JEE Kinematics Questions

The apparent (measured) weight inside an accelerating elevator equals the normal reaction $$N$$ on the object.

For upward acceleration $$a$$ (upward taken positive) the force equation is
$$N - mg = ma \; \Rightarrow \; N = m\,(g + a) \qquad -(1)$$

The elevator’s position as a function of time is given:
$$y = 8\left[1 + \sin\!\left(\frac{2\pi t}{T}\right)\right], \qquad T = 40\pi\;\text{s} \qquad -(2)$$

Differentiate twice to obtain the acceleration.

Angular frequency:
$$k = \frac{2\pi}{T} = \frac{2\pi}{40\pi} = \frac{1}{20}\;\text{s}^{-1}$$

Velocity:
$$v = \frac{dy}{dt} = 8\,k\,\cos(k t)$$

Acceleration:
$$a = \frac{dv}{dt} = -8\,k^{2}\,\sin(k t) \qquad -(3)$$

The sine term varies between $$+1$$ and $$-1$$, hence

Maximum upward acceleration (sin$$=-1$$):
$$a_{\max} = +8\,k^{2}$$

Maximum downward acceleration (sin$$=+1$$):
$$a_{\min} = -8\,k^{2}$$

Therefore the spread of acceleration values is
$$a_{\max} - a_{\min} = 8k^{2} - (-8k^{2}) = 16k^{2} \qquad -(4)$$

Insert $$k = 1/20$$:

$$k^{2} = \left(\frac{1}{20}\right)^{2} = \frac{1}{400}$$
$$16k^{2} = 16 \times \frac{1}{400} = \frac{16}{400} = 0.04\;\text{m\,s}^{-2}$$

Mass of the object: $$m = 50\;\text{kg}$$.
Using $$-(1)$$, the spread in the normal reaction (apparent weight) equals $$m$$ times the acceleration spread:

$$\Delta N = m\,(a_{\max} - a_{\min}) = 50 \times 0.04 = 2\;\text{N}$$

Hence, the maximum variation in the weight reading is

2 N.

r = 5t²i-5tj. v = 10ti-5j. At t=2: v = 20i-5j. |v| = √(400+25) = √425 = 5√17.

Direction: angle with -ve y-axis: tan θ = 20/5 = 4. So θ = tan⁻¹(4) with -ve y-axis.

The correct answer is Option 4: 5√17 m/s, making angle tan⁻¹4 with -ve Y-axis.

$$\text{Distance} = \text{Area of triangle (0 to 2 s)} + \text{Area of rectangle (2 to 4 s)}$$

$$\text{Distance} = \left(\frac{1}{2} \times 2 \times 10\right) + (2 \times 10) = 10 + 20 = 30\text{ m}$$

The river flows due east with speed $$v_r = 9 \text{ km h}^{-1}$$. The boat can move in still water with maximum speed $$v_b = 27 \text{ km h}^{-1}$$ and is steered at an angle $$150^{\circ}$$ to the river flow (angle measured counter-clockwise from east).

Take east as the $$+x$$-axis and north as the $$+y$$-axis.
Components of the boat’s velocity relative to water:

$$v_{bx} = v_b \cos 150^{\circ} = 27 \cos 150^{\circ} = 27 \left(-\frac{\sqrt{3}}{2}\right) = -\frac{27\sqrt{3}}{2} \text{ km h}^{-1}$$ $$-(1)$$
$$v_{by} = v_b \sin 150^{\circ} = 27 \sin 150^{\circ} = 27 \left(\frac{1}{2}\right) = \frac{27}{2} \text{ km h}^{-1}$$ $$-(2)$$

The river velocity has only an $$x$$-component:
$$v_{rx} = +9 \text{ km h}^{-1}, \qquad v_{ry}=0$$

Ground (resultant) velocity of the boat:
$$v_{gx} = v_{bx} + v_{rx} = -\frac{27\sqrt{3}}{2} + 9 \text{ km h}^{-1}$$ $$-(3)$$
$$v_{gy} = v_{by} + v_{ry} = \frac{27}{2} \text{ km h}^{-1}$$ $$-(4)$$

To cross the river we need the northward component $$v_{gy}$$ only. Hence the time taken $$t$$ and river width $$D$$ are related by
$$D = v_{gy}\, t$$ $$-(5)$$

The boat crosses in half a minute:
$$t = \frac{1}{2} \text{ min} = \frac{1}{2}\times\frac{1}{60} \text{ h} = \frac{1}{120} \text{ h}$$ $$-(6)$$

Substituting $$v_{gy}$$ from $$(4)$$ into $$(5)$$:
$$D = \frac{27}{2} \times \frac{1}{120} \text{ km}$$

Simplify:
$$D = \frac{27}{2 \times 120} \text{ km} = \frac{27}{240} \text{ km} = 0.1125 \text{ km}$$

Convert kilometres to metres:
$$D = 0.1125 \times 1000 = 112.5 \text{ m}$$

Therefore, the width of the river is $$112.5 \text{ m}$$, which corresponds to Option B.

