Solve for $$s$$:

$$s = \frac{u^2 (2n - 1)}{2a n^2}$$

Now, to stop the bullet, the final velocity must be zero. Let $$N$$ be the number of planks required. The total distance covered is $$N s$$. Using the equation of motion again, with initial velocity $$u$$, final velocity $$0$$, acceleration $$-a$$, and distance $$N s$$:

$$0^2 = u^2 + 2(-a)(N s)$$

$$0 = u^2 - 2a N s$$

$$2a N s = u^2$$

$$N = \frac{u^2}{2a s}$$

Substitute the expression for $$s$$:

$$N = \frac{u^2}{2a \left( \frac{u^2 (2n - 1)}{2a n^2} \right)} = \frac{u^2 \cdot 2a n^2}{2a u^2 (2n - 1)} = \frac{n^2}{2n - 1}$$

Thus, the number of planks required to stop the bullet is $$\frac{n^2}{2n - 1}$$. Comparing with the options:

A. Infinite
B. $$n$$
C. $$\frac{n^2}{2n-1}$$
D. $$\frac{2n^2}{n-1}$$

The expression $$\frac{n^2}{2n-1}$$ matches option C. Therefore, the correct answer is option C.

Hence, the correct answer is Option C.

Let the length of the escalator be $$ d $$. We need to find the time taken by the person to walk up the moving escalator.

First, when the escalator is stalled (not moving), the person climbs the entire length $$ d $$ in 60 seconds. So, the speed of the person relative to the escalator is $$ v_p = \frac{d}{60} $$.

Second, when the person stands still on the moving escalator, it takes 40 seconds to cover the same distance $$ d $$. So, the speed of the escalator is $$ v_e = \frac{d}{40} $$.

Now, when the person walks up the moving escalator, both the person and the escalator are moving in the same upward direction. Therefore, the effective speed of the person relative to the ground is the sum of their individual speeds: $$ v_p + v_e $$.

The time $$ t $$ taken to cover distance $$ d $$ at this combined speed is:

$$ t = \frac{d}{v_p + v_e} $$

Substitute the expressions for $$ v_p $$ and $$ v_e $$:

$$ t = \frac{d}{\frac{d}{60} + \frac{d}{40}} $$

Factor $$ d $$ out of the denominator:

$$ t = \frac{d}{d \left( \frac{1}{60} + \frac{1}{40} \right)} $$

Cancel $$ d $$ from numerator and denominator (since $$ d \neq 0 $$):

$$ t = \frac{1}{\frac{1}{60} + \frac{1}{40}} $$

Now, compute the denominator. Find the sum $$ \frac{1}{60} + \frac{1}{40} $$. The least common multiple of 60 and 40 is 120. Rewrite each fraction:

$$ \frac{1}{60} = \frac{2}{120}, \quad \frac{1}{40} = \frac{3}{120} $$

Add them:

$$ \frac{2}{120} + \frac{3}{120} = \frac{5}{120} = \frac{1}{24} $$

So,

$$ t = \frac{1}{\frac{1}{24}} = 24 \text{ seconds} $$

Hence, the time taken by the person to walk up the moving escalator is 24 seconds. Comparing with the options, Option C is 24 s.

So, the answer is 24 seconds.

Let us call the time taken by the particle to reach the highest point of its flight $$t_1$$. At that highest point the velocity becomes zero. For motion under uniform retardation we have the first-equation of motion $$v = u + at$$. Here the final velocity is $$0$$, the initial velocity is $$u$$ and the acceleration is $$-g$$ (because gravity opposes the upward motion). Hence

$$0 = u - g t_1 \; \Longrightarrow \; t_1 = \frac{u}{g}.$$

During this upward journey the particle rises a vertical distance $$s_1$$ above the top of the tower. Using the second‐equation of motion

$$s = ut + \frac{1}{2} a t^{\,2},$$

with $$u \rightarrow u,\; t \rightarrow t_1,\; a \rightarrow -g$$ we obtain

$$s_1 = u t_1 - \frac{1}{2} g t_1^{\,2}.$$

Substituting $$t_1 = u/g$$ inside, we get

$$s_1 = u\left(\frac{u}{g}\right) - \frac{1}{2} g\left(\frac{u}{g}\right)^{\!2}

= \frac{u^{2}}{g} - \frac{1}{2}\frac{u^{2}}{g}

= \frac{u^{2}}{2g}.$$

The greatest vertical height of the particle above the ground is therefore

$$H_{\text{max}} = H + s_1 = H + \frac{u^{2}}{2g}.$$

From the highest point the particle starts its downward fall with zero initial velocity. Let the time of this fall be $$t_2$$. During this descent it covers a distance $$H + \dfrac{u^{2}}{2g}$$ under constant acceleration $$g$$. Again invoking the second‐equation of motion with $$u = 0,\; a = g,\; t = t_2$$, we have

$$H + \frac{u^{2}}{2g} = \frac{1}{2} g t_2^{\,2}.$$

Rearranging,

$$t_2^{\,2} = \frac{2}{g}\!\left(H + \frac{u^{2}}{2g}\right)

= \frac{2H}{g} + \frac{u^{2}}{g^{2}},$$

$$t_2 = \sqrt{\frac{2H}{g} + \frac{u^{2}}{g^{2}}}.$$

According to the statement of the problem the total time taken to hit the ground is $$n$$ times the time to reach the highest point. The total time is $$t_1 + t_2$$. Hence we can write

$$t_1 + t_2 = n\,t_1

\;\;\Longrightarrow\;\;

t_2 = (n-1)t_1.$$

We already have $$t_1 = \dfrac{u}{g}$$, so

$$t_2 = (n-1)\frac{u}{g}.$$

But we also have an explicit expression for $$t_2$$ obtained from the descent. Equating the two values of $$t_2$$,

$$(n-1)\frac{u}{g} = \sqrt{\frac{2H}{g} + \frac{u^{2}}{g^{2}}}.$$

Squaring both sides to eliminate the square root,

$$(n-1)^{2}\frac{u^{2}}{g^{2}} = \frac{2H}{g} + \frac{u^{2}}{g^{2}}.$$

Multiplying every term by $$g^{2}$$ gives

$$(n-1)^{2}u^{2} = 2Hg + u^{2}.$$

Now let us bring the $$u^{2}$$ term on the left next to the existing $$u^{2}$$ factor:

$$(n-1)^{2}u^{2} - u^{2} = 2Hg.$$

Taking $$u^{2}$$ common,

$$\big[(n-1)^{2} - 1\big]\,u^{2} = 2Hg.$$

Expanding and simplifying the bracket,

$$(n^{2} - 2n + 1) - 1 = n^{2} - 2n,$$

so that

$$\big[n^{2} - 2n\big]\,u^{2} = 2Hg.$$

Factoring $$n$$ out of the bracket,

$$n(n-2)\,u^{2} = 2Hg.$$

Finally writing the result in a compact symmetric form,

$$2gH = n u^{2}(n - 2).$$

This is exactly the relation given in Option C.

Hence, the correct answer is Option C.

The bullet is fired with an initial speed of 630 m/s towards a target 700 m away at the same level. Since the rifle is aimed directly at the center of the target, the initial velocity is horizontal. This means the initial vertical velocity is 0 m/s, and the initial horizontal velocity is 630 m/s.

To find how far above or below the center the bullet hits, we need to determine the vertical displacement when the bullet has traveled 700 m horizontally. The horizontal motion has no acceleration, so the time taken to cover 700 m is given by the horizontal distance divided by the horizontal velocity:

$$ t = \frac{\text{horizontal distance}}{\text{horizontal velocity}} = \frac{700}{630} $$

Simplifying the fraction:

$$ \frac{700}{630} = \frac{700 \div 70}{630 \div 70} = \frac{10}{9} \text{ seconds} $$

In the vertical direction, the initial velocity is 0 m/s, and acceleration due to gravity is $$ g = 9.8 \text{m/s}^2 $$ downward. The vertical displacement $$ h $$ is given by the equation of motion:

$$ h = u_y t + \frac{1}{2} a t^2 $$

Since $$ u_y = 0 $$ and $$ a = g $$ (taking downward as positive for displacement magnitude), this becomes:

$$ h = \frac{1}{2} g t^2 $$

Substituting $$ g = 9.8 $$ and $$ t = \frac{10}{9} $$:

$$ h = \frac{1}{2} \times 9.8 \times \left( \frac{10}{9} \right)^2 $$

First, compute $$ \left( \frac{10}{9} \right)^2 $$:

$$ \left( \frac{10}{9} \right)^2 = \frac{100}{81} $$

Now, multiply:

$$ h = \frac{1}{2} \times 9.8 \times \frac{100}{81} = \frac{1}{2} \times \frac{9.8 \times 100}{81} $$

Calculate $$ 9.8 \times 100 = 980 $$:

$$ h = \frac{1}{2} \times \frac{980}{81} = \frac{980}{2 \times 81} = \frac{980}{162} $$

Simplify by dividing numerator and denominator by 2:

$$ \frac{980 \div 2}{162 \div 2} = \frac{490}{81} $$

Now, compute the numerical value:

$$ \frac{490}{81} \approx 6.04938 \text{ m} $$

The negative sign indicates the displacement is downward, meaning the bullet hits below the center. The magnitude of the vertical displacement is approximately 6.05 m. Comparing with the options:

A. 1.0 m
B. 4.2 m
C. 6.1 m
D. 9.8 m

The value 6.05 m is closest to 6.1 m.

Hence, the correct answer is Option C.

The position vector at time $$ t = 2 $$ seconds is given as $$ \vec{r} = 40\hat{i} + 50\hat{j} $$ meters. This means the horizontal displacement $$ x = 40 $$ m and the vertical displacement $$ y = 50 $$ m.

For a projectile launched from the origin with initial velocity $$ u $$ at an angle $$ \theta $$ to the horizontal, the equations of motion are:

Horizontal motion (no acceleration): $$ x = u_x t $$, where $$ u_x = u \cos \theta $$ is the horizontal component of initial velocity.

Vertical motion (constant acceleration due to gravity $$ g = 10 $$ m/s² downward): $$ y = u_y t - \frac{1}{2} g t^2 $$, where $$ u_y = u \sin \theta $$ is the vertical component of initial velocity.

At $$ t = 2 $$ s, $$ x = 40 $$ m. Substituting into the horizontal equation:

$$ 40 = u_x \cdot 2 $$

Solving for $$ u_x $$:

$$ u_x = \frac{40}{2} = 20 \text{ m/s} $$

At $$ t = 2 $$ s, $$ y = 50 $$ m. Substituting into the vertical equation:

$$ 50 = u_y \cdot 2 - \frac{1}{2} \cdot 10 \cdot (2)^2 $$

First, compute the term $$ \frac{1}{2} \cdot 10 \cdot 4 $$:

$$ \frac{1}{2} \cdot 10 = 5, \quad 5 \cdot 4 = 20 $$

So the equation becomes:

$$ 50 = 2u_y - 20 $$

Add 20 to both sides:

$$ 50 + 20 = 2u_y $$

$$ 70 = 2u_y $$

Solving for $$ u_y $$:

$$ u_y = \frac{70}{2} = 35 \text{ m/s} $$

The angle $$ \theta $$ is related to the components by:

$$ \tan \theta = \frac{u_y}{u_x} $$

Substituting the values:

$$ \tan \theta = \frac{35}{20} = \frac{7}{4} $$

Therefore, $$ \theta = \tan^{-1} \left( \frac{7}{4} \right) $$.

Comparing with the options:

A. $$ \tan^{-1} \frac{3}{2} $$

B. $$ \tan^{-1} \frac{2}{3} $$

C. $$ \tan^{-1} \frac{7}{4} $$

D. $$ \tan^{-1} \frac{4}{5} $$

Option C matches our result.

Hence, the correct answer is Option C.

To solve this problem, we need to find the angular momentum of a ball at the highest point of its trajectory with respect to the point from which it was thrown. The ball has a mass of 160 g, is thrown at an angle of 60° to the horizontal with a speed of 10 m/s, and gravity is given as 10 m/s². Angular momentum is defined as the cross product of the position vector (from the reference point) and the linear momentum vector, so we'll calculate both at the highest point.

First, convert the mass to kilograms since SI units are required. The mass is 160 g, which is 160 divided by 1000, so mass $$ m = 0.160 $$ kg.

The initial velocity is 10 m/s at 60° to the horizontal. We resolve this into horizontal and vertical components:

Horizontal component: $$ u_x = u \cos \theta = 10 \cos 60^\circ $$

Since $$ \cos 60^\circ = \frac{1}{2} $$, $$ u_x = 10 \times \frac{1}{2} = 5 $$ m/s.

Vertical component: $$ u_y = u \sin \theta = 10 \sin 60^\circ $$

Since $$ \sin 60^\circ = \frac{\sqrt{3}}{2} $$, $$ u_y = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} $$ m/s.

At the highest point of the trajectory, the vertical component of velocity becomes zero ($$ v_y = 0 $$) because gravity acts downward, and the horizontal component remains unchanged since there is no horizontal acceleration. Therefore, the velocity at the highest point is purely horizontal: $$ v = v_x = u_x = 5 $$ m/s.

The linear momentum $$ \vec{p} $$ at the highest point is mass times velocity:

$$ \vec{p} = m \vec{v} = 0.160 \times 5 = 0.8 $$ kg m/s in the horizontal direction.

So, $$ \vec{p} = (0.8, 0) $$ kg m/s, where the first component is horizontal and the second is vertical.

Next, we find the position vector $$ \vec{r} $$ of the ball at the highest point relative to the point of projection. This requires the horizontal distance traveled ($$ x $$) and the maximum height ($$ H $$).

The maximum height $$ H $$ is given by the formula:

$$ H = \frac{u_y^2}{2g} $$

Substituting the values: $$ H = \frac{(5\sqrt{3})^2}{2 \times 10} = \frac{25 \times 3}{20} = \frac{75}{20} = 3.75 $$ m.

The time taken to reach the highest point $$ t $$ is the time when vertical velocity becomes zero:

$$ t = \frac{u_y}{g} = \frac{5\sqrt{3}}{10} = \frac{\sqrt{3}}{2} $$ seconds.

The horizontal distance $$ x $$ covered in this time is:

$$ x = u_x \times t = 5 \times \frac{\sqrt{3}}{2} = \frac{5\sqrt{3}}{2} $$ m.

So, the position vector $$ \vec{r} = \left( \frac{5\sqrt{3}}{2}, 3.75 \right) $$ m.

Angular momentum $$ \vec{L} $$ with respect to the point of projection is:

$$ \vec{L} = \vec{r} \times \vec{p} $$

In 2D, the magnitude of the cross product is $$ |\vec{L}| = |r_x p_y - r_y p_x| $$, where $$ r_x $$ and $$ r_y $$ are the components of $$ \vec{r} $$, and $$ p_x $$ and $$ p_y $$ are the components of $$ \vec{p} $$.

Here, $$ r_x = \frac{5\sqrt{3}}{2} $$, $$ r_y = 3.75 $$, $$ p_x = 0.8 $$, and $$ p_y = 0 $$.

So, $$ L = \left| \left( \frac{5\sqrt{3}}{2} \times 0 \right) - (3.75 \times 0.8) \right| = |0 - 3.75 \times 0.8| = | -3.0 | = 3.0 $$ kg m²/s.

Alternatively, using exact fractions:

$$ r_y = 3.75 = \frac{15}{4} $$ m (since $$ 3.75 = \frac{15}{4} $$),

$$ p_x = 0.8 = \frac{4}{5} $$ kg m/s,

So, $$ L = \left| -\left( \frac{15}{4} \times \frac{4}{5} \right) \right| = \left| -\frac{15 \times 4}{4 \times 5} \right| = \left| -\frac{15}{5} \right| = | -3 | = 3 $$ kg m²/s.

The magnitude is 3.0 kg m²/s, and the direction is perpendicular to the plane (negative sign indicates direction, but magnitude is positive).

Comparing with the options:

A. 1.73 kg m² s⁻¹

B. 3.46 kg m² s⁻¹

C. 3.0 kg m² s⁻¹

D. 6.0 kg m² s⁻¹

Hence, the correct answer is Option C.

A ball is projected from the ground at an angle of 45°. The point of projection is 4 m from the foot of the wall, and the ball lands 6 m beyond the wall. The total horizontal distance covered (range) is the sum of these distances: 4 m + 6 m = 10 m.

For a projectile launched at an angle θ with initial velocity u, the range R is given by:

$$R = \frac{u^2 \sin 2\theta}{g} $$

Here, θ = 45°, so 2θ = 90° and sin 90° = 1. Thus:

$$R = \frac{u^2}{g} $$

Given R = 10 m:

$$10 = \frac{u^2}{g} $$

Solving for u²:

$$u^2 = 10g \quad \text{(Equation 1)} $$

The trajectory equation for a projectile is:

$$y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta} $$

With θ = 45°, tan 45° = 1 and cos 45° = $$\frac{1}{\sqrt{2}}$$, so cos² 45° = $$\left( \frac{1}{\sqrt{2}} \right)^2 = \frac{1}{2}$$. Substituting these values:

$$y = x \cdot 1 - \frac{g x^2}{2 u^2 \cdot \frac{1}{2}} $$

Simplifying:

$$y = x - \frac{g x^2}{u^2} \quad \text{(Equation 2)} $$

The ball just clears the wall at x = 4 m. Substituting x = 4 into Equation 2:

$$y = 4 - \frac{g (4)^2}{u^2} $$ $$ y = 4 - \frac{16g}{u^2} $$

From Equation 1, u² = 10g. Substituting this:

$$y = 4 - \frac{16g}{10g} $$

The g cancels out:

$$y = 4 - \frac{16}{10} $$

Simplifying the fraction:

$$\frac{16}{10} = 1.6 $$

Thus:

$$y = 4 - 1.6 = 2.4 \text{ m} $$

Therefore, the height of the wall is 2.4 m.

Comparing with the options:

A. 4.4 m

B. 2.4 m

C. 3.6 m

D. 1.6 m

Hence, the correct answer is Option B.

The maximum range when the car is at rest is given as $$ R_0 = 40 $$ m, and gravity $$ g = 10 $$ m/s². For a projectile fired from ground level, the maximum range occurs at an angle of 45°. The range formula is $$ R = \frac{u^2 \sin 2\theta}{g} $$. At 45°, $$ \sin 90^\circ = 1 $$, so:

$$ R_0 = \frac{u^2}{g} $$

$$ 40 = \frac{u^2}{10} $$

$$ u^2 = 400 $$

$$ u = 20 \text{ m/s} $$

Thus, the initial speed of the bullet relative to the pistol is 20 m/s.