Let’s check each graph using basic rules of 1D motion.

Key idea: for any physical motion, at a given time there must be a unique value of position/velocity/etc. Also, graphs must satisfy physical definitions (like velocity = dx/dt, distance can’t decrease, etc.)

(A) Phase vs Time (straight line)
Phase increasing linearly with time is possible (like uniform motion in SHM phase representation).
No contradiction → valid.

(B) Velocity vs Displacement (circle)
This can represent motion like SHM where

$$v^2\propto(A^2-x^2)$$

which gives a circular/elliptic relation.
For each displacement, two velocities (forward/backward) are allowed → physically valid.

(C) Velocity vs Time (circle)
This is not possible.
At a given time, velocity must have a single value.
But a circle gives two velocities for the same time → violates definition.
So invalid.

(D) Total distance vs Time
Total distance must:

• always increase (never decrease)
• can be constant temporarily (if particle stops)

The graph shown is non-decreasing → valid.

Final answer:
A, B and D only
(C is not possible because it gives multiple velocities at the same time)

The standard formulae for a projectile launched with speed $$u$$ at an angle $$\theta$$ above the horizontal are:

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

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

The condition given in the problem is that the range is three times the maximum height:

$$R = 3H$$

Substitute the expressions for $$R$$ and $$H$$:

$$\frac{u^{2}\sin 2\theta}{g} = 3\left(\frac{u^{2}\sin^{2}\theta}{2g}\right)$$

Cancel the common factors $$u^{2}$$ and $$g$$:

$$\sin 2\theta = \frac{3}{2}\sin^{2}\theta$$ $$-(1)$$

Use the trigonometric identity $$\sin 2\theta = 2\sin\theta\cos\theta$$ and substitute it into $$(1)$$:

$$2\sin\theta\cos\theta = \frac{3}{2}\sin^{2}\theta$$

Divide both sides by $$\sin\theta$$ (since $$\theta \neq 0^{\circ},90^{\circ}$$ for a projectile with non-zero range):

$$2\cos\theta = \frac{3}{2}\sin\theta$$

Rearrange to get the cotangent of the launch angle:

$$\frac{\cos\theta}{\sin\theta} = \frac{3}{4} \quad\Longrightarrow\quad \cot\theta = \frac{3}{4}$$

Hence
$$\tan\theta = \frac{4}{3}$$

From the right-triangle definition, this gives

$$\sin\theta = \frac{4}{5},\qquad \cos\theta = \frac{3}{5}$$

Now compute $$\sin 2\theta$$ using $$\sin 2\theta = 2\sin\theta\cos\theta$$:

$$\sin 2\theta = 2\left(\frac{4}{5}\right)\left(\frac{3}{5}\right) = \frac{24}{25}$$

Finally, substitute $$\sin 2\theta$$ into the range formula:

$$R = \frac{u^{2}\sin 2\theta}{g} = \frac{u^{2}}{g}\left(\frac{24}{25}\right) = \frac{24\,u^{2}}{25g}$$

The question writes the range as $$\dfrac{nu^{2}}{25g}$$. Comparing gives $$n = 24$$.

Therefore, the correct option is Option D (24).

First, remember:

• Slope of x-t graph = velocity
• Straight line → constant velocity
• Horizontal line → zero velocity
• Average velocity = (final position − initial position) / time

(A) Average velocity from 0 to 3 s

From the graph:

$$At\ t=0,x=0$$

$$At\ t=3,x=5$$

So,

$$v_{avg}=\frac{5-0}{3}=\frac{5}{3}\approx1.67\text{ m/s}$$

Given statement says $$10m/s,$$ which is clearly wrong.

So, (A) is false

(B) Average velocity from 3 to 5 s

From graph:

• $$At\ t=3,x=5$$
• $$At\ t=5,x=5$$

So displacement does not change

$$v_{avg}=\frac{5-5}{5-3}=0$$

This means the object is at rest.

So, (B) is true

(C) Instantaneous velocity at t=2 s

At t=2, the graph lies on the straight line between:

• $$(1,-5)and(3,5)$$

Slope (velocity) of this line:

$$v=\frac{5-(-5)}{3-1}=\frac{10}{2}=5\text{ m/s}$$

Since it’s a straight line, velocity is constant in that region.

So, (C) is true

(D) Compare velocities

Average velocity from 5 to 7:

• $$At\ t=5,x=5$$
• $$At t=7,x=0$$

$$v_{avg}=\frac{0-5}{7-5}=\frac{-5}{2}=-2.5$$

Instantaneous velocity at 6.5 s:

This lies between t=6 and t=7, where graph is a straight line from:

• (6,10)to (7,0)

$$v=\frac{0-10}{7-6}=-10$$

Clearly:

$$-2.5\ne-10$$

So, (D) is false

(E) Average velocity from 0 to 9 s

• At t=0, x=0
• At t=9, x=0

$$v_{avg}=\frac{0-0}{9-0}=0$$

So, (E) is true

Final Answer:

(B), (C), (E) only

Let the common initial speed of the two projectiles be $$u$$ and let the acceleration due to gravity be $$g$$ (acting downward).

Case 1: Projection angle $$\theta_1 = 45^\circ + \alpha$$
Time of flight formula for a projectile launched from ground level is
$$T = \frac{2u \sin\theta}{g}$$
Thus, $$T_1 = \frac{2u \sin (45^\circ + \alpha)}{g}$$ $$-(1)$$

Case 2: Projection angle $$\theta_2 = 45^\circ - \alpha$$
Similarly,
$$T_2 = \frac{2u \sin (45^\circ - \alpha)}{g}$$ $$-(2)$$

We require the ratio $$\dfrac{T_1}{T_2}$$. Using $$(1)$$ and $$(2)$$, the common factors $$\frac{2u}{g}$$ cancel out:

$$\frac{T_1}{T_2} = \frac{\sin (45^\circ + \alpha)}{\sin (45^\circ - \alpha)}$$ $$-(3)$$

Apply the sine addition-subtraction identities:
$$\sin (45^\circ \pm \alpha) = \sin 45^\circ \cos\alpha \pm \cos 45^\circ \sin\alpha$$

Since $$\sin 45^\circ = \cos 45^\circ = \frac{\sqrt{2}}{2}$$, we obtain
$$\sin (45^\circ + \alpha) = \frac{\sqrt{2}}{2}(\cos\alpha + \sin\alpha)$$
$$\sin (45^\circ - \alpha) = \frac{\sqrt{2}}{2}(\cos\alpha - \sin\alpha)$$

Substitute these into $$(3)$$:

$$\frac{T_1}{T_2} \;=\; \frac{\frac{\sqrt{2}}{2}(\cos\alpha + \sin\alpha)}{\frac{\sqrt{2}}{2}(\cos\alpha - \sin\alpha)}$$

The factor $$\frac{\sqrt{2}}{2}$$ cancels out, giving

$$\frac{T_1}{T_2} = \frac{\cos\alpha + \sin\alpha}{\cos\alpha - \sin\alpha}$$ $$-(4)$$

To convert the expression to a form involving $$\tan\alpha$$, divide numerator and denominator of $$(4)$$ by $$\cos\alpha$$:

$$\frac{T_1}{T_2} = \frac{1 + \tan\alpha}{1 - \tan\alpha}$$

Hence, the required ratio of times of flight is $$\dfrac{1 + \tan\alpha}{1 - \tan\alpha}$$, which corresponds to Option D.

The displacement of the particle is given as
$$x = c_0\,(t^{2}-2) + c\,(t-2)^{2} \qquad -(1)$$

Step 1 - Find velocity
Velocity is the time-derivative of displacement:
$$v = \frac{dx}{dt}$$

Differentiating $$-(1)$$ term by term:

• For $$c_0\,(t^{2}-2)$$: $$\frac{d}{dt}\big(c_0\,(t^{2}-2)\big)=c_0\,(2t)=2c_0t$$
• For $$c\,(t-2)^{2}$$: $$\frac{d}{dt}\big(c\,(t-2)^{2}\big)=c\,(2\,(t-2))=2c\,(t-2)$$

Therefore
$$v = 2c_0t + 2c\,(t-2) \qquad -(2)$$

Step 2 - Find acceleration
Acceleration is the time-derivative of velocity:
$$a = \frac{dv}{dt}$$

Differentiating $$-(2)$$:

• $$\frac{d}{dt}\big(2c_0 t\big)=2c_0$$
• $$\frac{d}{dt}\big(2c\,(t-2)\big)=2c$$

Hence
$$a = 2c_0 + 2c = 2\,(c_0 + c) \qquad -(3)$$

Acceleration is a constant equal to $$2(c_0 + c)$$.

Step 3 - Check other statements

Initial velocity (at $$t=0$$) from $$-(2)$$:
$$v(0)=2c_0\,(0) + 2c\,(0-2) = -4c$$

Thus the initial velocity is $$-4c$$, not $$4c$$.

Conclusion
From $$-(3)$$, the only correct statement is: “the acceleration of the particle is $$2(c + c_0)$$,” which corresponds to Option D.

Distance travelled :A TO B,B TO A, A TO B

A TO B =$$\pi\ $$r
B to A =$$\pi\ $$r
A TO B =$$\pi\ $$r

Total distance =3*$$\pi\ $$r
Displacement : A TO B = 2r

Let the common initial speed of both balls be $$u$$ and let the projection angles with the horizontal be $$\theta_1$$ and $$\theta_2$$ for the first and the second ball, respectively.