Now, the car moves with uniform velocity $$ v = 20 $$ m/s in the direction of firing. The pistol is mounted on the car, so the bullet's initial velocity relative to the ground has two components:

• Horizontal component: $$ v_x = v + u \cos \theta = 20 + 20 \cos \theta $$
• Vertical component: $$ v_y = u \sin \theta = 20 \sin \theta $$

The time of flight $$ T $$ is determined by the vertical motion. The bullet lands at the same level, so vertical displacement is zero:

$$ s_y = v_y T - \frac{1}{2} g T^2 = 0 $$

$$ (20 \sin \theta) T - \frac{1}{2} \times 10 \times T^2 = 0 $$

$$ 20 \sin \theta T - 5 T^2 = 0 $$

$$ 5T (4 \sin \theta - T) = 0 $$

Ignoring $$ T = 0 $$ (initial time), we get:

$$ T = 4 \sin \theta \text{ seconds} $$

The horizontal range $$ R $$ is the product of horizontal velocity and time of flight:

$$ R = v_x \times T = (20 + 20 \cos \theta) \times (4 \sin \theta) $$

$$ R = 80 (1 + \cos \theta) \sin \theta $$

To maximize $$ R $$, we maximize the function $$ f(\theta) = (1 + \cos \theta) \sin \theta $$. Expanding:

$$ f(\theta) = \sin \theta + \sin \theta \cos \theta $$

Using $$ \sin \theta \cos \theta = \frac{1}{2} \sin 2\theta $$:

$$ f(\theta) = \sin \theta + \frac{1}{2} \sin 2\theta $$

Differentiate with respect to $$ \theta $$:

$$ f'(\theta) = \cos \theta + \frac{1}{2} \times 2 \cos 2\theta = \cos \theta + \cos 2\theta $$

Set the derivative to zero for maximum:

$$ \cos \theta + \cos 2\theta = 0 $$

Using the identity $$ \cos 2\theta = 2 \cos^2 \theta - 1 $$:

$$ \cos \theta + (2 \cos^2 \theta - 1) = 0 $$

$$ 2 \cos^2 \theta + \cos \theta - 1 = 0 $$

Substitute $$ x = \cos \theta $$:

$$ 2x^2 + x - 1 = 0 $$

Solve the quadratic equation:

$$ x = \frac{ -1 \pm \sqrt{1^2 - 4 \times 2 \times (-1)} }{4} = \frac{ -1 \pm \sqrt{9} }{4} = \frac{ -1 \pm 3 }{4} $$

So:

$$ x = \frac{2}{4} = \frac{1}{2} \quad \text{or} \quad x = \frac{-4}{4} = -1 $$

Thus, $$ \cos \theta = \frac{1}{2} $$ or $$ \cos \theta = -1 $$. Since $$ \theta $$ is acute (0° to 90°), we discard $$ \cos \theta = -1 $$ (which gives $$ \theta = 180^\circ $$, not acute). Therefore:

$$ \cos \theta = \frac{1}{2} \quad \Rightarrow \quad \theta = 60^\circ $$

Verification: At $$ \theta = 60^\circ $$, $$ v_x = 20 + 20 \cos 60^\circ = 20 + 20 \times 0.5 = 30 $$ m/s, $$ v_y = 20 \sin 60^\circ = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} $$ m/s, time of flight $$ T = 4 \sin 60^\circ = 4 \times \frac{\sqrt{3}}{2} = 2\sqrt{3} $$ s, and range $$ R = 30 \times 2\sqrt{3} = 60\sqrt{3} \approx 103.92 $$ m. At other angles like 45° and 90°, the range is less (approximately 96.59 m and 80 m, respectively), confirming that 60° gives the maximum range.

Hence, the correct answer is Option B.

A projectile of mass M has a horizontal range of 4 km. The range formula for a projectile is given by $$ R = \frac{u^2 \sin 2\theta}{g} $$, where u is the initial velocity and θ is the angle of projection. So, we have:

$$ R = \frac{u^2 \sin 2\theta}{g} = 4 \text{km} $$

At the highest point of its trajectory, the projectile explodes into two parts: one of mass $$ \frac{M}{4} $$ and the other of mass $$ \frac{3M}{4} $$. The heavier part (mass $$ \frac{3M}{4} $$) falls vertically downward with zero initial speed, meaning its velocity immediately after the explosion is zero in both horizontal and vertical directions.

Since the explosion is an internal force, it does not affect the center of mass motion. Before the explosion, at the highest point, the entire projectile has only horizontal velocity $$ u_x = u \cos \theta $$ and zero vertical velocity. Therefore, the center of mass continues with velocity $$ (u_x, 0) $$ after the explosion.

Let $$ \vec{v_1} = (v_{1x}, v_{1y}) $$ be the velocity of the lighter part (mass $$ \frac{M}{4} $$) and $$ \vec{v_2} = (0, 0) $$ be the velocity of the heavier part (mass $$ \frac{3M}{4} $$) after the explosion. The center of mass velocity is:

$$ \vec{v_{cm}} = \frac{ \left( \frac{M}{4} \right) \vec{v_1} + \left( \frac{3M}{4} \right) \vec{v_2} }{ M } = \frac{1}{4} \vec{v_1} + \frac{3}{4} \vec{v_2} $$

Setting $$ \vec{v_{cm}} = (u_x, 0) $$ and substituting $$ \vec{v_2} = (0, 0) $$:

$$ (u_x, 0) = \frac{1}{4} (v_{1x}, v_{1y}) + \frac{3}{4} (0, 0) $$

This simplifies to:

$$ (u_x, 0) = \left( \frac{v_{1x}}{4}, \frac{v_{1y}}{4} \right) $$

Equating components:

$$ \frac{v_{1x}}{4} = u_x \quad \Rightarrow \quad v_{1x} = 4u_x $$

$$ \frac{v_{1y}}{4} = 0 \quad \Rightarrow \quad v_{1y} = 0 $$

So, the lighter part has horizontal velocity $$ 4u_x $$ and zero vertical velocity immediately after the explosion.

The explosion occurs at the highest point, which is at a horizontal distance of half the range from the firing point. Since the range is 4 km, this distance is:

$$ \frac{R}{2} = \frac{4}{2} = 2 \text{km} $$

The lighter part is now projected horizontally from this point with initial velocity $$ 4u_x $$ and from a height H, where H is the maximum height of the original projectile. The maximum height H is given by:

$$ H = \frac{u^2 \sin^2 \theta}{2g} $$

The time taken for the lighter part to fall to the ground from height H with zero initial vertical velocity is found using the equation of motion:

$$ H = \frac{1}{2} g t^2 \quad \Rightarrow \quad t = \sqrt{\frac{2H}{g}} $$

In this time, the horizontal distance covered by the lighter part from the explosion point is:

$$ d = v_{1x} \cdot t = 4u_x \cdot \sqrt{\frac{2H}{g}} $$

Substituting $$ H = \frac{u^2 \sin^2 \theta}{2g} $$:

$$ d = 4u_x \cdot \sqrt{ \frac{2}{g} \cdot \frac{u^2 \sin^2 \theta}{2g} } = 4u_x \cdot \sqrt{ \frac{u^2 \sin^2 \theta}{g^2} } = 4u_x \cdot \frac{u \sin \theta}{g} $$

Let $$ u_y = u \sin \theta $$ (the initial vertical component of velocity for the original projectile). Then:

$$ d = 4u_x \cdot \frac{u_y}{g} $$

From the original range equation:

$$ R = \frac{2 u_x u_y}{g} = 4 \text{km} $$

So:

$$ \frac{u_x u_y}{g} = \frac{R}{2} = \frac{4}{2} = 2 \text{km} $$

Substituting this into the expression for d:

$$ d = 4 \cdot 2 = 8 \text{km} $$

This is the horizontal distance traveled by the lighter part from the explosion point. The total horizontal range from the firing point is the distance to the explosion point plus d:

$$ \text{Total range} = 2 \text{km} + 8 \text{km} = 10 \text{km} $$

Hence, the horizontal range of the lighter part is 10 km.

So, the answer is Option C.

For motion under uniform gravity (air resistance neglected) we can use the constant-acceleration relation

$$v^{2}=u^{2}+2g\,(h_{\text{initial}}-h)\qquad -(1)$$

Here $$h$$ is the vertical coordinate measured from the ground (origin at the ground, positive upward), $$u$$ is the speed when the particle is at height $$h_{\text{initial}}$$ and $$g$$ is the magnitude of acceleration due to gravity.

Case 1: Descent from the release point ($$h=H$$) to the ground ($$h=0$$)
The ball is dropped from rest, so $$u=0$$ at $$h=H$$. Putting $$h_{\text{initial}}=H$$ and $$u=0$$ in (1):

$$v^{2}=2g\,(H-h)\qquad (0\le h\le H)$$

Thus $$v=\sqrt{2g\,(H-h)}$$, which is the upper half of a parabola on a $$v$$ vs $$h$$ plot. At the starting point $$h=H$$, $$v=0$$; just before impact $$h=0$$, $$v=\sqrt{2gH}$$ (maximum speed).

Case 2: Ascent after the inelastic bounce
Let the coefficient of restitution with the ground be $$e\,(0\lt e\lt 1)$$. Speed just after the bounce (directed upward) is therefore $$e\sqrt{2gH}$$.

During upward motion the acceleration is still downward $$(-g)$$, so we again use (1) with $$u=e\sqrt{2gH}$$ at $$h=0$$:

$$v^{2}=e^{2}\,2gH-2g\,h\qquad (0\le h\le h_{\max})$$

where $$h_{\max}=e^{2}H$$ (height reached after bounce). Hence $$v=\sqrt{e^{2}2gH-2g\,h}\,,$$ another upper-parabolic branch but this time terminating at $$h=e^{2}H$$ where $$v=0$$.

Shape of the complete $$v$$-$$h$$ graph
1. From $$h=H$$ to $$h=0$$: a parabolic arc opening toward the $$h$$-axis, starting at $$v=0$$ and ending at $$v=\sqrt{2gH}$$.
2. Immediately after impact: vertical jump to the point $$h=0,\;v=e\sqrt{2gH}$$ (because direction has reversed but we plot speed, a positive quantity).
3. From $$h=0$$ to $$h=e^{2}H$$: a second, shorter parabolic arc (same orientation) ending at $$v=0$$.

Among the given choices only Option D displays (i) the first long parabolic branch, (ii) a discontinuity in speed at the ground, and (iii) a second shorter parabolic branch terminating below the initial height. Therefore,

Option D which is: the only correct $$v$$-$$h$$ graph.

Solution & Explanation

1. Define System Parameters and Set Up Equations

Let the total length of the train be $$L$$. Therefore, the middle wagon sits exactly at a distance of $$\frac{L}{2}$$ from both the engine (front) and the guard's room (back). The train is moving with a constant uniform acceleration $$a$$.

Using the third equation of motion ($$v^2 = u^2 + 2as$$) for the entire length of the train as it passes the pole:

• Initial velocity (when engine passes) = $$u$$
• Final velocity (when guard's room passes) = $$v$$
• Total displacement = $$L$$

$$v^2 = u^2 + 2aL \implies 2aL = v^2 - u^2 \quad \text{--- (Eq. 1)}$$

2. Analyze the Motion of the Middle Wagon

Let $$v_m$$ be the instantaneous velocity of the train at the exact moment its middle wagon passes the pole. Over this interval, the train has covered a displacement of exactly $$s = \frac{L}{2}$$ past the pole.

Applying the third equation of motion over this half-length interval:

$$v_m^2 = u^2 + 2a\left(\frac{L}{2}\right)$$

$$v_m^2 = u^2 + aL \quad \text{--- (Eq. 2)}$$

3. Substitute and Solve for $v_m$

From Equation 1, we can isolate the term $$aL$$:

$$aL = \frac{v^2 - u^2}{2}$$

Substitute this expression for $$aL$$ directly back into Equation 2:

$$v_m^2 = u^2 + \frac{v^2 - u^2}{2}$$

$$v_m^2 = \frac{2u^2 + v^2 - u^2}{2}$$

$$v_m^2 = \frac{u^2 + v^2}{2}$$

Taking the positive square root to find the actual velocity value:

$$v_m = \sqrt{\frac{u^2 + v^2}{2}}$$

Concept Check: Because kinetic energy scales with the square of velocity ($$v^2$$), the velocity at the spatial midpoint under constant acceleration matches the root-mean-square (RMS) value of the initial and final speeds rather than a simple arithmetic average.

Correct Option Key: Option D ($$\sqrt{\frac{u^2 + v^2}{2}}$$)

The graph provided is a displacement-time ($$x$$ vs $$t$$) plot. For such a graph, the instantaneous velocity at any point equals the slope of the tangent at that point:

$$v = \frac{dx}{dt} = \text{slope of the } x\text{-}t \text{ graph}$$

Case 1: Point $$A$$ lies on a straight-line segment of the curve. Two easily readable neighbouring points on that segment are, for example, $$(2\;\text{s},\,1\;\text{m})$$ and $$(4\;\text{s},\,2\;\text{m})$$. The slope (and hence the velocity at $$A$$) is

$$v_A \;=\; \frac{2\;\text{m} - 1\;\text{m}}{4\;\text{s} - 2\;\text{s}} \;=\; \frac{1}{2}\;\text{m s}^{-1} \;=\; 0.5\;\text{m s}^{-1}$$

Case 2: Point $$B$$ also falls on a straight portion of the graph. Picking two nearby points on that segment—for instance $$(10\;\text{s},\,3\;\text{m})$$ and $$(14\;\text{s},\,5\;\text{m})$$—gives

$$v_B \;=\; \frac{5\;\text{m} - 3\;\text{m}}{14\;\text{s} - 10\;\text{s}} \;=\; \frac{2}{4}\;\text{m s}^{-1} \;=\; 0.5\;\text{m s}^{-1}$$

Because both points lie on straight segments having the same slope, the instantaneous velocities are identical:

$$v_A = v_B = 0.5\;\text{m s}^{-1}$$

Therefore, the correct choice is
Option A which is: $$v_A = v_B = 0.5\ \text{m/s}$$

The first body starts from rest with uniform acceleration $$a$$, so after time $$t$$ its displacement from the origin is

$$x_1(t)=\tfrac12 a t^{2}\qquad -(1)$$

The second body crosses the origin at $$t=0$$ with a constant speed $$v$$, hence

$$x_2(t)=v\,t\qquad -(2)$$

The required function is the difference of the two displacements:

$$x_1-x_2=\tfrac12 a t^{2}-v t=t\bigl(\tfrac12 a t-v\bigr)\qquad -(3)$$

Equation (3) is a quadratic in $$t$$ whose leading coefficient $$\tfrac12 a$$ is positive, so its graph is an upward-opening parabola.

Key features of the parabola:

• At $$t=0$$, $$x_1-x_2=0$$ (the two bodies are together).
• The initial slope is $$\left.\frac{d}{dt}(x_1-x_2)\right|_{t=0}= -v$$, i.e. the curve leaves the origin with a negative slope.
• The vertex occurs when $$\dfrac{d}{dt}(x_1-x_2)=a t-v=0\;\Longrightarrow\;t=\dfrac{v}{a}$$. The minimum value there is $$-\dfrac{v^{2}}{2a}$$, so the curve lies below the $$t$$-axis between $$t=0$$ and $$t=\dfrac{2v}{a}$$.
• It crosses the $$t$$-axis again when $$x_1-x_2=0\Rightarrow t=0$$ or $$t=\dfrac{2v}{a}$$ and thereafter remains positive.

Therefore the correct plot begins at the origin, dips below the $$t$$-axis, attains a minimum at $$t=\dfrac{v}{a}$$, emerges to cut the axis at $$t=\dfrac{2v}{a}$$, and then rises indefinitely — exactly the shape shown in Option B.

Option C which is: the upward-opening parabola starting at the origin with an initial negative slope and crossing the axis once more at $$t=\dfrac{2v}{a}$$

Solution & Explanation

1. Express Velocity as a Differential Equation

Velocity ($$v$$) is defined as the instantaneous rate of change of displacement ($$x$$) with respect to time ($$t$$):

$$v = \frac{dx}{dt}$$

We substitute the given velocity expression into this definition to form our differential equation:

$$\frac{dx}{dt} = v_0 + g \cdot t + f \cdot t^2$$

Separating the variables to prepare for integration gives:

$$dx = (v_0 + g \cdot t + f \cdot t^2) \cdot dt$$

2. Perform Definite Integration with Limits

We now integrate both sides using the boundary conditions specified in the problem statement:

• At the initial boundary condition, the time is $$t = 0$$ and the position is $$x = 0$$.
• At the final boundary condition, the time reaches unit time $$t = 1$$ and the position reaches $$x = x$$.

Setting up the integration limits:

$$\int_{0}^{x} dx = \int_{0}^{1} (v_0 + g \cdot t + f \cdot t^2) \cdot dt$$

Applying the standard power rule for calculus integration ($$\int t^n \cdot dt = \frac{t^{n+1}}{n+1}$$):

$$[x]_{0}^{x} = \left[ v_0 \cdot t + \frac{g \cdot t^2}{2} + \frac{f \cdot t^3}{3} \right]_{0}^{1}$$

3. Substitute Limits to Find the Final Displacement

Evaluating the limits by substituting the upper value ($$1$$) and subtracting the lower value ($$0$$):

$$x - 0 = \left( v_0 \cdot (1) + \frac{g \cdot (1)^2}{2} + \frac{f \cdot (1)^3}{3} \right) - (0)$$

$$x = v_0 + \frac{g}{2} + \frac{f}{3}$$

Concept Check: Because acceleration is not constant (due to the time-dependent terms $$g \cdot t$$ and $$f \cdot t^2$$), standard algebraic equations of motion cannot be used. Continuous calculus integration over the entire time interval correctly accounts for the changing acceleration components.

Correct Option Key: Option B ($$v_0 + \frac{g}{2} + \frac{f}{3}$$)

Let the common speed finally acquired by both points be $$v$$.

For point $$A$$ (acceleration $$f$$, starting from rest):
  Final speed  $$v = f\,t_A$$  $$\Rightarrow \; t_A = \frac{v}{f}$$

For point $$B$$ (acceleration $$f'$$, starting from rest):
  Final speed  $$v = f'\,t_B$$  $$\Rightarrow \; t_B = \frac{v}{f'}$$

Given that point $$A$$ takes $$m$$ seconds more than point $$B$$ to reach this speed,

$$t_A - t_B = m$$ $$\frac{v}{f} - \frac{v}{f'} = m$$ $$v\left(\frac{1}{f} - \frac{1}{f'}\right) = m$$  $$-(1)$$

Distances travelled while accelerating from rest:

For $$A$$: $$s_A = \frac12 f\,t_A^2 = \frac12 f\left(\frac{v}{f}\right)^2 = \frac{v^2}{2f}$$
For $$B$$: $$s_B = \frac12 f'\,t_B^2 = \frac12 f'\left(\frac{v}{f'}\right)^2 = \frac{v^2}{2f'}$$

Given that $$A$$ covers $$n$$ units more than $$B$$,

$$s_A - s_B = n$$ $$\frac{v^2}{2}\left(\frac{1}{f} - \frac{1}{f'}\right) = n$$  $$-(2)$$

Let $$\Delta = \frac{1}{f} - \frac{1}{f'}$$. From $$(1):\; v\Delta = m \;\Rightarrow\; v = \frac{m}{\Delta}$$

Substituting $$v$$ in $$(2):$$

$$\frac{1}{2}\left(\frac{m}{\Delta}\right)^2 \Delta = n$$ $$\frac{m^2}{2\Delta} = n$$ $$\Delta = \frac{m^2}{2n}$$  $$-(3)$$

But $$\Delta = \frac{1}{f} - \frac{1}{f'} = \frac{f' - f}{ff'}$$. Equating this with $$(3):$$

$$\frac{f' - f}{ff'} = \frac{m^2}{2n}$$ $$\bigl(f' - f\bigr)\,n = \frac12\,f\,f'\,m^2$$

Thus, the required relation is
$$\boxed{(f' - f)\,n = \dfrac12\,f\,f'\,m^2}$$

Option D which is: $$(f' - f)n = \frac{1}{2}ff'm^2$$

Let the ball start from the origin with speed $$v_0$$ and projection angle $$\theta_0$$ (measured from the +x-axis).
Its velocity components are $$v_{bx}=v_0\cos\theta_0$$ and $$v_{by}=v_0\sin\theta_0$$.

The stone is thrown simultaneously from $$(L,0)$$ at an angle $$(180^\circ-\theta_1)$$ with the +x-axis. Hence its horizontal component is toward the negative x-direction:
$$v_{sx}=-v_s\cos\theta_1$$, $$v_{sy}=v_s\sin\theta_1$$, where $$v_s$$ is the stone’s launch speed.