Step 1 - Write the expression for maximum height.
For a projectile launched with speed $$u$$ at an angle $$\theta$$, the maximum height reached is

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

Step 2 - Set up the given height ratio.
Given that the first ball rises eight times higher than the second,

$$\frac{H_1}{H_2}=8$$

Using the formula for $$H$$,

$$\frac{\dfrac{u^{2}\sin^{2}\theta_1}{2g}}{\dfrac{u^{2}\sin^{2}\theta_2}{2g}}=8$$

Simplifying (the factors $$u^{2}$$ and $$2g$$ cancel),

$$\frac{\sin^{2}\theta_1}{\sin^{2}\theta_2}=8$$

Taking square root on both sides,

$$\frac{\sin\theta_1}{\sin\theta_2}=2\sqrt{2}$$ $$-(1)$$

Step 3 - Write the expression for time of flight.
For the same projectile, the total time of flight is

$$T=\frac{2u\sin\theta}{g}$$

Step 4 - Form the required ratio of flight times.
Taking the ratio of times for the two balls,

$$\frac{T_1}{T_2}=\frac{\dfrac{2u\sin\theta_1}{g}}{\dfrac{2u\sin\theta_2}{g}} =\frac{\sin\theta_1}{\sin\theta_2}$$

Using result $$-(1)$$,

$$\frac{T_1}{T_2}=2\sqrt{2}$$

Final ratio.
Hence

$$T_1:T_2 = 2\sqrt{2}:1$$

So the correct choice is Option A.

Speed of the helicopter: $$v = 360 \text{ km h}^{-1} = 360 \times \frac{1000}{3600}\,\text{m s}^{-1} = 100 \,\text{m s}^{-1}$$

Altitude from which the object is released: $$h = 2 \text{ km} = 2000 \text{ m}$$

Time taken by the object to fall through $$h$$ (free-fall, neglecting air resistance) is obtained from $$h = \frac12 \, g t^{2} \; \Longrightarrow \; t = \sqrt{\frac{2h}{g}}$$

Substituting $$h = 2000 \text{ m}$$ and $$g = 10 \text{ m s}^{-2}$$, $$t = \sqrt{\frac{2 \times 2000}{10}} = \sqrt{400} = 20 \text{ s}$$

Thus the object reaches the ground exactly $$20 \text{ s}$$ after it is dropped.

Horizontal distance travelled in this time by either the helicopter or the object is $$x = v t = 100 \,\text{m s}^{-1} \times 20 \text{ s} = 2000 \text{ m} = 2 \text{ km}$$

Vertical distance descended by the object is the full altitude, i.e. $$y = 2 \text{ km}$$ downward.

Hence the displacement vector of the object, measured from the point of release on the helicopter, has components horizontal $$2 \text{ km}$$ and vertical $$2 \text{ km}$$. Its magnitude is $$\sqrt{(2\text{ km})^{2} + (2\text{ km})^{2}} = 2\sqrt{2}\ \text{km}$$

Therefore, the required displacement is $$2\sqrt{2}$$ km.
Option D.

Let us divide this problem into 2 parts:-

First 2 seconds where the velocity changes , which means acceleration is present

Second part from 2 seconds to 30.5 (28.5 seconds in total) seconds where Velocity is constant

We know that distance s:-

$$s=ut+\ \frac{\ 1}{2}at^2$$

For first 2 seconds :-

u=$$200\ \ \frac{\ m}{s}$$

a=$$\ \frac{\ v_f=v_i}{t}$$

a=$$\ \frac{\ 400-200}{2}$$

a=100 $$\ \frac{\ m}{s^2}$$

Now $$s=200\times\ 2\ +\ \ \frac{\ 1}{2}\times\ 100\times\ \left(2\right)^2$$

$$s=600\ m$$

For next 28.5 s a=0(constant velocity), Here u=400 $$\ \frac{\ m}{s}$$

so s=$$ut$$

s=$$400\times\ 28.5$$

s=11400 m

now adding both 

$$s_{total}$$ = 600+ 11400

$$s_{total}$$= 12000 m = 12 Km

given
angle φ is measured from vertical

so convert to usual projectile angle θ (from horizontal):

θ = 90° − φ

maximum height formula:

hₘ = (u² sin²θ) / (2g)

substitute θ:

hₘ = (u² sin²(90° − φ)) / (2g)

use identity:

sin(90° − φ) = cosφ

so,

hₘ = (u² cos²φ) / (2g)

so relation is:

hₘ ∝ cos²φ

now understand behavior of cos²φ

when φ = 0°
cosφ = 1 → hₘ is maximum

when φ increases
cosφ decreases → hₘ decreases

when φ = 90°
cosφ = 0 → hₘ = 0

so graph must:

start at maximum value at φ = 0
decrease continuously
reach zero at φ = 90°

shape:

cos²φ decreases slowly at first, then faster
so curve is downward bending (not straight)

check options:

A → increasing → wrong
B → goes up then down → wrong
C → starts high, smoothly decreases to zero → correct
D → decreases too sharply like exponential → wrong

final answer: C

Let the two position vectors be $$\vec{A} = 2\hat{i} + 3n\hat{j} + 2\hat{k}$$ and $$\vec{B} = 2\hat{i} - 2\hat{j} + 4p\hat{k}$$. Since both particles are equidistant from the origin, their magnitudes are equal, that is $$|\vec{A}| = |\vec{B}|$$.