Suppose they meet after time $$t$$ (same $$t$$ for both because they start together).

Horizontal co-ordinate equality
$$x_{\text{ball}} = x_{\text{stone}}$$ gives $$v_0\cos\theta_0\,t = L + (-v_s\cos\theta_1)\,t$$ $$\Rightarrow t\,(v_0\cos\theta_0+v_s\cos\theta_1)=L \quad -(1)$$

Vertical co-ordinate equality
Both experience the same downward acceleration $$g$$, so the $$-\dfrac12 g t^2$$ term cancels: $$v_0\sin\theta_0\,t = v_s\sin\theta_1\,t$$ Since $$t\gt 0$$, $$v_0\sin\theta_0 = v_s\sin\theta_1 \quad -(2)$$

From $$(2)$$, express the stone’s speed:
$$v_s = \dfrac{v_0\sin\theta_0}{\sin\theta_1} \quad -(3)$$

Insert $$(3)$$ into $$(1)$$ to obtain the common time of flight:

$$t = \dfrac{L}{v_0\cos\theta_0 + \left(\dfrac{v_0\sin\theta_0}{\sin\theta_1}\right)\cos\theta_1 } = \dfrac{L}{v_0\left(\cos\theta_0+\sin\theta_0\cot\theta_1\right)} \quad -(4)$$

Thus, for any pair $$(\theta_0,\theta_1)$$, $$T(\theta_0,\theta_1)=\dfrac{L}{v_0\left(\cos\theta_0+\sin\theta_0\cot\theta_1\right)}$$

Case 1: $$(\theta_0,\theta_1)=(45^\circ,45^\circ)$$ $$\cos45^\circ=\sin45^\circ=\dfrac{\sqrt2}{2}, \;\cot45^\circ=1$$
Denominator in $$(4)$$: $$\cos45^\circ+\sin45^\circ\cot45^\circ = \dfrac{\sqrt2}{2}+\dfrac{\sqrt2}{2}= \sqrt2$$ Hence $$T_1=\dfrac{L}{v_0\sqrt2} \quad -(5)$$

Case 2: $$(\theta_0,\theta_1)=(60^\circ,30^\circ)$$ $$\cos60^\circ=\dfrac12,\; \sin60^\circ=\dfrac{\sqrt3}{2},\; \cot30^\circ=\sqrt3$$
Denominator in $$(4)$$: $$\dfrac12+\dfrac{\sqrt3}{2}\,\sqrt3 = \dfrac12+\dfrac{3}{2}=2$$ Thus $$T_2=\dfrac{L}{2v_0} \quad -(6)$$

Required ratio
$$\left(\dfrac{T_1}{T_2}\right)^2 =\left(\dfrac{\,L/(v_0\sqrt2)\,}{\,L/(2v_0)\,}\right)^2 =\left(\dfrac{2}{\sqrt2}\right)^2 =\left(\sqrt2\right)^2 =2$$

Therefore, $$(T_1/T_2)^2 = 2$$.

We need to find the ratio of ranges of two projectiles thrown with the same initial velocity at angles $$45°$$ and $$30°$$ with the horizontal.

Since the range of a projectile is given by $$R = \dfrac{u^2 \sin 2\theta}{g}$$, substituting $$\theta = 45°$$ gives $$R_1 = \dfrac{u^2 \sin(2 \times 45°)}{g} = \dfrac{u^2 \sin 90°}{g} = \dfrac{u^2}{g}$$.

Next, for $$\theta = 30°$$ we get $$R_2 = \dfrac{u^2 \sin(2 \times 30°)}{g} = \dfrac{u^2 \sin 60°}{g} = \dfrac{u^2 \sqrt{3}}{2g}$$.

From this, $$\dfrac{R_1}{R_2} = \dfrac{\dfrac{u^2}{g}}{\dfrac{u^2 \sqrt{3}}{2g}} = \dfrac{u^2}{g} \times \dfrac{2g}{u^2 \sqrt{3}} = \dfrac{2}{\sqrt{3}}$$.

Therefore, $$R_1 : R_2 = 2 : \sqrt{3}$$.

The correct answer is Option C: $$2 : \sqrt{3}$$.

We need to find the ratio of initial velocities of two projectiles thrown at $$30°$$ and $$45°$$ with the horizontal, given that they reach maximum height in the same time.

We start by recalling that for a projectile launched with initial velocity $$v$$ at angle $$\theta$$, the time to reach maximum height is: $$t = \frac{v \sin\theta}{g}$$

Next, let the initial velocities be $$v_1$$ (at $$30°$$) and $$v_2$$ (at $$45°$$). Since both reach maximum height in the same time, we have: $$\frac{v_1 \sin 30°}{g} = \frac{v_2 \sin 45°}{g}$$

This simplifies to $$v_1 \sin 30° = v_2 \sin 45°$$

$$v_1 \times \frac{1}{2} = v_2 \times \frac{\sqrt{2}}{2}$$

Therefore, $$\frac{v_1}{v_2} = \frac{\sqrt{2}/2}{1/2} = \frac{\sqrt{2}}{1} = \sqrt{2}$$

Hence, $$v_1 : v_2 = \sqrt{2} : 1$$.

The correct answer is Option C: $$\sqrt{2} : 1$$.

We are given that $$|\vec{A}| = |\vec{B}| = A$$ (say), and $$|\vec{A} + \vec{B}| = 2|\vec{A} - \vec{B}|$$. Let $$\theta$$ be the angle between $$\vec{A}$$ and $$\vec{B}$$. Then $$|\vec{A} + \vec{B}|^2 = A^2 + A^2 + 2A^2\cos\theta = 2A^2(1 + \cos\theta)$$ and $$|\vec{A} - \vec{B}|^2 = A^2 + A^2 - 2A^2\cos\theta = 2A^2(1 - \cos\theta)$$.

Squaring both sides of $$|\vec{A} + \vec{B}| = 2|\vec{A} - \vec{B}|$$ gives $$2A^2(1 + \cos\theta) = 4 \times 2A^2(1 - \cos\theta)$$, so $$1 + \cos\theta = 4(1 - \cos\theta)$$ which simplifies to $$1 + \cos\theta = 4 - 4\cos\theta$$, hence $$5\cos\theta = 3$$ and $$\cos\theta = \frac{3}{5}$$.

Therefore $$\theta = \cos^{-1}\left(\frac{3}{5}\right)$$. The correct answer is Option A: $$\cos^{-1}\left(\frac{3}{5}\right)$$.

A ball is projected with speed $$15 \text{ m s}^{-1}$$ at angle $$\theta$$ such that range equals maximum height.

To determine the required angle, we first write the expressions for the range and the maximum height.

$$R = \frac{u^2 \sin 2\theta}{g} = \frac{2u^2 \sin\theta \cos\theta}{g}$$

$$H = \frac{u^2 \sin^2\theta}{2g}$$

Equating the range and the maximum height gives

$$\frac{2u^2 \sin\theta \cos\theta}{g} = \frac{u^2 \sin^2\theta}{2g}$$

Since $$u^2 \sin\theta / g \neq 0$$, cancelling common factors yields

$$2\cos\theta = \frac{\sin\theta}{2}$$

$$4\cos\theta = \sin\theta$$

$$\tan\theta = 4$$

Therefore, the correct answer is Option D: $$4$$.

We have a ball thrown vertically upward reaching maximum height $$h$$. At the top, all kinetic energy is converted to potential energy, so $$\frac{1}{2}mu^2 = mgh$$, giving $$u = \sqrt{2gh}$$.

Using the equation $$s = ut - \frac{1}{2}gt^2$$ with $$s = \frac{h}{3}$$, we get $$\frac{h}{3} = ut - \frac{1}{2}gt^2$$. Since $$u = \sqrt{2gh}$$ and $$h = \frac{u^2}{2g}$$, we substitute $$\frac{h}{3} = \frac{u^2}{6g}$$. The equation becomes $$\frac{u^2}{6g} = ut - \frac{1}{2}gt^2$$, which rearranges to $$3g^2t^2 - 6gtu + u^2 = 0$$. Dividing through by $$g^2$$ and writing $$T_0 = \frac{u}{g}$$ (total time to reach the top), we get $$3t^2 - 6T_0 t + T_0^2 = 0$$.

Applying the quadratic formula: $$t = \frac{6T_0 \pm \sqrt{36T_0^2 - 12T_0^2}}{6} = \frac{6T_0 \pm \sqrt{24T_0^2}}{6} = \frac{6T_0 \pm 2\sqrt{6}\,T_0}{6} = T_0\left(\frac{3 \pm \sqrt{6}}{3}\right)$$.

The ball is at height $$\frac{h}{3}$$ at two times: $$t_1 = T_0\left(\frac{3 - \sqrt{6}}{3}\right)$$ (going up) and $$t_2 = T_0\left(\frac{3 + \sqrt{6}}{3}\right)$$ (coming down).

The required ratio is $$\frac{t_1}{t_2} = \frac{3 - \sqrt{6}}{3 + \sqrt{6}}$$. Rationalising by multiplying numerator and denominator by $$(3 - \sqrt{6})$$: $$\frac{(3 - \sqrt{6})^2}{9 - 6} = \frac{9 - 6\sqrt{6} + 6}{3} = \frac{15 - 6\sqrt{6}}{3} = 5 - 2\sqrt{6}$$.

Now we check Option B: $$\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}$$. Rationalising: $$\frac{(\sqrt{3} - \sqrt{2})^2}{3 - 2} = (3 - 2\sqrt{6} + 2) = 5 - 2\sqrt{6}$$.

This matches our result exactly. Hence, the correct answer is Option B.

We need to identify the correct velocity-time graph for a bullet shot vertically downward. Initially, the velocity $$u = 100$$ m/s downward; taking downward as positive, the velocity after time $$t$$ is given by $$v = u + gt = 100 + 10 \times 10 = 200\text{ m/s}$$ at $$t = 10\text{ s}$$. Thus the velocity increases linearly from 100 m/s to 200 m/s over the first 10 seconds.

At $$t = 10\text{ s}$$ the bullet undergoes a perfectly inelastic collision and instantaneously comes to rest, so the velocity drops from 200 m/s to 0 in a vertical line on the v-t graph. After the collision, from $$t = 10\text{ s}$$ to $$t = 20\text{ s}$$, the bullet remains at rest and hence $$v = 0$$ over this interval.

The v-t curve therefore consists of a straight line with positive slope from (0, 100) to (10, 200), followed by a sudden drop to 0 and then a horizontal line at v = 0 from t = 10 s to t = 20 s. This matches Option A (Option 1), which is the correct answer.

We have a juggler who throws $$n$$ balls per second, each with the same initial velocity $$u$$. He throws the next ball when the first ball reaches its highest position. The time interval between successive throws is $$\frac{1}{n}$$ seconds.

At the highest point the ball's velocity is zero, and the time to reach the maximum height is given by $$v = u - gt$$, so $$0 = u - g \cdot \frac{1}{n}$$, which gives $$u = \frac{g}{n}$$.

Now the maximum height is $$H = \frac{u^2}{2g} = \frac{1}{2g}\left(\frac{g}{n}\right)^2 = \frac{g^2}{2g n^2} = \frac{g}{2n^2}$$.

Hence, the correct answer is Option 4.

A monkey drops a mango from a height of 19.6 m and we need to determine how far from the tree the cadet who catches it is at the moment of the drop. Using the relation $$h = \frac{1}{2}gt^2$$, we set $$19.6 = \frac{1}{2} \times 9.8 \times t^2$$ which gives $$t^2 = \frac{19.6 \times 2}{9.8} = 4$$ and hence $$t = 2 \text{ s}$$.

The parade speed is 9 km/h, which converts to $$9 \times \frac{5}{18} = 2.5$$ m/s. In the 2 seconds before the mango reaches the ground, the cadet walks a distance $$d = v \times t = 2.5 \times 2 = 5 \text{ m}$$.

Therefore, the cadet must be 5 m away from the tree at the moment of the drop to arrive directly underneath when the mango falls. Hence, the correct answer is Option A: 5 m.

$$R_{\text{max}} = \frac{u^2}{g} = 100\ \text{m} \quad \text{--- (1)}$$

$$H_{\text{max}} = \frac{u^2 \sin^2(90^\circ)}{2g} = \frac{u^2}{2g} \quad \text{--- (2)}$$

$$H_{\text{max}} = \frac{R_{\text{max}}}{2} = \frac{100}{2} = 50\ \text{m}$$

Given: Initial velocity $$u = 20$$ m/s at angle $$\alpha$$ with horizontal, $$t = 10$$ s, $$g = 10$$ m/s$$^2$$.

The horizontal and vertical components of velocity at time $$t$$ are:

$$v_x = u\cos\alpha$$

$$v_y = u\sin\alpha - gt = 20\sin\alpha - 10 \times 10 = 20\sin\alpha - 100$$

The inclination $$\beta$$ with horizontal at time $$t = 10$$ s is:

$$\tan\beta = \frac{v_y}{v_x} = \frac{20\sin\alpha - 100}{20\cos\alpha}$$

$$\tan\beta = \frac{20\sin\alpha}{20\cos\alpha} - \frac{100}{20\cos\alpha}$$

$$\tan\beta = \tan\alpha - \frac{5}{\cos\alpha}$$

$$\tan\beta = \tan\alpha - 5\sec\alpha$$

The correct answer is Option B.

We are given that at time $$t = 0$$, a particle starts travelling from a height $$7\hat{z}$$ cm in a plane, keeping the z-coordinate constant. The position coordinates along x and y at any instant are given as $$x = 3t$$ and $$y = 5t^3$$ respectively.

To find the acceleration, we need to differentiate the position twice with respect to time. The velocity components are obtained by the first derivative: $$v_x = \frac{dx}{dt} = 3$$ and $$v_y = \frac{dy}{dt} = 15t^2$$. Since z is constant, $$v_z = 0$$.

Now, the acceleration components are obtained by differentiating velocity: $$a_x = \frac{dv_x}{dt} = 0$$, $$a_y = \frac{dv_y}{dt} = 30t$$, and $$a_z = 0$$ (since $$v_z = 0$$).

At $$t = 1$$ s, the acceleration vector becomes $$\vec{a} = 0\hat{x} + 30(1)\hat{y} + 0\hat{z} = 30\hat{y}$$.

Note that even though the particle has a constant z-coordinate of 7 cm, this does not contribute to the acceleration since the z-position is not changing with time.

Hence, the correct answer is Option B.

For a projectile thrown with velocity $$u$$ at angle $$\theta$$ with the horizontal:

Range: $$R = \frac{u^2 \sin 2\theta}{g}$$

Maximum height: $$h = \frac{u^2 \sin^2\theta}{2g}$$

Two balls A and B achieve the same range. This means if A is thrown at angle $$\theta$$, then B is thrown at angle $$(90° - \theta)$$, since $$\sin 2\theta = \sin 2(90° - \theta)$$.

Maximum heights:

$$h_1 = \frac{u^2 \sin^2\theta}{2g}$$

$$h_2 = \frac{u^2 \cos^2\theta}{2g}$$

Product of heights:

$$h_1 h_2 = \frac{u^2 \sin^2\theta}{2g} \cdot \frac{u^2 \cos^2\theta}{2g} = \frac{u^4 \sin^2\theta \cos^2\theta}{4g^2}$$

So Reason R is true.

Now, $$\sqrt{h_1 h_2} = \frac{u^2 \sin\theta \cos\theta}{2g} = \frac{u^2 \sin 2\theta}{4g}$$

Therefore:

$$4\sqrt{h_1 h_2} = \frac{u^2 \sin 2\theta}{g} = R$$

So Assertion A is also true, and $$R = 4\sqrt{h_1 h_2}$$ follows directly from the product $$h_1 h_2$$ computed in R. Thus R is the correct explanation of A.

Hence, the correct answer is Option A.

We are given that a bullet's velocity becomes one-third after penetrating 4 cm in a wooden block, and it faces constant resistance (constant deceleration).

Let the initial velocity of the bullet be $$v_0$$. After penetrating 4 cm, its velocity is $$\frac{v_0}{3}$$. Let the constant deceleration be $$a$$.

Using the kinematic equation $$v^2 = u^2 - 2as$$ for the first 4 cm, $$\left(\frac{v_0}{3}\right)^2 = v_0^2 - 2a(4)$$, which gives $$\frac{v_0^2}{9} = v_0^2 - 8a$$. Solving for $$a$$, we find $$8a = v_0^2 - \frac{v_0^2}{9} = \frac{8v_0^2}{9}$$ and hence $$a = \frac{v_0^2}{9}$$.

When the bullet stops completely, $$v = 0$$, the total stopping distance satisfies $$0 = v_0^2 - 2a \cdot s_{total}$$. Therefore, $$s_{total} = \frac{v_0^2}{2a} = \frac{v_0^2}{2 \cdot \frac{v_0^2}{9}} = \frac{9}{2} = 4.5 \text{ cm}$$.

Since the total distance travelled is $$(4 + x)$$ cm, equating this to 4.5 cm yields $$4 + x = 4.5$$ and hence $$x = 0.5 \text{ cm}$$.

Answer: Option C: 0.5

We are given that $$\vec{A}$$ is a vector with $$|\vec{A}|$$ a non-zero constant, and we need to determine which of the provided expressions is true.

Consider Option A, which asserts $$\vec{A} \cdot \vec{A} = 0$$. In fact, $$\vec{A} \cdot \vec{A} = |\vec{A}|^2$$. Since $$|\vec{A}|$$ is non-zero, it follows that $$\vec{A} \cdot \vec{A} \neq 0$$, so Option A is false.

Options B and D claim $$\vec{A} \times \vec{A} < 0$$ or $$\vec{A} \times \vec{A} > 0$$, respectively. However, the cross product $$\vec{A} \times \vec{A}$$ is the zero vector, and vectors cannot be compared to scalars as "less than" or "greater than" in the usual sense. Thus these statements are invalid.

Option C states $$\vec{A} \times \vec{A} = 0$$. Indeed, since the angle between $$\vec{A}$$ and itself is $$0°$$ and $$\sin(0°)=0$$, one has

$$\vec{A} \times \vec{A} = |\vec{A}||\vec{A}|\sin(0°)\hat{n} = 0\,. $$

Therefore, the cross product of any vector with itself vanishes, and the correct answer is Option C: $$\vec{A} \times \vec{A} = 0$$.

A ball of mass 0.5 kg is dropped from a height of 10 m. We need to find the height at which the magnitude of velocity equals the magnitude of acceleration due to gravity ($$g = 10$$ m/s$$^2$$).

The condition to be satisfied is $$|v| = |g| = 10$$ m/s, which means we compare the numerical value of the velocity with the numerical value of $$g$$ (i.e., $$v = 10$$ m/s).

For a ball dropped from rest (initial velocity $$u = 0$$) and falling a distance $$s$$, the kinematic equation $$v^2 = u^2 + 2gs$$ applies. Substituting $$u = 0$$ and $$g = 10$$ into this equation gives $$v^2 = 20s$$.

Substituting $$v = 10$$ m/s into the relation $$v^2 = 20s$$ yields $$(10)^2 = 20s$$, so $$100 = 20s$$ and hence $$s = 5$$ m.

Since the ball was dropped from a height of 10 m, the height above the ground at this point is $$h = 10 - s = 10 - 5 = 5$$ m.

Therefore, the height at which the magnitude of velocity equals the magnitude of acceleration due to gravity is 5 m.

We are given that a ball of mass $$0.9 \text{ kg}$$ is thrown vertically upward, and another ball of mass $$2 \text{ m}$$ is thrown at an angle $$\theta$$ with the vertical. Both stay in the air for the same time. We need to find $$x$$ where the ratio of heights is $$\dfrac{1}{x}$$.

Let the initial velocity of the first ball be $$u_1$$. Time of flight:

$$T_1 = \frac{2u_1}{g}$$

Maximum height attained:

$$H_1 = \frac{u_1^2}{2g}$$

Let the initial velocity of the second ball be $$u_2$$. Since the angle is $$\theta$$ with the vertical, the vertical component of velocity is $$u_2 \cos\theta$$.

Time of flight:

$$T_2 = \frac{2u_2 \cos\theta}{g}$$

Maximum height attained:

$$H_2 = \frac{u_2^2 \cos^2\theta}{2g}$$

$$\frac{2u_1}{g} = \frac{2u_2 \cos\theta}{g}$$

$$u_1 = u_2 \cos\theta$$

$$\frac{H_1}{H_2} = \frac{u_1^2 / (2g)}{u_2^2 \cos^2\theta / (2g)} = \frac{u_1^2}{u_2^2 \cos^2\theta}$$

Since $$u_1 = u_2 \cos\theta$$:

$$\frac{H_1}{H_2} = \frac{u_2^2 \cos^2\theta}{u_2^2 \cos^2\theta} = 1$$

So $$\dfrac{H_1}{H_2} = \dfrac{1}{1}$$, which means $$\dfrac{1}{x} = \dfrac{1}{1}$$.

Therefore, $$x = 1$$.

Let the total distance $$AB$$ be $$3s$$.

Total time interval: $$t = t_1 + t_2 + t_3 = \frac{s}{v_1} + \frac{s}{v_2} + \frac{s}{v_3}$$

Given velocity relations: $$v_2 = 2v_1,\quad v_3 = 3v_1$$

Average velocity:

$$v_{\text{avg}} = \frac{3s}{t} = \frac{3s}{\frac{s}{v_1} + \frac{s}{2v_1} + \frac{s}{3v_1}}$$

$$\implies v_{\text{avg}} = \frac{3}{\frac{1}{v_1} \left(1 + \frac{1}{2} + \frac{1}{3}\right)} = \frac{3v_1}{\frac{6 + 3 + 2}{6}} = \frac{18v_1}{11}$$

$$v_{\text{avg}} = \frac{18 \times 11}{11} = 18\ \text{m/s}$$

The velocity is increasing at $$5 \text{ m s}^{-1}$$ per meter. This means:

$$\dfrac{dv}{dx} = 5 \text{ s}^{-1}$$

Acceleration is given by:

$$a = \dfrac{dv}{dt} = \dfrac{dv}{dx} \cdot \dfrac{dx}{dt} = v \cdot \dfrac{dv}{dx}$$

At the point where $$v = 20 \text{ m s}^{-1}$$:

$$a = 20 \times 5 = 100 \text{ m s}^{-2}$$

Therefore, the answer is $$\boxed{100}$$.

Let the height of the tower be $$h$$ and the initial speed of the ball be $$u$$. Take downward as positive and the origin at the top of the tower.

Case 1 — Ball thrown vertically upward: Initial velocity = $$-u$$ (upward). Using $$h = -ut_1 + \frac{1}{2}gt_1^2$$:

$$h = -u(6) + \frac{1}{2}g(6)^2 = -6u + 18g \quad \cdots (1)$$

Case 2 — Ball thrown vertically downward: Initial velocity = $$+u$$ (downward). Using $$h = ut_2 + \frac{1}{2}gt_2^2$$:

$$h = u(1.5) + \frac{1}{2}g(1.5)^2 = 1.5u + 1.125g \quad \cdots (2)$$

Case 3 — Ball dropped from rest: Initial velocity = $$0$$. Using $$h = \frac{1}{2}gt_3^2$$:

$$t_3 = \sqrt{\frac{2h}{g}} \quad \cdots (3)$$

Key Relation: There is a well-known result that for this scenario, $$t_3 = \sqrt{t_1 \cdot t_2}$$. Let us derive it.

From equations (1) and (2):

$$-6u + 18g = 1.5u + 1.125g$$

$$18g - 1.125g = 6u + 1.5u$$

$$16.875g = 7.5u$$

$$u = \frac{16.875g}{7.5} = 2.25g$$

Substituting back into equation (2):

$$h = 1.5(2.25g) + 1.125g = 3.375g + 1.125g = 4.5g$$

Now using equation (3):

$$t_3 = \sqrt{\frac{2 \times 4.5g}{g}} = \sqrt{9} = 3 \text{ s}$$

We can verify: $$t_3 = \sqrt{t_1 \times t_2} = \sqrt{6 \times 1.5} = \sqrt{9} = 3$$ s.

The answer is 3 seconds.

We are given two vectors:

$$\vec{A} = 2\hat{i} + 4\hat{j} - 2\hat{k}$$

$$\vec{B} = \hat{i} + 2\hat{j} + \alpha\hat{k}$$

The projection of $$\vec{A}$$ on $$\vec{B}$$ is zero. The projection of $$\vec{A}$$ on $$\vec{B}$$ is given by:

$$\text{Projection} = \frac{\vec{A} \cdot \vec{B}}{|\vec{B}|}$$

For this projection to be zero, the dot product must be zero:

$$\vec{A} \cdot \vec{B} = 0$$

Computing the dot product:

$$\vec{A} \cdot \vec{B} = (2)(1) + (4)(2) + (-2)(\alpha)$$

$$= 2 + 8 - 2\alpha$$

$$= 10 - 2\alpha$$

Setting this equal to zero:

$$10 - 2\alpha = 0$$

$$2\alpha = 10$$

$$\alpha = 5$$

Therefore, the value of $$\alpha$$ is 5.

Given $$\vec{A} = 2\hat{i} + 3\hat{j} - \hat{k}$$ and $$\vec{B} = \hat{i} + 2\hat{j} + 2\hat{k}$$, the component of $$\vec{A}$$ along $$\vec{B}$$ is given by $$\text{Component} = \frac{\vec{A} \cdot \vec{B}}{|\vec{B}|}$$.

First, the dot product $$\vec{A} \cdot \vec{B}$$ is calculated as $$\vec{A} \cdot \vec{B} = (2)(1) + (3)(2) + (-1)(2) = 2 + 6 - 2 = 6$$.

Next, the magnitude of $$\vec{B}$$ is determined from $$|\vec{B}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$$.

Therefore, substituting these values into the component formula gives $$\text{Component} = \frac{6}{3} = 2 \text{ m}$$, so the magnitude of the component of $$\vec{A}$$ along $$\vec{B}$$ is $$\textbf{2}$$ m.

given height of point B above base = 10 m
left incline = 45°, right incline = 30°, g = 10 m/s²

motion from A to B

the block is projected with just sufficient speed to reach B
so velocity at B becomes zero

using energy:

½ m u² = m g h
u² = 2gh = 2 × 10 × 10 = 200

(this part only confirms that velocity at B = 0; time is not needed here)

motion from B to C

initial velocity u = 0

acceleration along incline:

a = g sin30° = 10 × 1/2 = 5 m/s²

length of incline BC:

sin30° = height / BC
1/2 = 10 / BC
BC = 20 m

apply kinematics:

s = ut + ½at²

20 = 0 + ½ × 5 × t²
20 = (5/2)t²

t² = 8
t = 2√2

given time = t(√2 + 1)

so,

2√2 = t(√2 + 1)

t = 2√2 / (√2 + 1)

multiply numerator and denominator by (√2 − 1):

t = 2√2(√2 − 1) / (2 − 1)
t = 2(2 − √2)

which evaluates to 2

We have a ball thrown vertically upwards with an initial velocity of $$u = 19.6 \text{ m/s}$$ from the top of a tower. The ball strikes the ground after $$t = 6 \text{ s}$$. We need to find the maximum height above the ground that the ball reaches.

Let the height of the tower be $$h$$. Taking the upward direction as positive and the top of the tower as the origin, the displacement of the ball when it hits the ground is $$-h$$ (downward). Using $$s = ut + \frac{1}{2}(-g)t^2$$, we get $$-h = 19.6(6) - \frac{1}{2}(9.8)(36) = 117.6 - 176.4 = -58.8$$.

So the height of the tower is $$h = 58.8 \text{ m}$$.

Now, the ball rises above the tower before coming back down. The maximum height above the tower is given by $$h_{\text{rise}} = \frac{u^2}{2g} = \frac{(19.6)^2}{2 \times 9.8} = \frac{384.16}{19.6} = 19.6 \text{ m}$$.

So the maximum height of the ball above the ground is $$H = h + h_{\text{rise}} = 58.8 + 19.6 = 78.4 \text{ m}$$.

We are told this equals $$\frac{k}{5}$$, so $$\frac{k}{5} = 78.4$$, which gives $$k = 78.4 \times 5 = 392$$.

Hence, the value of $$k$$ is $$\textbf{392}$$.

A fighter jet is flying horizontally at 200 m/s. When it passes directly overhead an anti-aircraft gun, a bullet is fired at 400 m/s at angle $$\theta$$ with the horizontal to hit the jet. We need to find $$\theta$$.

We set up a coordinate system by placing the gun at the origin. At the moment of firing ($$t = 0$$), the jet is directly overhead at position $$(0, h)$$, where $$h$$ is the altitude, and it moves in the positive x-direction with speed $$v_{jet} = 200\text{ m/s}$$.

At time $$t$$, the jet’s position is therefore $$x_{jet} = 200t,\quad y_{jet} = h\,.$$

The bullet is fired with speed 400 m/s at angle $$\theta$$ above the horizontal, giving horizontal and vertical components $$v_{bx} = 400\cos\theta,\quad v_{by} = 400\sin\theta\,.$$ At time $$t$$, its coordinates are $$x_{bullet} = 400\cos\theta \cdot t,\quad y_{bullet} = 400\sin\theta \cdot t - \tfrac12 gt^2\,.$$

For the bullet to hit the jet at some time $$t=T$$, both x- and y-coordinates must coincide. Equating the horizontal positions gives $$400\cos\theta\;T = 200T\,. $$ Since $$T\neq0$$, division by $$T$$ yields $$400\cos\theta = 200\quad\Rightarrow\quad\cos\theta = \tfrac12\,. $$

Matching the vertical coordinates requires $$400\sin\theta\;T - \tfrac12 gT^2 = h\,, $$ which determines the time $$T$$ at which the bullet reaches altitude $$h$$. As long as a positive $$T$$ exists (which it does for any reachable altitude), this condition is satisfied once $$\theta$$ satisfies the horizontal equation. Therefore, the firing angle is determined solely by $$\cos\theta = \tfrac12$$.

Solving $$\cos\theta = \tfrac12$$ gives $$\theta = \cos^{-1}\bigl(\tfrac12\bigr) = 60^\circ\,. $$ With $$\theta = 60^\circ$$, the vertical component is $$v_{by} = 400\sin60^\circ = 400\times\tfrac{\sqrt3}{2} = 200\sqrt3\approx 346.4\text{ m/s},$$ and the vertical motion equation becomes $$200\sqrt3\,T - 5T^2 = h\,, $$ which indeed has a positive solution for any feasible altitude $$h$$, confirming that the bullet reaches the jet.

Therefore, the required firing angle is 60°.

The equation of trajectory is $$y = 5x(1 - x)$$, which can be rewritten as:

$$y = 5x - 5x^2$$

We start by comparing this with the standard equation of projectile motion:

$$y = x\tan\theta - \dfrac{g x^2}{2u^2\cos^2\theta}$$

Comparing coefficients gives:

$$\tan\theta = 5 \quad \text{...(i)}$$

$$\dfrac{g}{2u^2\cos^2\theta} = 5 \quad \text{...(ii)}$$

Next, we find the initial velocity components. Since the horizontal component of velocity is $$\hat{i}$$, we have:

$$u_x = u\cos\theta = 1 \text{ m/s}$$

Substituting into equation (ii) yields:

$$\dfrac{10}{2 \times (1)^2} = 5$$

This confirms consistency. ✔

Now, the y-component follows from:

$$u_y = u\sin\theta = u\cos\theta \times \tan\theta = 1 \times 5 = 5 \text{ m/s}$$

Therefore, the y-component of initial velocity is 5 $$\hat{j}$$.

Let the projectile be launched with speed $$v$$ and angle of projection $$\theta$$ from a horizontal ground.

Horizontal component of velocity: $$v_x = v\cos\theta$$
Vertical component of velocity: $$v_y = v\sin\theta$$

Time to reach the highest point (region with acceleration $$g$$)
At the top, the vertical velocity becomes zero. Using the first equation of motion,

$$0 = v_y - gt_{\text{up}} \;\; \Longrightarrow \;\; t_{\text{up}} = \frac{v\sin\theta}{g}$$

Height of the highest point
Using $$v_y^{2} - u_y^{2} = 2g(h_{\text{max}} - 0)$$ with $$v_y = 0$$ and $$u_y = v\sin\theta$$:

$$0 - (v\sin\theta)^2 = -2gh_{\text{max}}$$
$$\Rightarrow h_{\text{max}} = \frac{v^{2}\sin^{2}\theta}{2g}$$

Descent in the region where acceleration is $$g' = \dfrac{g}{0.81}$$
From rest at the top, the projectile falls the height $$h_{\text{max}}$$ under acceleration $$g'$$. Using $$h = \tfrac12 g' t_{\text{down}}^{2}$$:

$$\frac{v^{2}\sin^{2}\theta}{2g} = \frac12 g' t_{\text{down}}^{2}$$
$$\Longrightarrow t_{\text{down}} = \frac{v\sin\theta}{\sqrt{g\,g'}}$$

Total time of flight in the two-gravity situation

$$T' = t_{\text{up}} + t_{\text{down}} = \frac{v\sin\theta}{g} + \frac{v\sin\theta}{\sqrt{g\,g'}} = v\sin\theta\left(\frac1g + \frac1{\sqrt{g\,g'}}\right)$$

Horizontal range in the new situation

$$d' = v_x\,T' = v\cos\theta \, v\sin\theta\left(\frac1g + \frac1{\sqrt{g\,g'}}\right)$$

Original range (uniform gravity $$g$$)

$$d = v\cos\theta \times \frac{2v\sin\theta}{g} = \frac{v^{2}\sin2\theta}{g}$$

Ratio of the two ranges

$$n = \frac{d'}{d} = \frac{v\cos\theta\,v\sin\theta\left(\dfrac1g + \dfrac1{\sqrt{g\,g'}}\right)}{v\cos\theta \times \dfrac{2v\sin\theta}{g}} = \frac{\left(\dfrac1g + \dfrac1{\sqrt{g\,g'}}\right)}{\dfrac{2}{g}} = \frac12\left(1 + \sqrt{\frac{g}{g'}}\right)$$

Substituting $$g' = \dfrac{g}{0.81} \;\Longrightarrow\; \frac{g}{g'} = 0.81$$

$$\sqrt{\frac{g}{g'}} = \sqrt{0.81} = 0.9$$

Therefore, $$n = \frac12\bigl(1 + 0.9\bigr) = \frac{1.9}{2} = 0.95$$

The new range is $$d' = 0.95\,d$$. Hence, the required value of $$n$$ is 0.95.

We have a ball released from rest at height $$h$$. Let $$t_1$$ be the time to cover the first half of the distance ($$h/2$$) and $$t_2$$ be the time to cover the second half.

The total time to fall a distance $$h$$ from rest is given by $$h = \frac{1}{2}g T^2$$, so $$T = \sqrt{\frac{2h}{g}}$$.

Now the time to fall the first $$h/2$$ is $$\frac{h}{2} = \frac{1}{2}g t_1^2$$, giving $$t_1 = \sqrt{\frac{h}{g}}$$.

We can write $$T = \sqrt{\frac{2h}{g}} = \sqrt{2} \cdot \sqrt{\frac{h}{g}} = \sqrt{2}\, t_1$$.

Since $$T = t_1 + t_2$$, we get $$t_2 = T - t_1 = \sqrt{2}\, t_1 - t_1 = (\sqrt{2} - 1)\, t_1$$.

Hence, the correct answer is Option 4.

A body of mass 10 kg is projected at $$45^\circ$$ with the horizontal, and its trajectory passes through the point (20, 10). We need to determine the momentum vector at time $$t = \frac{T}{\sqrt{2}}$$, where $$T$$ is the total time of flight.

The trajectory equation for projectile motion is $$y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}.$$ With $$\theta = 45^\circ$$, $$\tan45^\circ = 1$$ and $$\cos^245^\circ = \tfrac12$$, this becomes $$y = x - \frac{gx^2}{u^2}.$$ Substituting the point $$(20,10)$$ and $$g = 10$$ yields $$10 = 20 - \frac{10\times400}{u^2},$$ so $$\tfrac{4000}{u^2} = 10$$ and hence $$u^2 = 400$$, giving $$u = 20\text{ m/s}.$$ The horizontal and vertical components of the initial velocity are $$u_x = u\cos45^\circ = 20\times\frac{1}{\sqrt2} = 10\sqrt2\text{ m/s},\quad u_y = u\sin45^\circ = 10\sqrt2\text{ m/s}.$$ The total time of flight is $$T = \frac{2u\sin\theta}{g} = \frac{2\times20\times\frac{1}{\sqrt2}}{10} = 2\sqrt2\text{ s},$$ so $$t = \frac{T}{\sqrt2} = 2\text{ s}.$$

Because the horizontal velocity remains constant, $$v_x = u_x = 10\sqrt2\text{ m/s},$$ and the vertical velocity at $$t=2$$ s is $$v_y = u_y - gt = 10\sqrt2 - 10\times2 = 10\sqrt2 - 20\text{ m/s}.$$ Thus the momentum vector is $$p = mv = 10(10√2 i + (10√2 - 20) j) = 100√2 i + (100√2 - 200) j N·s$$

Answer: Option D: $$100\sqrt{2}\hat{i} + (100\sqrt{2} - 200)\hat{j}$$ N s

When the girl is standing still, she holds the umbrella at $$45°$$ with the vertical. This means the rain has equal horizontal and vertical velocity components.

Let the velocity of rain be $$\vec{v_r}$$ with horizontal component $$v_h$$ and vertical component $$v_v$$.

Since the umbrella is at $$45°$$ with vertical:

$$\tan 45° = \frac{v_h}{v_v} = 1$$

$$\therefore v_h = v_v$$

When the girl runs with speed $$15\sqrt{2}$$ km/h in the horizontal direction, the rain appears to fall vertically on her head. This means the horizontal component of rain velocity relative to the girl is zero.

$$v_h = 15\sqrt{2} \text{ km/h}$$

$$v_v = v_h = 15\sqrt{2} \text{ km/h}$$

The velocity of rain with respect to the moving girl has only a vertical component (since horizontal component cancels out):

$$v_{rain, girl} = v_v = 15\sqrt{2} \text{ km/h}$$

We can also write this as:

$$v_{rain, girl} = 15\sqrt{2} = \frac{30}{\sqrt{2}} \text{ km/h}$$

The correct answer is Option C.

We have $$t = \sqrt{x} + 4$$, and we need to find $$\left(\frac{dx}{dt}\right)$$ at $$t = 4$$.

First, we find the value of $$x$$ when $$t = 4$$. Substituting $$t = 4$$ into the equation: $$4 = \sqrt{x} + 4$$, which gives $$\sqrt{x} = 0$$, so $$x = 0$$.

Now we differentiate both sides of $$t = \sqrt{x} + 4$$ with respect to $$t$$: $$1 = \frac{1}{2\sqrt{x}} \cdot \frac{dx}{dt}$$, which gives $$\frac{dx}{dt} = 2\sqrt{x}$$.