Thus we have $$ 4 + 9n^2 + 4 = 4 + 4 + 16p^2 $$ which simplifies to $$ 9n^2 = 16p^2 \quad \cdots (1).$$

Moreover, the perpendicularity condition $$\vec{A}\cdot\vec{B}=0$$ gives $$ (2)(2) + (3n)(-2) + (2)(4p) = 0 $$ or $$ 4 - 6n + 8p = 0, $$ hence $$ 8p = 6n - 4 $$ and $$ p = \frac{6n - 4}{8} = \frac{3n - 2}{4} \quad \cdots (2).$$

Substituting (2) into (1) yields $$ 9n^2 = 16 \left(\frac{3n - 2}{4}\right)^2 = 16 \cdot \frac{(3n-2)^2}{16} = (3n-2)^2, $$ so $$ 9n^2 = 9n^2 - 12n + 4, $$ which leads to $$ 0 = -12n + 4 $$ and $$ n = \frac{1}{3}. $$

Therefore, $$n^{-1} = 3$$, and the answer is 3.

A particle is projected at $$30°$$ from horizontal at 60 m/s. We need to find the ratio $$h_0 : h_1$$.

$$u_y = u \sin 30° = 60 \times \frac{1}{2} = 30$$ m/s

At maximum height, $$v_y = 0$$:

$$0 = u_y - gt \Rightarrow t = \frac{u_y}{g} = \frac{30}{10} = 3$$ s

$$h_0 = u_y(1) - \frac{1}{2}g(1)^2 = 30 - 5 = 25$$ m

The last second is from t = 2 to t = 3.

Height at t = 2: $$y_2 = 30(2) - \frac{1}{2}(10)(4) = 60 - 20 = 40$$ m

Height at t = 3: $$y_3 = 30(3) - \frac{1}{2}(10)(9) = 90 - 45 = 45$$ m

$$h_1 = y_3 - y_2 = 45 - 40 = 5$$ m

$$h_0 : h_1 = 25 : 5 = 5 : 1$$

The answer is 5.

Total distance covered  $$S = x + \frac{3}{2}x = \frac{5}{2}x$$.

Time taken in the first part  $$t_1 = \frac{\text{distance}}{\text{speed}} = \frac{x}{v_1}$$.

Given  $$v_1 = 5 \text{ m/s}$$, so  $$t_1 = \frac{x}{5}$$.

Time taken in the second part  $$t_2 = \frac{\frac{3}{2}x}{v_2} = \frac{3x}{2v_2}$$.

Total time  $$T = t_1 + t_2 = \frac{x}{5} + \frac{3x}{2v_2}$$.

The average velocity formula is
$$\text{Average velocity} = \frac{\text{Total distance}}{\text{Total time}}$$.

According to the question,
$$\frac{\frac{5}{2}x}{\frac{x}{5} + \frac{3x}{2v_2}} = \frac{50}{7}$$.

Cancel the common factor $$x$$ from numerator and denominator:
$$\frac{\frac{5}{2}}{\frac{1}{5} + \frac{3}{2v_2}} = \frac{50}{7}$$.

Cross-multiply:
$$\frac{5}{2} \times \frac{7}{50} = \frac{1}{5} + \frac{3}{2v_2}$$.

Simplify the left side:
$$\frac{5}{2} \times \frac{7}{50} = \frac{35}{100} = \frac{7}{20}$$.

Thus,
$$\frac{7}{20} = \frac{1}{5} + \frac{3}{2v_2}$$.

Convert $$\frac{1}{5}$$ to a denominator of 20:
$$\frac{1}{5} = \frac{4}{20}$$.

Subtract $$\frac{4}{20}$$ from both sides:
$$\frac{7}{20} - \frac{4}{20} = \frac{3}{2v_2}$$
$$\frac{3}{20} = \frac{3}{2v_2}$$.

Cancel the common factor 3:
$$\frac{1}{20} = \frac{1}{2v_2}$$.

Cross-multiply to find $$v_2$$:
$$2v_2 = 20 \; \Longrightarrow \; v_2 = 10 \text{ m/s}.$$

Hence, the required velocity $$v_2$$ is 10 m/s.

We need to find $$|\vec{A} + \vec{B}|$$ for two vectors of equal magnitude $$R$$ inclined at angle $$\theta$$.

Since the magnitude of the sum of two vectors is given by the parallelogram law, $$|\vec{A} + \vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}|\cos\theta$$.

Substituting $$|\vec{A}| = |\vec{B}| = R$$ into this expression gives $$|\vec{A} + \vec{B}|^2 = R^2 + R^2 + 2R^2\cos\theta = 2R^2(1 + \cos\theta)$$.