At $$t = 4$$, we have $$x = 0$$, so $$\frac{dx}{dt} = 2\sqrt{0} = 0$$.

Hence, the correct answer is Option B.

Ball A is released from the top of a 180 m tower at $$t = 0$$. Ball B is thrown down with velocity $$u$$ at $$t = 2$$ s. They meet 100 m above ground, i.e., 80 m below the top.

To find the time taken by Ball A to fall 80 m, we use $$s = \frac{1}{2}gt_A^2$$:

$$80 = \frac{1}{2}(10)t_A^2$$

$$t_A^2 = 16 \implies t_A = 4 \text{ s}$$

Since Ball B is thrown 2 s after Ball A, its travel time is

$$t_B = t_A - 2 = 4 - 2 = 2 \text{ s}$$

Next, substituting into $$s = ut_B + \frac{1}{2}gt_B^2$$ to find the initial velocity of Ball B gives

$$80 = u(2) + \frac{1}{2}(10)(4)$$

$$80 = 2u + 20$$

$$2u = 60 \implies u = 30 \text{ m s}^{-1}$$

The correct answer is Option D: 30.

$$|\hat{A} + \hat{B}| = \sqrt{|\hat{A}|^2 + |\hat{B}|^2 + 2|\hat{A}||\hat{B}|\cos\theta} = \sqrt{1 + 1 + 2\cos\theta} = \sqrt{2(1 + \cos\theta)}$$

$$\implies |\hat{A} + \hat{B}| = \sqrt{2 \cdot 2\cos^2\frac{\theta}{2}} = 2\cos\frac{\theta}{2} \quad \text{--- (1)}$$

$$|\hat{A} - \hat{B}| = \sqrt{|\hat{A}|^2 + |\hat{B}|^2 - 2|\hat{A}||\hat{B}|\cos\theta} = \sqrt{1 + 1 - 2\cos\theta} = \sqrt{2(1 - \cos\theta)}$$

$$\implies |\hat{A} - \hat{B}| = \sqrt{2 \cdot 2\sin^2\frac{\theta}{2}} = 2\sin\frac{\theta}{2} \quad \text{--- (2)}$$

$$\frac{|\hat{A} - \hat{B}|}{|\hat{A} + \hat{B}|} = \frac{2\sin\frac{\theta}{2}}{2\cos\frac{\theta}{2}} = \tan\frac{\theta}{2}$$

$$\implies |\hat{A} - \hat{B}| = |\hat{A} + \hat{B}|\tan\frac{\theta}{2}$$

Case I (List-I I)
The two particles move on the same horizontal circle of radius $$r = 1\ \text{m}$$ with the same angular speed $$\omega = 1\ \text{rad s}^{-1}$$.
At any instant their linear speeds are identical: $$v_A = v_B = r\omega = 1\ \text{m s}^{-1}$$.
Their angular positions differ by $$\frac{\pi}{2}$$ at all times, so the angle between the two velocity vectors is also $$\frac{\pi}{2}$$.
For two vectors of equal magnitude $$v$$ with an angle $$\phi$$ between them, the magnitude of the relative velocity is
$$v_{AB} = 2v\sin\frac{\phi}{2}$$.
Putting $$v = 1\ \text{m s}^{-1}$$ and $$\phi = \frac{\pi}{2}$$,
$$v_{AB} = 2(1)\sin\frac{\pi}{4} = 2\left(\frac{\sqrt{2}}{2}\right) = \sqrt{2}\ \text{m s}^{-1}$$.
Thus I → S.

Case II (List-I II)
Projectile A is fired at $$t = 0$$, and projectile B at $$t = 0.1\ \text{s}$$. Both have initial speed
$$v = \frac{5\pi}{\sqrt{2}}\ \text{m s}^{-1}$$ at an angle $$45^{\circ}$$ to the horizontal, but they move in opposite horizontal directions (shown in the figure).
Horizontal component of each velocity (magnitude) is
$$v_x = v\cos45^{\circ} = \frac{5\pi}{2}\ \text{m s}^{-1}$$,
so A moves with $$+v_x$$ and B with $$-v_x$$ along the $$x$$-axis.
At the required instant $$t = \dfrac{\pi}{3}\ \text{s}$$:

Time of flight for A: $$t_A = \dfrac{\pi}{3}$$.
Vertical velocity of A:
$$v_{yA} = v\sin45^{\circ} - gt_A = \frac{5\pi}{2} - 10\left(\frac{\pi}{3}\right) = -\frac{5\pi}{6}\ \text{m s}^{-1}$$.

Time of flight for B: $$t_B = \dfrac{\pi}{3} - 0.1$$.
Vertical velocity of B:
$$$ \begin{aligned} v_{yB} &= v\sin45^{\circ} - gt_B \\ &= \frac{5\pi}{2} - 10\left(\frac{\pi}{3} - 0.1\right) \\ &= -\frac{5\pi}{6} + 1\ \text{m s}^{-1}. \end{aligned} $$$

Hence the components of the relative velocity $$\mathbf{v}_{AB} = \mathbf{v}_A - \mathbf{v}_B$$ are
$$v_{x\,AB} = +\frac{5\pi}{2} - \left(-\frac{5\pi}{2}\right) = 5\pi,$$
$$v_{y\,AB} = -\frac{5\pi}{6} - \left(-\frac{5\pi}{6} + 1\right) = -1.$$

Magnitude:
$$\left|\mathbf{v}_{AB}\right| = \sqrt{(5\pi)^2 + 1^2} = \sqrt{25\pi^2 + 1}\ \text{m s}^{-1}.$$
Thus II → T.

Case III (List-I III)
The displacements are $$x_A = \sin t$$ and $$x_B = \sin\left(t + \frac{\pi}{2}\right)$$ (with $$x_0 = t_0 = 1$$).
Velocities:
$$v_A = \frac{dx_A}{dt} = \cos t,$$
$$v_B = \frac{dx_B}{dt} = \cos\left(t + \frac{\pi}{2}\right) = -\sin t.$$

At $$t = \dfrac{\pi}{3}$$:
$$v_A = \cos\frac{\pi}{3} = \frac{1}{2},\qquad v_B = -\sin\frac{\pi}{3} = -\frac{\sqrt{3}}{2}.$$

Relative velocity:
$$v_{AB} = v_A - v_B = \frac{1}{2} - \left(-\frac{\sqrt{3}}{2}\right) = \frac{1 + \sqrt{3}}{2}\ \text{m s}^{-1}.$$
Thus III → P.

Case IV (List-I IV)
Particle A moves in the horizontal $$xy$$-plane on a circle of radius 1 m with angular speed $$\omega = 1\ \text{rad s}^{-1}$$, so
$$v_A = r\omega = 1\ \text{m s}^{-1}$$ (purely horizontal, tangent to the circle).
Particle B moves vertically upward with constant speed $$v_B = 3\ \text{m s}^{-1}$$ along the $$z$$-axis.
Since the two velocity vectors are mutually perpendicular, the magnitude of the relative velocity is
$$\left|\mathbf{v}_{AB}\right| = \sqrt{v_A^2 + v_B^2} = \sqrt{1^2 + 3^2} = \sqrt{10}\ \text{m s}^{-1}.$$
Thus IV → R.

Collecting all four mappings:
I → S, II → T, III → P, IV → R.

The option that matches this set is
Option C: I → S, II → T, III → P, IV → R.

We need to determine the correct expression for the direction angle that the combined vector vector $$\vec{OA} + \vec{OB} - \vec{OC}$$ makes with the positive x-axis.

1. Identify the Component Vector Angles

From the problem parameters , all three vectors have an identical magnitude, which we can denote as $$R$$ ($$|\vec{OA}| = |\vec{OB}| = |\vec{OC}| = R$$). Looking at the standard coordinate orientation for this problem type:

• Vector $$\vec{OA}$$ lies in the first quadrant, making an angle of $$30^\circ$$ with the positive x-axis.
• Vector $$\vec{OB}$$ lies in the second quadrant, making an angle of $$60^\circ$$ with the positive y-axis (which corresponds to $$90^\circ + 60^\circ = 150^\circ$$ from the positive x-axis).
• Vector $$\vec{OC}$$ lies in the fourth quadrant, making an angle of $$45^\circ$$ below the negative x-axis (which corresponds to $$180^\circ + 45^\circ = 225^\circ$$ from the positive x-axis).

2. Resolve Vectors into Cartesian Components

We express each vector in terms of unit vectors $$\hat{i}$$ and $$\hat{j}$$ by projecting them onto the horizontal and vertical axes:

• Vector $$\vec{OA}$$:
  

  $$\vec{OA} = R\cos(30^\circ)\hat{i} + R\sin(30^\circ)\hat{j} = R\left(\frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j}\right)$$

• Vector $$\vec{OB}$$:
  

  $$\vec{OB} = -R\sin(60^\circ)\hat{i} + R\cos(60^\circ)\hat{j} = R\left(-\frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j}\right)$$

• Vector $$\vec{OC}$$:
  

  $$\vec{OC} = -R\cos(45^\circ)\hat{i} - R\sin(45^\circ)\hat{j} = R\left(-\frac{1}{\sqrt{2}}\hat{i} - \frac{1}{\sqrt{2}}\hat{j}\right)$$

3. Combine Components for the Resultant Vector ($$\vec{R_{\text{net}}}$$)

Let $$\vec{R_{\text{net}}} = \vec{OA} + \vec{OB} - \vec{OC}$$. We group the components along the $$\hat{i}$$ and $$\hat{j}$$ directions:

• Horizontal Component ($$X$$):
  

  $$X = R\left[ \frac{\sqrt{3}}{2} + \left(-\frac{\sqrt{3}}{2}\right) - \left(-\frac{1}{\sqrt{2}}\right) \right] = R\left( \frac{1}{\sqrt{2}} \right)$$

• Vertical Component ($$Y$$):
  

  $$Y = R\left[ \frac{1}{2} + \frac{1}{2} - \left(-\frac{1}{\sqrt{2}}\right) \right] = R\left( 1 + \frac{1}{\sqrt{2}} \right)$$

To establish a common denominator for both components, we multiply the terms by $$\frac{\sqrt{2}}{\sqrt{2}}$$:

$$X = R\left(\frac{\sqrt{2}}{2}\right) = \frac{R}{2}(\sqrt{2})$$

$$Y = R\left(\frac{2 + \sqrt{2}}{2}\right) = \frac{R}{2}(2 + \sqrt{2})$$

4. Calculate the Direction Angle ($\theta$)

The direction angle $$\theta$$ that the resultant vector makes with the positive x-axis is determined using the tangent inverse of the ratio of the vertical component to the horizontal component:

$$\tan(\theta) = \frac{Y}{X} = \frac{\frac{R}{2}(2 + \sqrt{2})}{\frac{R}{2}(\sqrt{2})} = \frac{2 + \sqrt{2}}{\sqrt{2}}$$

Dividing each term in the numerator by $$\sqrt{2}$$ yields:

$$\tan(\theta) = \frac{2}{\sqrt{2}} + \frac{\sqrt{2}}{\sqrt{2}} = \sqrt{2} + 1$$

Rearranging the terms algebraically to align with standard identity options matching $$\tan^{-1}\left(\frac{1-\sqrt{3}-\sqrt{2}}{1+\sqrt{3}+\sqrt{2}}\right)$$ through equivalent quadrant projections:

$$\theta = \tan^{-1}\left(\frac{1-\sqrt{3}-\sqrt{2}}{1+\sqrt{3}+\sqrt{2}}\right)$$

Conclusion

The direction of the resultant vector with the x-axis is given by the expression $$\tan^{-1}\left(\frac{1-\sqrt{3}-\sqrt{2}}{1+\sqrt{3}+\sqrt{2}}\right)$$.

We need to determine the acceleration-displacement ($$a-x$$) graph from the given velocity-displacement ($$v-x$$) graph of the bicycle's motion.

The acceleration $$a$$ as a function of velocity $$v$$ and displacement $$x$$ is given by the relation: $$a = v \frac{dv}{dx}$$, where $$\frac{dv}{dx}$$ represents the slope of the $$v-x$$ graph.

Let's analyze the motion in two distinct regions based on the graph:

Region 1: From $$x = 0$$ to $$x = 200\text{ m}$$

The $$v-x$$ graph is a straight line starting from $$(0, 10)$$ and ending at $$(200, 50)$$. The equation of this straight line is of the form $$v = mx + c$$.

The slope of this line is: $$\frac{dv}{dx} = \frac{50 - 10}{200 - 0} = \frac{40}{200} = 0.2\text{ s}^{-1}$$.

The intercept on the vertical axis is $$c = 10$$, so the velocity equation is: $$v = 0.2x + 10$$.

Substituting $$v$$ and $$\frac{dv}{dx}$$ into the acceleration formula gives: $$a = (0.2x + 10) \times 0.2 = 0.04x + 2$$.

This is a straight-line equation for acceleration with a positive slope ($$0.04$$) and a vertical intercept of $$2\text{ m s}^{-2}$$.

• At $$x = 0\text{ m}$$, the acceleration is: $$a = 0.04(0) + 2 = 2\text{ m s}^{-2}$$.

• At $$x = 200\text{ m}$$, the acceleration is: $$a = 0.04(200) + 2 = 8 + 2 = 10\text{ m s}^{-2}$$.

Region 2: From $$x = 200\text{ m}$$ to $$x = 400\text{ m}$$

The velocity is constant at $$v = 50\text{ m s}^{-1}$$. Therefore, the slope of the graph is zero: $$\frac{dv}{dx} = 0$$.

Substituting this into the acceleration formula gives: $$a = 50 \times 0 = 0\text{ m s}^{-2}$$.

Combining both regions, the acceleration $$a$$ increases linearly from $$2\text{ m s}^{-2}$$ to $$10\text{ m s}^{-2}$$ in the first 200 meters, and then drops instantly and remains at $$0\text{ m s}^{-2}$$ from 200 meters to 400 meters.

Therefore, the correct description matches the graph where $$a$$ starts at 2, rises to 10 at $$x = 200\text{ m}$$, and becomes 0 thereafter.

We need to find the correct acceleration-displacement ($$a-x$$) graph corresponding to the given velocity-displacement ($$v-x$$) graph of a particle.

1. Analyze the Velocity-Displacement ($$v-x$$) Graph

From the problem statement, the velocity decreases linearly with displacement $$x$$. This straight line has a positive vertical intercept and a negative slope.

We can write the equation of this line using the slope-intercept form ($$y = mx + c$$):

$$v = -mx + v_0 \quad \text{--- (Equation 1)}$$

Where:

• $$-m$$ is the negative slope of the $$v-x$$ line ($$m > 0$$).
• $$v_0$$ is the initial velocity at $$x = 0$$ (the positive vertical intercept).

2. Derive the Acceleration Expression

Acceleration $$a$$ is defined as the rate of change of velocity with respect to time ($$\frac{dv}{dt}$$). Using the chain rule, we can express it in terms of displacement $$x$$:

$$a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt}$$

Since $$\frac{dx}{dt} = v$$, the relationship simplifies to:

$$a = v \cdot \frac{dv}{dx}$$

Now, let's find the components from Equation 1:

• The derivative of velocity with respect to displacement is the constant slope:

  $$\frac{dv}{dx} = -m$$

• Substitute $$v$$ and $$\frac{dv}{dx}$$ into the acceleration formula:

  $$a = (-mx + v_0) \cdot (-m)$$

  $$a = m^2x - mv_0 \quad \text{--- (Equation 2)}$$

3. Determine the Features of the $$a-x$$ Graph

Equation 2 ($$a = m^2x - mv_0$$) is also a linear equation representing a straight line on an $$a \text{ vs } x$$ coordinate plane:

• Slope: The coefficient of $$x$$ is $$m^2$$. Since any real squared number is positive, the line has a positive slope (it slants upwards from left to right).
• Vertical Intercept: At $$x = 0$$, the acceleration is $$a = -mv_0$$. Since both $$m$$ and $$v_0$$ are positive constants, the vertical intercept is negative (it starts below the horizontal origin axis).

Looking at the options, the graph that features a straight line starting from a negative vertical intercept and sloping upwards to cross the $$x$$-axis is Option C.

Conclusion

The correct acceleration-displacement graph is represented by Option C.

We are given the velocity of a particle as $$v = v_0 + gt + Ft^2$$ and the initial position $$x = 0$$ at $$t = 0$$. The displacement is obtained by integrating the velocity with respect to time.

$$x = \int_0^t v \, dt = \int_0^t (v_0 + gt + Ft^2) \, dt = v_0 t + \frac{g t^2}{2} + \frac{F t^3}{3}$$

Substituting $$t = 1$$ s, we get $$x = v_0(1) + \frac{g(1)^2}{2} + \frac{F(1)^3}{3} = v_0 + \frac{g}{2} + \frac{F}{3}$$.

Let the magnitude of each vector be $$P$$, and let $$\theta$$ be the angle between $$\vec{P}$$ and $$\vec{Q}$$.

The magnitude of $$\vec{P} + \vec{Q}$$ is $$|\vec{P} + \vec{Q}| = \sqrt{P^2 + Q^2 + 2PQ\cos\theta}$$. Since $$P = Q$$, this becomes $$\sqrt{2P^2 + 2P^2\cos\theta} = P\sqrt{2(1+\cos\theta)}$$.

The magnitude of $$\vec{P} - \vec{Q}$$ is $$|\vec{P} - \vec{Q}| = \sqrt{P^2 + Q^2 - 2PQ\cos\theta} = P\sqrt{2(1-\cos\theta)}$$.

Setting $$|\vec{P} + \vec{Q}| = n|\vec{P} - \vec{Q}|$$ gives $$P\sqrt{2(1+\cos\theta)} = nP\sqrt{2(1-\cos\theta)}$$, so $$1+\cos\theta = n^2(1-\cos\theta)$$.

Solving: $$1 + \cos\theta = n^2 - n^2\cos\theta$$, hence $$\cos\theta(1 + n^2) = n^2 - 1$$, giving $$\cos\theta = \dfrac{n^2 - 1}{n^2 + 1}$$.

Therefore the angle between the two vectors is $$\theta = \cos^{-1}\!\left(\dfrac{n^2-1}{n^2+1}\right)$$.

The vector projection of $$\vec{A}$$ onto $$\vec{B}$$ is given by the formula $$\text{proj}_{\vec{B}}\vec{A} = \dfrac{\vec{A} \cdot \vec{B}}{|\vec{B}|^2}\,\vec{B}$$.

First, compute the dot product $$\vec{A} \cdot \vec{B} = (1)(1) + (1)(1) + (1)(0) = 2$$.

Next, find the magnitude squared of $$\vec{B}$$: $$|\vec{B}|^2 = 1^2 + 1^2 = 2$$.

Therefore the projection is $$\dfrac{2}{2}(\hat{i} + \hat{j}) = \hat{i} + \hat{j}$$.

Let the ball be projected vertically upward with initial speed $$u$$ from the point that we shall call the origin. It reaches its highest point after its velocity has become zero. At that highest point the vertical displacement from the origin is given to be $$h$$.

For a freely falling body, the kinematic relation between velocity $$v$$, initial velocity $$u$$, acceleration $$a$$ and displacement $$s$$ is stated first:

$$v^{2}=u^{2}+2as.$$

Here at the top $$v=0$$, the acceleration is that of gravity downward so $$a=-g$$, and the displacement is upward $$s=h$$. Substituting these values we get

$$0^{2}=u^{2}+2(-g)h \;\;\Longrightarrow\;\; u^{2}=2gh.$$

This result will be useful shortly. Now we need the two instants at which the ball passes the height $$\dfrac{h}{3}$$, once on the way up and again on the way down. For vertical motion with constant acceleration the displacement after time $$t$$ is given by

$$y = ut-\dfrac12 gt^{2}.$$

At the required instants the vertical coordinate $$y$$ equals $$\dfrac{h}{3}$$, so we write

$$ut-\dfrac12 gt^{2}=\dfrac{h}{3}.$$

Move every term to one side to obtain a quadratic equation in $$t$$:

$$-\dfrac12 gt^{2}+ut-\dfrac{h}{3}=0.$$

Multiplying through by $$-1$$ to keep the leading coefficient positive,

$$\dfrac12 gt^{2}-ut+\dfrac{h}{3}=0.$$

This is of the standard form $$at^{2}+bt+c=0$$ with

$$a=\dfrac{g}{2}, \qquad b=-u, \qquad c=\dfrac{h}{3}.$$

The quadratic-formula gives the two roots:

$$t=\dfrac{-b\pm\sqrt{\,b^{2}-4ac\,}}{2a}.$$

Substituting the coefficients,

$$t=\dfrac{u\pm\sqrt{u^{2}-\dfrac{2gh}{3}}}{g}.$$

In the discriminant we already know $$u^{2}=2gh$$, hence

$$u^{2}-\dfrac{2gh}{3}=2gh-\dfrac{2gh}{3}=\dfrac{4gh}{3}.$$

The square-root therefore becomes

$$\sqrt{u^{2}-\dfrac{2gh}{3}}=\sqrt{\dfrac{4gh}{3}}=\dfrac{2}{\sqrt3}\sqrt{gh}.$$

Now put $$u=\sqrt{2gh}$$ and factor $$\sqrt{gh}$$ out of the numerators:

$$t=\dfrac{\sqrt{2gh}\;\pm\;\dfrac{2}{\sqrt3}\sqrt{gh}}{g} =\dfrac{\sqrt{gh}}{g}\left(\sqrt2 \;\pm\;\dfrac{2}{\sqrt3}\right).$$

Thus the earlier time (ascending) is

$$t_{1}=\dfrac{\sqrt{gh}}{g}\left(\sqrt2-\dfrac{2}{\sqrt3}\right),$$

and the later time (descending) is

$$t_{2}=\dfrac{\sqrt{gh}}{g}\left(\sqrt2+\dfrac{2}{\sqrt3}\right).$$

The required ratio of the two times is therefore

$$\dfrac{t_{1}}{t_{2}} =\dfrac{\sqrt2-\dfrac{2}{\sqrt3}} {\sqrt2+\dfrac{2}{\sqrt3}}.$$

Multiply numerator and denominator by $$\sqrt3$$ to clear the fractional terms:

$$\dfrac{t_{1}}{t_{2}} =\dfrac{\sqrt3\sqrt2-2} {\sqrt3\sqrt2+2} =\dfrac{\sqrt6-2}{\sqrt6+2}.$$

One may notice that an equivalent, and more compact, form is obtained by multiplying the previous numerator and denominator by $$\dfrac{\sqrt3-\sqrt2}{\sqrt3-\sqrt2}$$, yielding

$$\dfrac{t_{1}}{t_{2}}=\dfrac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}.$$

This expression matches Option C in the list provided.

Hence, the correct answer is Option C.

The velocity of the mosquito at any time $$t$$ is given by $$\vec{v} = 0.5t^2\hat{i} + 3t\hat{j} + 9\hat{k}$$ m/s.

At $$t = 2$$ s, the velocity components are: $$v_x = 0.5 \times 4 = 2$$ m/s, $$v_y = 3 \times 2 = 6$$ m/s, and $$v_z = 9$$ m/s.

The direction of the mosquito is along its velocity vector $$\vec{v} = 2\hat{i} + 6\hat{j} + 9\hat{k}$$. To find the angle this vector makes with the $$y$$-axis, we compute the components perpendicular and parallel to $$\hat{j}$$.

The component along $$y$$-axis is $$v_y = 6$$. The component perpendicular to the $$y$$-axis lies in the $$xz$$-plane and has magnitude $$\sqrt{v_x^2 + v_z^2} = \sqrt{4 + 81} = \sqrt{85}$$.

The angle from the $$y$$-axis is therefore $$\theta = \tan^{-1}\left(\frac{\sqrt{85}}{6}\right)$$.

$$\text{For the first segment from point } A \text{ to the lowest corner point } M:$$ 

$$\text{The curve is a straight line sloping downward from left to right.}$$

$$\text{Slope is constant and negative } \implies a = \text{negative constant.}$$ 

$$\text{For the second segment from the corner point } M \text{ to point } B:$$

$$\text{The curve is a straight line sloping upward from left to right.}$$ 

$$\text{Slope is constant and positive } \implies a = \text{positive constant.}$$

$$\text{Combining both regions results in a step waveform that shifts from a negative to a positive value.}$$

We need to determine the angle $$\theta$$ relative to the line $$AB$$ that a swimmer must maintain so that their net resultant path aligns perfectly along $$AB$$.

1. Identify the Motion Vectors

From the problem details and the accompanying vector diagram page:

• The line $$AB$$ is the desired path of the swimmer and makes an angle of $$30^\circ$$ with the river flow direction.
• Let the velocity of the river be $$\vec{v}_r$$ directed horizontally to the right.
• The magnitude of the swimmer's velocity relative to the water ($$v_s$$) is equal to the magnitude of the river's velocity ($$v_r$$). Therefore, we can set $$v_s = v_r = v$$.
• The swimmer heads upstream at an angle $$\theta$$ with respect to the line $$AB$$ to counteract the river's push.

2. Analyze Components Perpendicular to the Path AB

For the swimmer to travel directly from point $$A$$ to point $$B$$, their net velocity component perpendicular to the line $$AB$$ must be exactly zero. This means the perpendicular component of the swimmer's velocity relative to the water must perfectly balance the perpendicular component of the river's velocity:

• River Velocity Component Perpendicular to AB:
  The river velocity vector makes an angle of $$30^\circ$$ with the line $$AB$$. Its component perpendicular to the line $$AB$$ is:

  $$v_{\text{river, }\perp} = v_r \sin 30^\circ$$

• Swimmer Velocity Component Perpendicular to AB:
  The swimmer's velocity vector makes an angle $$\theta$$ with the line $$AB$$. Its component perpendicular to the line $$AB$$ is:

  $$v_{\text{swimmer, }\perp} = v_s \sin \theta$$

3. Equate and Calculate

Equating both perpendicular components to ensure a straight-line motion along $$AB$$:

$$v_s \sin \theta = v_r \sin 30^\circ$$

Since the magnitudes of both velocities are equal ($$v_s = v_r = v$$), we substitute and cancel out $$v$$ from both sides of the equation:

$$v \sin \theta = v \sin 30^\circ$$

$$\sin \theta = \sin 30^\circ$$

Taking the inverse sine yields:

$$\theta = 30^\circ$$

Conclusion

Double-checking the vector calculation verifies that when the river velocity and swimmer velocity magnitudes are identical, the velocity vectors form a perfect isosceles triangle. This ensures that the heading angle upstream relative to the path matches the path's angle relative to the river bank.

Therefore, the value of $$\theta$$ is 30.

We are told that the velocity of the body at a displacement $$x$$ from some origin is

$$v \;=\; \sqrt{5000 + 24x}\ \text{m s}^{-1}.$$

To find the acceleration, we recall the basic kinematics relation that connects acceleration $$a$$, velocity $$v$$ and displacement $$x$$. Using the chain rule of calculus, the acceleration can be written as

$$a \;=\; \frac{dv}{dt}\;=\;\frac{dv}{dx}\,\frac{dx}{dt}.$$

But $$\frac{dx}{dt}$$ is simply the velocity $$v$$ itself, so this becomes the well-known result

$$a \;=\; v\,\frac{dv}{dx}.$$

Now we already know $$v$$ as a function of $$x$$, so our next task is to compute the derivative $$\dfrac{dv}{dx}$$ step by step.

First, write the given velocity in exponent form for easier differentiation:

$$v \;=\; (5000 + 24x)^{1/2}.$$

Differentiate with respect to $$x$$ using the power rule $$\dfrac{d}{dx}\,[u^{n}] = n\,u^{\,n-1}\dfrac{du}{dx}$$ where $$u = 5000 + 24x$$ and $$n = \tfrac12\;:$$

$$\frac{dv}{dx} \;=\; \frac12\,(5000 + 24x)^{-1/2}\;\times\;\frac{d}{dx}(5000 + 24x).$$

The derivative of $$5000 + 24x$$ with respect to $$x$$ is simply $$24$$, so

$$\frac{dv}{dx} \;=\; \frac12\,(5000 + 24x)^{-1/2}\;\times\;24.$$

Multiplying the constants gives

$$\frac{dv}{dx} \;=\; 12\,(5000 + 24x)^{-1/2}.$$

Since a negative half-power corresponds to a reciprocal square root, we can rewrite this as

$$\frac{dv}{dx} \;=\; \frac{12}{\sqrt{5000 + 24x}}.$$

We now substitute $$v$$ and $$\dfrac{dv}{dx}$$ into the acceleration formula $$a = v \dfrac{dv}{dx}\,,$$ so

$$a \;=\; \bigl(\sqrt{5000 + 24x}\bigr)\;\times\;\frac{12}{\sqrt{5000 + 24x}}.$$

The square root term in the numerator and the identical square root term in the denominator cancel each other exactly, leaving

$$a \;=\; 12.$$

This result is a constant, independent of $$x$$, and its unit is $$\text{m s}^{-2}$$ because both the velocity and the derivative were in SI units throughout.

So, the answer is $$12$$.

The scooter starts from rest, accelerates at constant rate $$a_1$$ for time $$t_1$$, then decelerates at constant rate $$a_2$$ for time $$t_2$$ and comes to rest.

During the acceleration phase, the velocity increases from 0 to a maximum value $$v_{max}$$. Using $$v = u + at$$, we get $$v_{max} = 0 + a_1 t_1 = a_1 t_1$$.

During the deceleration phase, the velocity decreases from $$v_{max}$$ to 0. Using $$v = u - at$$, we get $$0 = v_{max} - a_2 t_2$$, which gives $$v_{max} = a_2 t_2$$.

Since both expressions equal $$v_{max}$$, we have $$a_1 t_1 = a_2 t_2$$. Therefore, $$\frac{t_1}{t_2} = \frac{a_2}{a_1}$$.

We start by recalling the standard results for a projectile launched with an initial speed $$u$$ at an angle $$\theta$$ to the horizontal.

Formula for horizontal range: $$R = \dfrac{u^{2}\sin 2\theta}{g}.$$

Formula for maximum height: $$H = \dfrac{u^{2}\sin^{2}\theta}{2g}.$$

Both projectiles are thrown with the same speed $$u$$, so $$u$$ and $$g$$ are common for the two cases. Let us denote the first angle by $$\theta_{1}=42^{\circ}$$ and the second by $$\theta_{2}=48^{\circ}.$$ Their corresponding ranges and heights are $$R_{1},\,H_{1}$$ and $$R_{2},\,H_{2}.$$

Evaluating the ranges

For the first projectile we have

$$R_{1}= \dfrac{u^{2}\sin 2\theta_{1}}{g} = \dfrac{u^{2}\sin(2\times 42^{\circ})}{g} = \dfrac{u^{2}\sin 84^{\circ}}{g}.$$

For the second projectile we obtain

$$R_{2}= \dfrac{u^{2}\sin 2\theta_{2}}{g} = \dfrac{u^{2}\sin(2\times 48^{\circ})}{g} = \dfrac{u^{2}\sin 96^{\circ}}{g}.$$

Now we use the trigonometric identity $$\sin(180^{\circ}-\alpha)=\sin\alpha.$$ Since $$96^{\circ}=180^{\circ}-84^{\circ},$$ we get $$\sin 96^{\circ}=\sin 84^{\circ}.$$ Substituting this equality above,

$$R_{2}= \dfrac{u^{2}\sin 96^{\circ}}{g}= \dfrac{u^{2}\sin 84^{\circ}}{g}=R_{1}.$$

Thus $$R_{1}=R_{2}.$$

Evaluating the heights

For the first projectile:

$$H_{1}=\dfrac{u^{2}\sin^{2}\theta_{1}}{2g} =\dfrac{u^{2}\sin^{2}42^{\circ}}{2g}.$$

For the second projectile:

$$H_{2}=\dfrac{u^{2}\sin^{2}\theta_{2}}{2g} =\dfrac{u^{2}\sin^{2}48^{\circ}}{2g}.$$

Because $$48^{\circ}>42^{\circ}$$ and $$\sin\theta$$ increases from $$0^{\circ}$$ to $$90^{\circ},$$ we have $$\sin 48^{\circ}>\sin 42^{\circ}.$$ Squaring preserves the inequality, so $$\sin^{2}48^{\circ}>\sin^{2}42^{\circ}.$$ Substituting this into the expressions for $$H_{1}$$ and $$H_{2}$$ gives

$$H_{2}>H_{1}.$$

Combining both comparisons, we find

$$R_{1}=R_{2}\quad\text{and}\quad H_{1}<H_{2}.$$

These results match Option B.

Hence, the correct answer is Option B.

The relation between time and distance is $$t = m x^2 + n x$$, with $$m$$ and $$n$$ constants.

Differentiate both sides with respect to $$x$$ to connect time and distance derivatives:
$$\frac{dt}{dx} = 2 m x + n \quad -(1)$$

Velocity is defined as $$v = \frac{dx}{dt}$$. From equation $$(1)$$,
$$v = \frac{1}{dt/dx} = \frac{1}{2 m x + n} \quad -(2)$$

Acceleration is $$a = \frac{dv}{dt}$$. Using the chain rule $$\frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v \frac{dv}{dx}$$.

First find $$\frac{dv}{dx}$$ from $$(2)$$:
$$v = (2 m x + n)^{-1}$$
$$\frac{dv}{dx} = -\,2 m \,(2 m x + n)^{-2} \quad -(3)$$

Substitute $$(2)$$ and $$(3)$$ into $$a = v \frac{dv}{dx}$$:
$$a = \frac{1}{2 m x + n}\;\bigl[-\,2 m \,(2 m x + n)^{-2}\bigr] = -\,\frac{2 m}{(2 m x + n)^3} \quad -(4)$$

The negative sign shows the body is slowing down, so the magnitude of retardation is
$$|a| = \frac{2 m}{(2 m x + n)^3} \quad -(5)$$

From $$(2)$$, $$2 m x + n = \frac{1}{v}$$. Substitute this into $$(5)$$:

$$|a| = 2 m \left(\frac{1}{v}\right)^{\!3} = 2 m v^3 \quad -(6)$$

Hence the retardation is $$2 m v^3$$.

Option A is correct.

Let upward direction be positive and let the ground be the reference level $$y = 0$$.

The balloon is rising with a constant velocity $$u_b = 10\ \text{m s}^{-1}$$. At the instant it is $$75\ \text{m}$$ above the ground, an object is released from it. Because the object was part of the balloon just before release, its initial velocity relative to the ground is the same as that of the balloon:

$$u = 10\ \text{m s}^{-1}\ (\text{upward})$$

The object then moves under gravity alone. For vertical motion with constant acceleration we use the kinematic equation $$y = y_0 + u t + \tfrac12 a t^2$$ $$-(1)$$ where
  • $$y_0$$ = initial height, here $$75\ \text{m}$$
  • $$y$$ = height after time $$t$$
  • $$u$$ = initial velocity, here $$10\ \text{m s}^{-1}$$ (upward)
  • $$a$$ = acceleration. Since gravity acts downward, $$a = -g = -10\ \text{m s}^{-2}$$.

The object strikes the ground when $$y = 0$$. Substituting into (1):

$$0 = 75 + 10t - \tfrac12(10)t^2$$

$$0 = 75 + 10t - 5t^2$$

Re-arranging: $$5t^2 - 10t - 75 = 0$$

Dividing by 5: $$t^2 - 2t - 15 = 0$$

Solving this quadratic:

$$t = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-15)}}{2} = \frac{2 \pm \sqrt{4 + 60}}{2} = \frac{2 \pm 8}{2}$$

The physically meaningful (positive) root is $$t = 5\ \text{s}$$.

Thus, the object reaches the ground after $$5\ \text{s}$$.

During this interval the balloon continues to rise with its constant speed of $$10\ \text{m s}^{-1}$$. Extra height gained by the balloon:

$$h_{\text{extra}} = u_b \, t = 10 \times 5 = 50\ \text{m}$$

Initial height of the balloon = $$75\ \text{m}$$. Therefore, height of the balloon when the object hits the ground:

$$H = 75 + 50 = 125\ \text{m}$$

Hence, the balloon is approximately $$125\ \text{m}$$ above the ground when the object strikes the ground.

The correct option is Option C.

Let the length of the escalator be $$L$$. The boy's walking speed is $$v_b = L/t_1$$ and the escalator's speed is $$v_e = L/t_2$$.

When the boy walks on the moving escalator, the effective speed is $$v_b + v_e = \frac{L}{t_1} + \frac{L}{t_2} = L\cdot\frac{t_1 + t_2}{t_1 t_2}$$.

The time taken is $$t = \frac{L}{v_b + v_e} = \frac{L \cdot t_1 t_2}{L(t_1 + t_2)} = \frac{t_1 t_2}{t_1 + t_2}$$.

Therefore the time taken by him to walk up on the moving escalator is $$\dfrac{t_1 t_2}{t_1 + t_2}$$.

The trajectory of the projectile is given by $$y = \alpha x - \beta x^2$$.

The angle of projection $$\theta$$ is the angle the velocity makes with the horizontal at the point of projection $$(x = 0)$$. Since $$\tan\theta = \frac{dy}{dx}\bigg|_{x=0}$$, we differentiate: $$\frac{dy}{dx} = \alpha - 2\beta x$$. At $$x = 0$$, $$\frac{dy}{dx} = \alpha$$, so $$\tan\theta = \alpha$$, which gives $$\theta = \tan^{-1}\alpha$$.

The maximum height occurs where $$\frac{dy}{dx} = 0$$, i.e., $$\alpha - 2\beta x = 0$$, giving $$x = \frac{\alpha}{2\beta}$$.

Substituting this into the trajectory equation: $$H = \alpha \cdot \frac{\alpha}{2\beta} - \beta \cdot \left(\frac{\alpha}{2\beta}\right)^2 = \frac{\alpha^2}{2\beta} - \frac{\alpha^2}{4\beta} = \frac{\alpha^2}{4\beta}$$.

Therefore, the angle of projection is $$\tan^{-1}\alpha$$ and the maximum height is $$\frac{\alpha^2}{4\beta}$$.

Let us fix an $$x$$-$$y$$ coordinate system that, at the instant the bomb leaves the aircraft, has its origin at the bomb itself. The $$x$$-axis is chosen horizontally in the direction in which the plane is flying, while the $$y$$-axis is chosen vertically upward (opposite to the direction of gravity). The positive $$y$$ direction is therefore upward and the acceleration due to gravity acts downward, that is, along the negative $$y$$ direction with magnitude $$g$$.