Using the double angle identity $$1 + \cos\theta = 2\cos^2\left(\frac{\theta}{2}\right)$$, it follows that $$|\vec{A} + \vec{B}|^2 = 2R^2 \times 2\cos^2\left(\frac{\theta}{2}\right) = 4R^2\cos^2\left(\frac{\theta}{2}\right)$$.

Now, taking the positive square root yields $$|\vec{A} + \vec{B}| = 2R\cos\left(\frac{\theta}{2}\right)$$.

For verification, one can similarly compute $$|\vec{A} - \vec{B}|^2 = 2R^2(1-\cos\theta) = 4R^2\sin^2\left(\frac{\theta}{2}\right)$$, giving $$|\vec{A}-\vec{B}| = 2R\sin\left(\frac{\theta}{2}\right)$$, which matches Option 1's formula but with coefficient 2 (not $$\sqrt{2}$$) and therefore Option 1 is incorrect.

The correct answer is Option 3: $$|\vec{A} + \vec{B}| = 2R\cos\frac{\theta}{2}$$.

We need to find the angle between $$\vec{Q}$$ and the resultant of $$(2\vec{Q} + 2\vec{P})$$ and $$(2\vec{Q} - 2\vec{P})$$.

Find the resultant.

$$ \vec{R} = (2\vec{Q} + 2\vec{P}) + (2\vec{Q} - 2\vec{P}) = 4\vec{Q} $$

The resultant is $$4\vec{Q}$$, which is in the same direction as $$\vec{Q}$$.

Therefore, the angle between $$\vec{Q}$$ and $$\vec{R} = 4\vec{Q}$$ is $$0°$$.

The correct answer is Option (2): $$0°$$.

Let the tower height be $$h$$, initial speed be $$u$$, and $$g$$ be acceleration due to gravity.

Projected upward: Taking downward as positive:

$$h = -ut_1 + \frac{1}{2}gt_1^2$$ ... (1)

Projected downward:

$$h = ut_2 + \frac{1}{2}gt_2^2$$ ... (2)

Dropped (u = 0):

$$h = \frac{1}{2}gt^2$$ ... (3)

From (1): $$h = \frac{1}{2}gt_1^2 - ut_1$$
From (2): $$h = \frac{1}{2}gt_2^2 + ut_2$$

Adding (1) and (2): $$2h = \frac{1}{2}g(t_1^2 + t_2^2) + u(t_2 - t_1)$$

Subtracting (1) from (2): $$0 = \frac{1}{2}g(t_2^2 - t_1^2) + u(t_1 + t_2)$$

$$u = \frac{g(t_1^2 - t_2^2)}{2(t_1 + t_2)} = \frac{g(t_1 - t_2)}{2}$$

Substituting back into (2): $$h = \frac{g(t_1-t_2)}{2} \cdot t_2 + \frac{1}{2}gt_2^2 = \frac{g}{2}(t_1 t_2 - t_2^2 + t_2^2) = \frac{g t_1 t_2}{2}$$

From (3): $$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \cdot \frac{g t_1 t_2}{2}}{g}} = \sqrt{t_1 t_2}$$.

The correct answer is Option A: $$\sqrt{t_1 t_2}$$.

$$S_1 = \frac{1}{2}a(p-1)^2$$, $$S_2 = \frac{1}{2}ap^2$$.

$$S_1 + S_2 = \frac{a}{2}[(p-1)^2 + p^2] = \frac{a}{2}(2p^2 - 2p + 1)$$

If this equals $$\frac{1}{2}at^2$$: $$t^2 = 2p^2 - 2p + 1$$, $$t = \sqrt{2p^2-2p+1}$$.

The answer corresponds to Option (2).

Let the initial velocity be $$u$$. After travelling 4 cm, velocity = $$\frac{2u}{3}$$.

Using $$v^2 = u^2 - 2as$$ (deceleration $$a$$):

$$\frac{4u^2}{9} = u^2 - 2a(0.04)$$

$$2a(0.04) = u^2 - \frac{4u^2}{9} = \frac{5u^2}{9}$$

$$a = \frac{5u^2}{9 \times 0.08} = \frac{5u^2}{0.72}$$

For the bullet to come to rest from $$v = \frac{2u}{3}$$:

$$0 = \frac{4u^2}{9} - 2a \cdot D'$$

$$D' = \frac{4u^2}{18a} = \frac{4u^2}{18 \times \frac{5u^2}{0.72}} = \frac{4 \times 0.72}{18 \times 5} = \frac{2.88}{90} = 0.032 \text{ m} = 32 \times 10^{-3} \text{ m}$$

So $$D = 32$$.

The answer is $$32$$, which corresponds to Option (1).

We need to find the velocity of a particle when its acceleration is zero, given $$x = t^3 - 6t^2 + 20t + 15$$.