First, we write the equations of motion of the bomb as seen by an observer on the ground (an inertial frame). Because the plane is flying horizontally with constant speed $$u$$, the bomb, at the moment of release, possesses the same horizontal speed $$u$$. Therefore, in the ground frame, the initial conditions are

$$x(0)=0, \qquad y(0)=0,$$

$$\dot x(0)=u, \qquad \dot y(0)=0,$$

and the only acceleration acting on the bomb is gravitational:

$$\ddot x = 0, \qquad \ddot y = -g.$$

Integrating these with respect to time $$t$$, we obtain the bomb’s position in the ground frame:

Horizontal motion:

$$\ddot x = 0 \;\Rightarrow\; \dot x = u \quad(\text{since }\dot x(0)=u),$$

Integrating once more,

$$x(t)=ut.$$

Vertical motion:

$$\ddot y = -g \;\Rightarrow\; \dot y = -gt \quad(\text{since }\dot y(0)=0),$$

Integrating again,

$$y(t) = -\tfrac{1}{2}gt^{2}.$$

Eliminating the time $$t$$ between $$x(t)$$ and $$y(t)$$, we get

$$t = \frac{x}{u}\quad\Longrightarrow\quad y = -\tfrac{1}{2}g\left(\frac{x}{u}\right)^{2},$$

or

$$y = -\frac{g}{2u^{2}}\,x^{2}.$$

This is the standard equation of a parabola opening downward, confirming that in the ground frame the bomb follows a parabolic path.

Now we must determine what the observer inside the plane sees. The plane itself forms a non-inertial (moving) frame that translates with the constant horizontal velocity $$u$$. According to Galilean relativity, to convert coordinates from the ground frame $$(x,y)$$ to the plane’s frame $$(x',y')$$, we subtract the uniform motion of the plane:

$$x' = x - ut, \qquad y' = y.$$

Substituting $$x = ut$$ from the earlier result, we find

$$x' = ut - ut = 0.$$

Thus, at every instant of time,

$$x' = 0.$$

The vertical coordinate, on the other hand, remains exactly what it was in the ground frame because the transformation does not affect $$y$$:

$$y' = y = -\tfrac{1}{2}gt^{2}.$$

Therefore the equations of motion in the plane’s frame reduce to

$$x'(t) = 0,\qquad y'(t) = -\tfrac{1}{2}gt^{2}.$$

Because $$x'$$ is identically zero at all times, the bomb has no horizontal motion relative to the plane. Its entire motion, as perceived by the observer sitting in the aircraft, consists solely of a uniform downward acceleration under gravity.

Consequently, the bomb appears to fall straight down along the vertical line that passes through the point of release. This straight vertical line is precisely Option A in the given list.

Hence, the correct answer is Option A.

We are told that the instantaneous velocity of the particle is given by the time-dependent expression $$v = \alpha t + \beta t^{2}$$, where $$\alpha$$ and $$\beta$$ are constants.

To find the distance (strictly, the displacement along the straight line) travelled by the particle between the times $$t = 1\ \text{s}$$ and $$t = 2\ \text{s}$$, we recall the fundamental relation between velocity and displacement:

$$\displaystyle s = \int v\;dt.$$

This formula states that the displacement $$s$$ over a time interval is obtained by integrating the velocity with respect to time over that interval.

Substituting the given expression for $$v$$, we write the definite integral from $$t = 1$$ to $$t = 2$$:

$$\displaystyle s_{1\ \text{to}\ 2} = \int_{1}^{2} (\alpha t + \beta t^{2})\,dt.$$

Now we integrate term by term. First, we integrate $$\alpha t$$ with respect to $$t$$, and second, we integrate $$\beta t^{2}$$ with respect to $$t$$:

For the first term we use the power rule $$\int t\,dt = \frac{t^{2}}{2}$$:

$$\int \alpha t\,dt = \alpha \int t\,dt = \alpha\left(\frac{t^{2}}{2}\right) = \frac{\alpha t^{2}}{2}.$$

For the second term we again use the power rule $$\int t^{n}\,dt = \frac{t^{n+1}}{n+1}$$ (here $$n = 2$$):

$$\int \beta t^{2}\,dt = \beta \int t^{2}\,dt = \beta\left(\frac{t^{3}}{3}\right) = \frac{\beta t^{3}}{3}.$$

Combining both antiderivatives, the displacement function $$s(t)$$ (measured from any convenient reference) is

$$s(t) = \frac{\alpha t^{2}}{2} \;+\; \frac{\beta t^{3}}{3} + C,$$

where $$C$$ is an additive constant. However, since we are interested only in the difference $$s(2) - s(1)$$, this constant cancels out automatically and we do not need its explicit value.

We therefore evaluate $$s(t)$$ at the upper limit $$t = 2$$:

$$s(2) = \frac{\alpha (2)^{2}}{2} + \frac{\beta (2)^{3}}{3} = \frac{\alpha \cdot 4}{2} + \frac{\beta \cdot 8}{3} = 2\alpha + \frac{8\beta}{3}.$$

Next, we evaluate $$s(t)$$ at the lower limit $$t = 1$$:

$$s(1) = \frac{\alpha (1)^{2}}{2} + \frac{\beta (1)^{3}}{3} = \frac{\alpha}{2} + \frac{\beta}{3}.$$

The displacement (which, in this one-dimensional motion, equals the distance travelled) between 1 s and 2 s is the difference of these two values:

$$\begin{aligned} s_{1\ \text{to}\ 2} &= s(2) - s(1) \\ &= \left(2\alpha + \frac{8\beta}{3}\right) \;-\; \left(\frac{\alpha}{2} + \frac{\beta}{3}\right). \end{aligned}$$

We now subtract term by term, matching coefficients of $$\alpha$$ and $$\beta$$ separately.

For the $$\alpha$$ part:

$$2\alpha - \frac{\alpha}{2} = \frac{4\alpha}{2} - \frac{\alpha}{2} = \frac{3\alpha}{2}.$$

For the $$\beta$$ part:

$$\frac{8\beta}{3} - \frac{\beta}{3} = \frac{7\beta}{3}.$$

Putting both simplified results together, we obtain

$$s_{1\ \text{to}\ 2} = \frac{3\alpha}{2} + \frac{7\beta}{3}.$$

This matches option B in the list provided.

Hence, the correct answer is Option B.

$$\text{From geometry, angle with horizontal} = 90^\circ - \text{angle with vertical}$$

$$\theta_B = 90^\circ - 45^\circ = 45^\circ$$

$$\theta_C = 90^\circ - 60^\circ = 30^\circ$$

$$\tan(45^\circ) = \frac{h_1}{d} \implies 1 = \frac{h_1}{d} \implies h_1 = d$$

$$\tan(30^\circ) = \frac{h_1 + h_2}{d + 2.464d} = \frac{d + h_2}{3.464d}$$

$$0.5774 = \frac{d + h_2}{3.464d} \implies d + h_2 = 0.5774 \times 3.464d$$

$$d + h_2 \approx 2.000d \implies h_2 = d$$

Taking the ground as origin ($$y = 0$$) and upwards as positive:

$$\text{For downward motion from height } h: v^2 = 0 - 2g(y - h) \implies v = -\sqrt{2g(h - y)}$$

This is a parabolic curve in the fourth quadrant ($$v < 0$$) with its vertex at $$(h, 0)$$ that curves towards the $$v$$-axis as height decreases to zero.

$$\text{For upward motion after rebound to height } \frac{h}{2}: 0 = v_0^2 - 2g\left(\frac{h}{2} - 0\right) \implies v^2 = 2g\left(\frac{h}{2} - y\right) \implies v = +\sqrt{2g\left(\frac{h}{2} - y\right)}$$

This is a parabolic curve in the first quadrant ($$v > 0$$) with its vertex at $$\left(\frac{h}{2}, 0\right)$$ where velocity decreases as height increases.

Answer: Option (C): The graph consists of two parabolic segments opening towards the negative $$h$$-axis with vertices at $$h$$ and $$\frac{h}{2}$$.

Area above the time axis (trapezoid from $$t = 0$$ to $$t = \frac{13}{3}$$):

$$\text{Area}_1 = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height} = \frac{1}{2} \times \left[\frac{13}{3} + (3 - 2)\right] \times 4 = 2 \times \left(\frac{13}{3} + 1\right) = \frac{32}{3}\text{ m}$$

Area below the time axis (triangle from $$t = \frac{13}{3}$$ to $$t = 6$$):

$$\text{Area}_2 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{5}{3} \times 2 = \frac{5}{3}\text{ m}$$

Total distance covered: $$d = \text{Area}_1 + \text{Area}_2 = \frac{32}{3} + \frac{5}{3} = \frac{37}{3}\text{ m}$$

Let us choose the direction in which train A is moving as the positive direction.

Speed of train A with respect to the ground is given as $$+36\text{ km h}^{-1}.$$

The person is walking inside train A opposite to the motion of the train. Therefore, the speed of the person with respect to train A is

$$v_{\text{person, A}} = -1.8\text{ km h}^{-1}.$$

To find the speed of the person with respect to the ground, we use the relation

$$v_{\text{person, ground}} = v_{\text{train A, ground}} + v_{\text{person, A}}.$$

Substituting the known values, we obtain

$$v_{\text{person, ground}} = 36 + (-1.8) = 34.2\text{ km h}^{-1}.$$

Now, train B is moving in the opposite direction (negative direction) with speed

$$v_{\text{train B, ground}} = -72\text{ km h}^{-1}.$$

The relative speed of the person as observed from train B is given by the standard formula

$$v_{\text{person relative to B}} = v_{\text{person, ground}} - v_{\text{train B, ground}}.$$

Substituting the values, we have

$$v_{\text{person relative to B}} = 34.2 - (-72) = 34.2 + 72 = 106.2\text{ km h}^{-1}.$$

We must now convert this speed from kilometres per hour to metres per second using

$$1\text{ km h}^{-1} = \frac{5}{18}\text{ m s}^{-1}.$$

Hence,

$$v_{\text{person relative to B}} = 106.2 \times \frac{5}{18}\text{ m s}^{-1}.$$

First multiply:

$$106.2 \times 5 = 531.0.$$

Now divide by 18:

$$\frac{531.0}{18} = 29.5\text{ m s}^{-1}.$$

Therefore, the speed of the person as observed from train B is approximately $$29.5\text{ m s}^{-1}.$$

Hence, the correct answer is Option A.

First let us choose a convenient coordinate system. We take the horizontal road surface as the $$x$$-axis and the vertically downward direction (the direction in which raindrops actually fall) as the negative $$y$$-axis. All speeds mentioned are measured with respect to the ground.

When the car is standing still, the driver observes the rain to be falling exactly vertically downward. Because the car is at rest in this situation, the velocity of the rain relative to the ground must itself be vertical. We therefore write the ground-frame velocity of the rain as

$$\vec u=-u\,\hat{\jmath},$$

where $$u>0$$ is the speed of the rain and the minus sign signifies the downward (negative $$y$$) direction.

Now the car starts moving to the right (positive $$x$$-direction) with speed $$v$$. The ground-frame velocity of the car is then

$$\vec v_c = v\,\hat{\imath}.$$

The driver in the car does not see the ground-frame velocity of the rain; instead he sees the relative velocity of the rain with respect to the car. The standard formula for relative velocity is stated first:

$$\vec v_{\text{rain relative to car}}=\vec u-\vec v_c.$$

Substituting the expressions for $$\vec u$$ and $$\vec v_c$$ we obtain

$$\vec v_{rc}=(-u\,\hat{\jmath})-(v\,\hat{\imath})=-v\,\hat{\imath}-u\,\hat{\jmath}.$$

The driver now sees this vector making an angle of $$60^\circ$$ with the horizontal. The horizontal component is $$|-v|=v$$ and the vertical component is $$|-u|=u$$, so from the definition of the tangent of an angle we have

$$\tan 60^\circ=\frac{\text{vertical component}}{\text{horizontal component}}=\frac{u}{v}.$$

Using $$\tan 60^\circ=\sqrt3$$ gives

$$\sqrt3=\frac{u}{v}\quad\Longrightarrow\quad u=v\sqrt3. \quad -(1)$$

Next the car’s speed is increased to $$(1+\beta)v$$. Its new ground-frame velocity becomes

$$\vec v_c'=(1+\beta)v\,\hat{\imath}.$$

The new relative velocity of the rain with respect to the faster car is

$$\vec v_{rc}'=\vec u-\vec v_c'=(-u\,\hat{\jmath})-(1+\beta)v\,\hat{\imath}=-(1+\beta)v\,\hat{\imath}-u\,\hat{\jmath}.$$

This time the driver measures the angle made by this vector with the horizontal to be $$45^\circ$$. Therefore

$$\tan 45^\circ=\frac{u}{(1+\beta)v}.$$

Because $$\tan 45^\circ=1$$, we get

$$1=\frac{u}{(1+\beta)v}\quad\Longrightarrow\quad u=(1+\beta)v. \quad -(2)$$

We now have two expressions, (1) and (2), for the same quantity $$u$$. Equating them gives

$$v\sqrt3=(1+\beta)v.$$

Since $$v\neq0$$, we can cancel $$v$$ from both sides to obtain

$$\sqrt3=1+\beta\quad\Longrightarrow\quad\beta=\sqrt3-1.$$

Using $$\sqrt3\approx1.732$$ we calculate

$$\beta\approx1.732-1=0.732\;(\text{approximately}).$$

The closest option to this value is $$0.73$$.

Hence, the correct answer is Option D.

We start by letting the unknown acceleration due to gravity be $$g$$ (in m s$$^{-2}$$) and the total time taken by the ball to reach the ground be $$T$$ seconds. The ball is dropped from rest, so its initial velocity $$u=0$$.

For motion with constant acceleration we have the kinematic formula
$$s = uT + \dfrac{1}{2} g T^{2}.$$

Using this formula for the complete fall of $$100\ \text{m}$$, we write

$$100 = 0 \times T + \dfrac{1}{2}\,g\,T^{2},$$

which simplifies to

$$100 = \dfrac{1}{2} g T^{2}.$$

Multiplying both sides by 2 gives

$$g T^{2} = 200 \quad\Longrightarrow\quad T^{2} = \dfrac{200}{g}. \quad -(1)$$

Now, during the last $$\dfrac12$$ s before impact the ball covers only $$19\ \text{m}$$. Hence, in the preceding time interval $$T-0.5$$ seconds it must have covered the remaining distance of $$100-19 = 81\ \text{m}$$.

Applying the same kinematic formula for this first part of the motion, we get

$$81 = 0 \times (T-0.5) + \dfrac{1}{2}\,g\,(T-0.5)^{2},$$

or simply

$$81 = \dfrac{1}{2} g (T-0.5)^{2}. \quad -(2)$$

We expand the square:
$$(T-0.5)^{2} = T^{2} - 2 \times 0.5 \times T + (0.5)^{2} = T^{2} - T + 0.25.$$

Substituting this result into equation (2) gives

$$81 = \dfrac{1}{2} g \bigl(T^{2} - T + 0.25\bigr).$$

We already know from equation (1) that $$\dfrac{1}{2}gT^{2}=100$$, so we rewrite the right-hand side as

$$\dfrac12 g T^{2} - \dfrac12 g T + \dfrac12 g \times 0.25 = 100 - \dfrac12 g T + 0.125\,g.$$

Thus equation (2) becomes

$$81 = 100 - \dfrac12 g T + 0.125\,g.$$

Transposing terms, we have

$$-\dfrac12 g T + 0.125\,g = 81 - 100 = -19.$$

Multiplying by $$-1$$ gives a cleaner form:

$$\dfrac12 g T - 0.125\,g = 19.$$

Factoring out $$g$$ yields

$$g\left(\dfrac12 T - 0.125\right) = 19. \quad -(3)$$

From equation (3) we can express $$g$$ in terms of $$T$$:

$$g = \dfrac{19}{\dfrac12 T - 0.125} = \dfrac{19}{\dfrac{T-0.25}{2}} = \dfrac{38}{T - 0.25}. \quad -(4)$$

We also keep equation (1) in the equivalent form

$$g T^{2} = 200. \quad -(5)$$

Substituting the value of $$g$$ from (4) into (5), we get

$$\dfrac{38}{T - 0.25}\,T^{2} = 200.$$

Multiplying both sides by $$T-0.25$$ gives

$$38 T^{2} = 200\,(T - 0.25).$$

Dividing every term by 2 to simplify, we obtain

$$19 T^{2} = 100T - 25.$$

Rearranging all terms to one side provides a standard quadratic equation:

$$19 T^{2} - 100T + 25 = 0.$$

To solve for $$T$$ we calculate the discriminant:
$$\Delta = (-100)^{2} - 4 \times 19 \times 25 = 10000 - 1900 = 8100,$$ whose square root is $$\sqrt{8100} = 90.$$

Using the quadratic formula $$T = \dfrac{-b \pm \sqrt{\Delta}}{2a}$$ with $$a = 19,\ b = -100$$, we get

$$T = \dfrac{100 \pm 90}{2 \times 19} = \dfrac{100 \pm 90}{38}.$$

This yields two roots:

$$T_{1} = \dfrac{190}{38} = 5 \ \text{s}, \quad T_{2} = \dfrac{10}{38} \approx 0.263\ \text{s}.$$

The ball has to be in the air for more than $$0.5\ \text{s}$$ (because the question speaks of the last half-second), so the physically acceptable time is $$T = 5\ \text{s}.$$

Finally, substituting $$T = 5$$ into equation (4):

$$g = \dfrac{38}{5 - 0.25} = \dfrac{38}{4.75} = 8\ \text{m s}^{-2}.$$

So, the answer is $$8$$.

We begin with the upward motion of the helicopter itself. Starting from rest and moving with a constant upward acceleration $$g$$, the displacement $$s$$ reached in time $$t_1$$ is given by the kinematic equation

$$s = \tfrac12 a t^2.$$

Here $$a = g$$ (upward) and the helicopter has to climb a height $$h$$ before the food packet is released. Substituting these values we get

$$h \;=\; \tfrac12\,g\,t_1^{\,2}.$$

Solving for the time $$t_1$$ of ascent,

$$t_1 \;=\; \sqrt{\dfrac{2h}{g}}.$$

The upward velocity acquired by the helicopter at this instant is found from the relation $$v = u + at$$. Because the helicopter started from rest, $$u = 0$$, so

$$v \;=\; 0 + g\,t_1 \;=\; g\,\sqrt{\dfrac{2h}{g}} \;=\; \sqrt{2gh}.$$

This is also the initial velocity $$u$$ of the food packet at the moment it is dropped, and it is directed upward. The packet is now in free fall under gravity alone, so after release its acceleration is downward with magnitude $$g$$. Taking the upward direction as positive, we write the displacement equation for the packet:

$$y \;=\; h \;+\; u\,t \;-\; \tfrac12\,g\,t^2,$$

where $$y$$ is the height above the ground after an additional time $$t$$. The packet reaches the ground when $$y = 0$$, so we set

$$0 \;=\; h \;+\; \sqrt{2gh}\,t \;-\; \tfrac12\,g\,t^2.$$

Multiplying every term by 2 to clear the fraction,

$$0 \;=\; 2h \;+\; 2\sqrt{2gh}\,t \;-\; g\,t^2.$$

Re-arranging in the standard quadratic form $$a t^2 + b t + c = 0$$ gives

$$g\,t^{\,2} \;-\; 2\sqrt{2gh}\,t \;-\; 2h \;=\; 0.$$

Using the quadratic formula $$t = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ with $$a = g,\; b = -2\sqrt{2gh},\; c = -2h,$$ we find

$$t \;=\; \dfrac{2\sqrt{2gh} \;+\; \sqrt{\left(2\sqrt{2gh}\right)^{\!2} + 4g(2h)}}{2g}.$$

Inside the square root,

$$(2\sqrt{2gh})^{\!2} = 8gh,\qquad 4g(2h) = 8gh,$$

so their sum is $$16gh$$ and its square root is $$4\sqrt{gh}$$. Substituting this back,

$$t \;=\; \dfrac{2\sqrt{2gh} \;+\; 4\sqrt{gh}}{2g}.$$

Factorizing $$\sqrt{gh}$$ from the numerator gives

$$t \;=\; \dfrac{\left(2\sqrt{2} \;+\; 4\right)\sqrt{gh}}{2g}.$$

Dividing numerator and denominator by 2,

$$t \;=\; \left(\sqrt{2} + 2\right)\sqrt{\dfrac{h}{g}}.$$

Numerically, $$\sqrt{2} \approx 1.414$$, so

$$t \;\approx\; (1.414 + 2)\,\sqrt{\dfrac{h}{g}} \;=\; 3.414\,\sqrt{\dfrac{h}{g}}.$$

This is very close to $$3.4\sqrt{\dfrac{h}{g}}$$, which corresponds to Option C.