Find the velocity.

$$v = \frac{dx}{dt} = 3t^2 - 12t + 20$$

Find the acceleration.

$$a = \frac{dv}{dt} = 6t - 12$$

Find when acceleration is zero.

$$6t - 12 = 0 \implies t = 2$$ s

Find velocity at $$t = 2$$.

$$v(2) = 3(4) - 12(2) + 20 = 12 - 24 + 20 = 8$$ m/s

The correct answer is Option (2): 8 m/s.

Let the total distance be $$2d$$.

First half (distance $$d$$): Speed = 6 m/s, so time = $$\frac{d}{6}$$.

Second half (distance $$d$$): Covered in two equal time intervals. Let each interval be $$t$$.

Distance in first interval = $$9t$$, distance in second = $$15t$$.

Total: $$9t + 15t = 24t = d$$, so $$t = \frac{d}{24}$$.

Total time for second half = $$2t = \frac{d}{12}$$.

Total time = $$\frac{d}{6} + \frac{d}{12} = \frac{2d + d}{12} = \frac{3d}{12} = \frac{d}{4}$$.

Average speed = $$\frac{\text{Total distance}}{\text{Total time}} = \frac{2d}{d/4} = 8$$ m/s.

The correct answer is Option 2: 8 m/s.

The position is given by $$\vec{S} = 2t^2 \hat{j} + 5\hat{k}$$.

Velocity is the time derivative of position:

$$\vec{v} = \frac{d\vec{S}}{dt} = 4t\hat{j} + 0\hat{k} = 4t\hat{j}$$

At $$t = 1$$ s:

$$\vec{v} = 4(1)\hat{j} = 4\hat{j}$$ m/s

The magnitude is $$4$$ m/s and the direction is along the y-direction.

The answer is $$4 \text{ m s}^{-1}$$ in y-direction, which corresponds to Option (4).

Given,

Two Projectiles A and B  are thrown with an angle  of $$45°$$ and $$60°$$ with vertical

In the question it is specified that the angle is with respect to the vertical 

For the Time of flight (T) to be same their vertical velocities must be equal

So

As it is with respect to vertical the vertical velocities are:-

For A $$v_A=v_a\cos\theta\ _A$$ ($$\cos\theta\ \ as\ \ \theta\ \ is\ with\ respect\ to\ vertical$$)

and For B $$v_B=v_b\cos\theta\ _A$$

Now these both should be equal
so ,

$$v_a\cos\left(45^{\circ\ }\right)=v_b\cos\left(60^{\circ\ }\right)$$

$$v_a\times\ \frac{1}{\sqrt{\ 2}}=v_b\times\ \frac{1}{2}$$

$$\frac{v_a}{v_b}=\ \frac{\sqrt{\ 2}}{2}=\frac{1}{\sqrt{\ 2}}$$

R = u²sin2θ/g, H = u²sin²θ/(2g). R=H: sin2θ = sin²θ/2 → 2sinθcosθ = sin²θ/2 → 4cosθ = sinθ → tanθ = 4.

Option (1): tan⁻¹(4).

First, convert the speeds from km/h to m/s using the conversion factor: $$1 \text{ km/h} = \frac{5}{18} \text{ m/s}$$.

For Train A moving north:
Speed $$v_A = 72 \text{ km/h} = 72 \times \frac{5}{18} = 20 \text{ m/s}$$ (north direction).

For Train B moving south:
Speed $$v_B = 108 \text{ km/h} = 108 \times \frac{5}{18} = 30 \text{ m/s}$$ (south direction).

Assign directions: Let north be positive and south be negative.
Thus, $$v_A = +20 \text{ m/s}$$ and $$v_B = -30 \text{ m/s}$$.

Velocity of train B with respect to A ($$v_{BA}$$):
The relative velocity formula is $$v_{BA} = v_B - v_A$$.
Substitute values: $$v_{BA} = (-30) - (20) = -50 \text{ m/s}$$.

Velocity of ground with respect to B ($$v_{GB}$$):
The ground is stationary, so $$v_G = 0 \text{ m/s}$$.
The relative velocity formula is $$v_{GB} = v_G - v_B$$.
Substitute values: $$v_{GB} = 0 - (-30) = +30 \text{ m/s}$$.

Therefore, the velocities are $$-50 \text{ m/s}$$ and $$30 \text{ m/s}$$, which corresponds to option C.

Two cars travel towards each other at 20 m/s each, initially 300 m apart, both braking with retardation 2 m/s$$^2$$. We need the distance between them when they stop.

Using the kinematic equation $$v^2 = u^2 - 2as$$ (where $$v = 0$$ at stop):

$$ 0 = (20)^2 - 2(2)s $$

$$ 0 = 400 - 4s $$

$$ s = \frac{400}{4} = 100 \text{ m} $$

Each car travels 100 m before stopping.

Both cars move toward each other, so the total distance closed between them is:

$$ s_{total} = 100 + 100 = 200 \text{ m} $$

Initial separation was 300 m. After both cars stop:

$$ d = 300 - 200 = 100 \text{ m} $$

The distance between them when they come to rest is 100 m.

The correct answer is Option (2): 100 m.