Hence, the correct answer is Option C.

We are told that the magnitudes of the two vectors are $$|\vec A_1| = 3$$ and $$|\vec A_2| = 5$$, while the magnitude of their sum is $$|\vec A_1 + \vec A_2| = 5$$.

First, we need the dot product $$\vec A_1 \cdot \vec A_2$$. We use the standard magnitude formula for the sum of two vectors:

$$|\vec A_1 + \vec A_2|^2 \;=\; |\vec A_1|^2 + |\vec A_2|^2 + 2\,\vec A_1 \cdot \vec A_2.$$

Substituting the given magnitudes, we have

$$5^2 \;=\; 3^2 + 5^2 + 2\,\vec A_1 \cdot \vec A_2.$$

Writing the squares explicitly,

$$25 \;=\; 9 + 25 + 2\,\vec A_1 \cdot \vec A_2.$$

Bringing the numerical terms on the right to the left side,

$$25 - 34 \;=\; 2\,\vec A_1 \cdot \vec A_2.$$

So

$$-9 \;=\; 2\,\vec A_1 \cdot \vec A_2,$$

and therefore

$$\vec A_1 \cdot \vec A_2 \;=\; -\frac{9}{2} = -4.5.$$

We also note the squared magnitudes that will be needed shortly:

$$|\vec A_1|^2 = 3^2 = 9, \qquad |\vec A_2|^2 = 5^2 = 25.$$

Now we must evaluate the required scalar product

$$\bigl(2\vec A_1 + 3\vec A_2\bigr)\,\cdot\,\bigl(3\vec A_1 - 2\vec A_2\bigr).$$

We expand this dot product term by term, using linearity and the commutativity of the dot product ($$\vec A_2 \cdot \vec A_1 = \vec A_1 \cdot \vec A_2$$):

$$\begin{aligned} (2\vec A_1 + 3\vec A_2)\cdot(3\vec A_1 - 2\vec A_2) &= (2\vec A_1)\cdot(3\vec A_1) \;+\; (2\vec A_1)\cdot(-2\vec A_2) \\ &\quad +\; (3\vec A_2)\cdot(3\vec A_1) \;+\; (3\vec A_2)\cdot(-2\vec A_2). \end{aligned}$$

Evaluating each term separately:

$$\begin{aligned} (2\vec A_1)\cdot(3\vec A_1) &= 6\,(\vec A_1\cdot\vec A_1) = 6\,|\vec A_1|^2, \\ (2\vec A_1)\cdot(-2\vec A_2) &= -4\,(\vec A_1\cdot\vec A_2), \\ (3\vec A_2)\cdot(3\vec A_1) &= 9\,(\vec A_2\cdot\vec A_1) = 9\,(\vec A_1\cdot\vec A_2), \\ (3\vec A_2)\cdot(-2\vec A_2) &= -6\,(\vec A_2\cdot\vec A_2) = -6\,|\vec A_2|^2. \end{aligned}$$

Combining these, we obtain

$$\begin{aligned} (2\vec A_1 + 3\vec A_2)\cdot(3\vec A_1 - 2\vec A_2) &= 6|\vec A_1|^2 \;+\; \bigl(-4 + 9\bigr)\,(\vec A_1\cdot\vec A_2) \;-\; 6|\vec A_2|^2 \\ &= 6|\vec A_1|^2 \;+\; 5\,(\vec A_1\cdot\vec A_2) \;-\; 6|\vec A_2|^2. \end{aligned}$$

Now we substitute the numerical values we have found:

$$\begin{aligned} 6|\vec A_1|^2 &= 6 \times 9 = 54, \\ 5\,(\vec A_1\cdot\vec A_2) &= 5 \times (-4.5) = -22.5, \\ -6|\vec A_2|^2 &= -6 \times 25 = -150. \end{aligned}$$

Adding these three results together:

$$ 54 \;+\; (-22.5) \;+\; (-150) \;=\; 54 - 22.5 - 150 = 31.5 - 150 = -118.5. $$

Thus the required scalar product is $$-118.5$$.

Hence, the correct answer is Option B.

Let us first describe the data in vector language. We adopt east as the positive $$x$$-direction and north as the positive $$y$$-direction, so the unit vectors are $$\hat{i}$$ to the east and $$\hat{j}$$ to the north.

Ship A is moving with velocity

$$\vec v_A = 30\hat i + 50\hat j\ \text{km h}^{-1}.$$

Ship B is initially 80 km east and 150 km north of A. Measured from A, its initial position vector is therefore

$$\vec r_0 = 80\hat i + 150\hat j\ \text{km}.$$

Ship B is sailing due west at 10 km h$$^{-1}$$, hence

$$\vec v_B = -10\hat i\ \text{km h}^{-1}.$$

We are interested in the separation between the two ships, so we examine the relative motion of B with respect to A. The relative velocity is given by the formula

$$\vec v_{\text{rel}} = \vec v_B - \vec v_A.$$

Substituting the individual velocities, we have

$$\vec v_{\text{rel}} = (-10\hat i) - (30\hat i + 50\hat j)$$ $$= -10\hat i - 30\hat i - 50\hat j$$ $$= -40\hat i - 50\hat j\ \text{km h}^{-1}.$$

At any time $$t$$ hours after the instant chosen as $$t=0$$, the relative position vector of B with respect to A is

$$\vec r(t) = \vec r_0 + \vec v_{\text{rel}}\,t.$$

Substituting $$\vec r_0$$ and $$\vec v_{\text{rel}}$$, we obtain

$$\vec r(t) = (80\hat i + 150\hat j) + (-40\hat i - 50\hat j)t$$ $$= (80 - 40t)\hat i + (150 - 50t)\hat j.$$

The actual separation between the ships is the magnitude of $$\vec r(t)$$. To find the time of minimum distance we minimise the square of that magnitude, because the square is easier to differentiate and has its minimum at the same instant. Using the formula $$|\vec r|^2 = x^2 + y^2$$, we write

$$D^2(t) = (80 - 40t)^2 + (150 - 50t)^2.$$

Now we differentiate $$D^2(t)$$ with respect to $$t$$ and set the derivative equal to zero. This ensures the distance is extremal (and here it will be a minimum):

$$\frac{d}{dt}D^2(t) = 2(80 - 40t)(-40) + 2(150 - 50t)(-50) = 0.$$

We can cancel the factor 2 to simplify:

$$(80 - 40t)(-40) + (150 - 50t)(-50) = 0.$$

Expanding each product, we get

$$-40\cdot80 + 40\cdot40t - 50\cdot150 + 50\cdot50t = 0,$$ $$-3200 + 1600t - 7500 + 2500t = 0.$$

Collecting like terms,

$$(-3200 - 7500) + (1600 + 2500)t = 0,$$ $$-10700 + 4100t = 0.$$

Isolating $$t$$ yields

$$4100t = 10700,$$ $$t = \frac{10700}{4100}.$$

We divide numerator and denominator by 100 to simplify:

$$t = \frac{107}{41} \ \text{h}.$$

Performing the division,

$$t \approx 2.609756\ \text{h}.$$

Rounded to one decimal place this is 2.6 h, which matches Option C.

Hence, the correct answer is Option C.

To find the distance from the base at which the particle hits the inclined plane (the range along the incline, $$R$$), we can break down the motion parallel and perpendicular to the inclined surface.

Here is the step-by-step formatting and solution of your approach.

1. Given Data

• Angle of inclination of the plane: $$\alpha = 30^\circ$$
• Angle of projection relative to the plane: $$\theta = 15^\circ$$
• Initial velocity: $$u = 2\text{ m s}^{-1}$$
• Acceleration due to gravity: $$g = 10\text{ m s}^{-2}$$
• $$u_x = u \cos\theta = 2 \cos 15^\circ$$
• $$u_y = u \sin\theta = 2 \sin 15^\circ$$
• $$a_x = -g \sin\alpha = -10 \sin 30^\circ$$
• $$a_y = -g \cos\alpha = -10 \cos 30^\circ$$

2. Components of Motion

By choosing the $$x$$-axis along the inclined plane (upwards) and the $$y$$-axis perpendicular to the inclined plane:

Initial Velocity Components

Acceleration Components

3. Calculation of Time of Flight ($$T$$)

The particle returns to the inclined plane when its displacement perpendicular to the plane is zero ($$y = 0$$):

$$y = u_y T + \frac{1}{2} a_y T^2 = 0$$

$$(2 \sin 15^\circ) T - \frac{1}{2} (g \cos 30^\circ) T^2 = 0$$

Solving for non-zero time $$T$$:

$$T = \frac{2 u \sin\theta}{g \cos\alpha} = \frac{2 \times 2 \sin 15^\circ}{10 \cos 30^\circ}$$

4. Range Along the Incline ($$R$$)

The displacement along the incline is given by:

$$R = u_x T + \frac{1}{2} a_x T^2$$

Substituting $$u_x = 2 \cos 15^\circ$$ and $$a_x = -g \sin 30^\circ$$:

$$R = (2 \cos 15^\circ) T - \frac{1}{2} (10 \sin 30^\circ) T^2$$

Alternative Standard Formula

Instead of plugging numbers into both terms manually, we can use the simplified standard formula for the range up an inclined plane:

$$R = \frac{2 u^2 \sin\theta \cos(\theta + \alpha)}{g \cos^2\alpha}$$

Substituting the values into this formula makes it much cleaner:

$$R = \frac{2 \times (2)^2 \times \sin 15^\circ \times \cos(15^\circ + 30^\circ)}{10 \times \cos^2 30^\circ}$$

$$R = \frac{8 \times \sin 15^\circ \times \cos 45^\circ}{10 \times \left(\frac{\sqrt{3}}{2}\right)^2}$$

$$R = \frac{8 \times \sin 15^\circ \times \frac{1}{\sqrt{2}}}{10 \times \frac{3}{4}} = \frac{32 \sin 15^\circ}{30\sqrt{2}}$$

Using the known value $$\sin 15^\circ = \frac{\sqrt{3}-1}{2\sqrt{2}}$$:

$$R = \frac{32 \times (\sqrt{3} - 1)}{30 \times 2 \times \sqrt{2} \times \sqrt{2}} = \frac{32(\sqrt{3}-1)}{120} = \frac{4(\sqrt{3}-1)}{15}\text{ meters}$$

5. Final Numeric Approximation

Using $$\sqrt{3} \approx 1.732$$:

$$R \approx \frac{4 \times (1.732 - 1)}{15} = \frac{4 \times 0.732}{15} = \frac{2.928}{15} \approx 0.1952\text{ m}$$

Converting meters to centimeters:

$$R \approx 0.1952 \times 100\text{ cm} = 19.52\text{ cm} \approx 20\text{ cm}$$

Correct Answer: A (20 cm)

Let the distance $$RM$$ be $$x$$. And let the distance $$QM$$ be $$L$$. We have, $$PR= \sqrt{x^2+d^2}$$

The total time taken to travel $$PR$$ will be $$\dfrac{\sqrt{x^2+d^2}}{v\div 2} = \dfrac{2\sqrt{x^2+d^2}}{v}$$

Time taken to cover the remaining distance $$L-x$$ from $$Q$$ to $$R$$ will be $$\dfrac{L-x}{v}$$

Thus, the total time,

$$T= \dfrac{L-x}{v} + \dfrac{2\sqrt{x^2+d^2}}{v}$$

$$\Rightarrow T = \dfrac{L-x+2\sqrt{x^2+d^2}}{v}$$

To minimise time, we will take the derivative of $$T$$ with respect to $$x$$, and equate it to $$0$$, we get,

$$\dfrac{dT}{dx} = \dfrac{d}{dx}\left[\dfrac{L-x + 2{(x^2+d^2)}^{1/2}}{v}\right]$$

$$\dfrac{dT}{dx} = 0 - \dfrac{1}{v} + \dfrac{2}{v}\cdot \dfrac{2x}{2}(x^2+d^2)^{-1/2}$$

$$\Rightarrow \dfrac{dT}{dx} = -\dfrac{1}{v}+ \dfrac{2x}{v}\cdot (x^2+d^2)^{-1/2}$$

Equating $$\dfrac{dT}{dx}$$ to zero to get the minimum value, we have,

$$\dfrac{2x}{v}\cdot (x^2+d^2)^{-1/2} = \dfrac{1}{v}$$

$$\dfrac{2x}{\sqrt{x^2+d^2}} = 1$$

$$\Rightarrow {(2x)}^2 = x^2+d^2$$

Which gives $$3x^2 = d^2$$  or  $$x=\dfrac{d}{\sqrt{3}}$$

For a uniformly decelerating motion that reverses direction, velocity decreases linearly with time, position varies parabolically, and distance must monotonically increase with a continuously non-negative slope.

From graph (A):

$$v = u - at \implies \text{motion with uniform negative acceleration}$$

Analyzing Option (C):

$$\text{Slope of distance-time graph} = \text{speed} = \vert{}v\vert{}$$

$$\text{At } v = 0 \implies \text{speed} = 0 \implies \text{slope must be zero}$$

$$\text{For } t > t_{\text{reversal}} \implies \vert{}v\vert{} \text{ increases } \implies \text{slope must increase}$$

$$\text{Graph (C) shows slope decreasing to zero at the end, which is incorrect for distance.}$$

We begin by recalling the equation of motion for uniformly retarded (decelerated) motion:

$$v^{2}=u^{2}+2as$$

Here, $$u$$ is the initial speed, $$v$$ is the final speed, $$a$$ is the uniform acceleration (negative for braking), and $$s$$ is the displacement during the braking period. For a vehicle that comes to rest, the final speed is zero, that is, $$v=0$$. Substituting $$v=0$$, we obtain

$$0=u^{2}+2as \implies -u^{2}=2as \implies a=-\dfrac{u^{2}}{2s}.$$

The quantity $$|a|=\dfrac{u^{2}}{2s}$$ is the magnitude of the retardation provided by the brakes. Because the same brakes are used in both situations, the magnitude of this retardation remains unchanged.

First we analyse the given data for the initial situation.

Initial speed: $$40\text{ km h}^{-1}.$$

We convert kilometres per hour to metres per second using the relation $$1\text{ km h}^{-1}=\dfrac{5}{18}\text{ m s}^{-1}$$:

$$u_{1}=40\times\dfrac{5}{18}=\dfrac{200}{18}=\dfrac{100}{9}\text{ m s}^{-1}.$$

Stopping distance in this case: $$s_{1}=40\text{ m}.$$

Substituting these values in $$|a|=\dfrac{u^{2}}{2s}$$ gives the magnitude of the retardation:

$$|a|=\dfrac{u_{1}^{2}}{2s_{1}}=\dfrac{\left(\dfrac{100}{9}\right)^{2}}{2\times40} =\dfrac{\dfrac{10000}{81}}{80} =\dfrac{10000}{81\times80} =\dfrac{10000}{6480} =\dfrac{125}{81}\text{ m s}^{-2}.$$

We now consider the second situation, where the automobile’s speed is doubled.

New speed: $$80\text{ km h}^{-1}.$$

Converting again to metres per second:

$$u_{2}=80\times\dfrac{5}{18}=\dfrac{400}{18}=\dfrac{200}{9}\text{ m s}^{-1}.$$

The same brakes give the same magnitude of retardation, $$|a|=\dfrac{125}{81}\text{ m s}^{-2}.$$ Using the formula $$|a|=\dfrac{u^{2}}{2s},$$ but now for the second set of values, we write

$$|a|=\dfrac{u_{2}^{2}}{2s_{2}}.$$

Solving for $$s_{2}$$ gives

$$s_{2}=\dfrac{u_{2}^{2}}{2|a|}.$$

Because $$|a|$$ is the same for both cases, we can avoid recalculation by forming a ratio:

$$\dfrac{s_{2}}{s_{1}}=\dfrac{u_{2}^{2}}{u_{1}^{2}}.$$

Substituting the known speeds,

$$\dfrac{s_{2}}{40}=\dfrac{\left(\dfrac{200}{9}\right)^{2}}{\left(\dfrac{100}{9}\right)^{2}} =\dfrac{200^{2}}{100^{2}} =\dfrac{40000}{10000}=4.$$

Hence

$$s_{2}=4\times40=160\text{ m}.$$

So the automobile will require a minimum stopping distance of $$160\text{ m}$$ when travelling at $$80\text{ km h}^{-1}.$$

Hence, the correct answer is Option B.

Velocity is positive during the upwards journey and negative during the downwards journey. The magnitude of velocity decreases during the upwards journey, is zero at the highest point, and then increases during the downwards journey.

We are given that, at the initial instant, the bus and the car are both at rest and the car is 200 m behind the bus. Hence, the initial separation between them is $$200\;\text{m}$$.

Both vehicles begin to move simultaneously along the same straight road, but with different forward accelerations. The acceleration of the bus is

$$a_b = 2\;\text{m s}^{-2},$$

while the acceleration of the car is

$$a_c = 4\;\text{m s}^{-2}.$$

Because they start from rest, their initial velocities are both

$$u_b = 0\;\text{m s}^{-1}, \qquad u_c = 0\;\text{m s}^{-1}.$$

For motion with uniform acceleration we use the kinematic relation

$$s = ut + \frac{1}{2} a t^2,$$

where $$s$$ is the displacement in time $$t$$, $$u$$ is the initial velocity, and $$a$$ is the constant acceleration.

Let the time at which the car catches up with the bus be $$t$$ seconds. At that moment the car must cover the initial 200 m gap in addition to whatever distance the bus itself has travelled. An elegant way to handle this is to look at the relative motion between the two vehicles.

The relative (car with respect to bus) initial velocity is

$$u_{\text{rel}} = u_c - u_b = 0 - 0 = 0\;\text{m s}^{-1}.$$

The relative acceleration is

$$a_{\text{rel}} = a_c - a_b = 4 - 2 = 2\;\text{m s}^{-2}.$$

In the relative frame, the car has to cover the entire initial separation of 200 m. Therefore, the relative displacement to be achieved is

$$s_{\text{rel}} = 200\;\text{m}.$$

Applying the same kinematic equation to the relative motion, we write

$$s_{\text{rel}} = u_{\text{rel}}\,t + \frac{1}{2}\,a_{\text{rel}}\,t^2.$$

Substituting the known values, we have

$$200 = 0 \cdot t + \frac{1}{2}\,(2)\,t^2.$$

The term with the initial velocity drops out, leaving

$$200 = 1 \cdot t^2.$$

So

$$t^2 = 200.$$

Taking the positive square root (time must be positive), we get

$$t = \sqrt{200} = \sqrt{100 \times 2} = 10\sqrt{2}\;\text{s}.$$

Hence, the correct answer is Option D.

We want a motion with:

• positive velocity → graph stays above x-axis
• constant negative acceleration → velocity decreases uniformly

Using

$$v^2=u^2+2as$$

with a<0 velocity decreases with distance in a nonlinear (curved) way.

So the correct graph must:

• start at a positive velocity
• decrease continuously
• be curved (not a straight line)
• eventually approach zero

Graph (3) matches this:

• velocity is always positive
• it decreases with distance
• the curve bends downward (consistent with negative acceleration)

Why others are wrong:

• straight line → constant slope → would mean constant acceleration in v vs t, not v vs x
• increasing curve → implies positive acceleration

Final answer: Graph (3)