Let the ball leave the edge of the topmost step with only a horizontal velocity $$u$$. Choose origin at this edge: horizontal forward direction is $$+x$$, vertical downward is $$+y$$.

The riser (height) and tread (width) of every step are each $$h = 0.1\text{ m}$$. Hence the corner (outer edge) of step $$n$$ is located at $$x_n = n\,h$$ and $$y_n = n\,h$$  for  $$n = 1,2,3,\dots$$

For horizontal projection, the co-ordinates of the projectile after time $$t$$ are $$x = u\,t$$,  $$y = \tfrac12 g t^{2}$$. Eliminating $$t$$ gives the trajectory equation

$$y = \frac{g\,x^{2}}{2u^{2}} \qquad -(1)$$

The problem asks for the minimum $$u$$ such that the ball clears steps 1 to 4 and first lands on step 5. Step 5 has

vertical drop  $$y_5 = 5h = 0.5\text{ m}$$, horizontal range  $$0.4\text{ m} \le x \le 0.5\text{ m}$$.

Condition at landing (step 5)
Let the ball meet step 5 at some $$x = a$$, where $$0.4 \le a \le 0.5$$. Substitute $$y = 0.5$$ and $$x = a$$ into (1):

$$0.5 = \frac{g\,a^{2}}{2u^{2}}$$ $$\Rightarrow \; u^{2} = g\,a^{2} \qquad -(2)$$ $$\Rightarrow \; u = a\,\sqrt{g}$$

Clearance over the earlier steps
At the corner of step $$n$$ (for $$n = 1,2,3,4$$) the co-ordinates are $$(x_n , y_n) = (nh , nh).$$ The ball must be above each such corner, i.e.

$$y(x_n) < y_n$$ $$\Longrightarrow \frac{g\,x_n^{2}}{2u^{2}} < nh$$ Using $$(2)$$ (replace $$u^{2}$$ by $$g\,a^{2}$$) and $$x_n = nh$$:

$$\frac{g\,(nh)^{2}}{2g\,a^{2}} < nh$$ $$\Rightarrow \frac{n^{2}h^{2}}{2a^{2}} < nh$$ $$\Rightarrow \frac{nh}{2a^{2}} < 1$$ $$\Rightarrow a^{2} > \frac{nh}{2} \qquad -(3)$$

Compute the right-hand side of (3) for $$h = 0.1\text{ m}$$:

n = 1: $$a^{2} > 0.05$$ → $$a > 0.223$$
n = 2: $$a^{2} > 0.10$$ → $$a > 0.316$$
n = 3: $$a^{2} > 0.15$$ → $$a > 0.387$$
n = 4: $$a^{2} > 0.20$$ → $$a > 0.447$$

The most stringent requirement comes from step 4: $$a \ge \sqrt{0.20} = 0.447\text{ m}$$.

Choosing the landing point for minimum $$u$$
Equation (2) shows $$u = a\sqrt{g}$$, so smaller $$a$$ means smaller $$u$$. Subject to both conditions $$0.4 \le a \le 0.5$$ (must be on step 5) and $$a \ge 0.447\text{ m}$$ (must clear step 4), the minimum admissible value is

$$a_{\min} = \sqrt{0.20}\text{ m} = 0.447\text{ m}$$.

Insert this $$a_{\min}$$ into (2):

$$u_{\min} = a_{\min}\sqrt{g} = \sqrt{0.20\,g}$$ With $$g = 10\text{ m s}^{-2}$$,

$$u_{\min} = \sqrt{0.20 \times 10} = \sqrt{2}\text{ m s}^{-1}$$.

Thus the minimum horizontal velocity is $$\sqrt{2}\text{ m s}^{-1}$$, which corresponds to $$x = 2$$ in the given form $$\sqrt{x}$$.

Answer: $$x = 2$$

Let A be at distance $$s$$ from the starting point. The body covers 80 m from A to B in 2 seconds.

Distance from start to A: $$s = \frac{1}{2}gt_1^2$$ where $$t_1$$ is time to reach A.

Distance from start to B: $$s + 80 = \frac{1}{2}g(t_1 + 2)^2$$

$$80 = \frac{1}{2}g[(t_1+2)^2 - t_1^2] = \frac{1}{2}(10)(4t_1 + 4) = 5(4t_1 + 4) = 20t_1 + 20$$

$$20t_1 = 60$$

$$t_1 = 3$$ s

$$s = \frac{1}{2}(10)(9) = 45$$ m

The answer is $$\boxed{45}$$ m.

A body starts from rest on a frictionless plane. We are given the formula for the ratio $$\frac{S_{n-1}}{S_n} = \left(1 - \frac{2}{x}\right)$$ for $$n = 10$$, and we need to find $$x$$.

We first recall the distance covered in the $$n$$-th second under uniform acceleration.

For a body starting from rest (initial velocity $$u = 0$$) with uniform acceleration $$a$$, the distance covered from $$t = 0$$ to $$t = n$$ is:

$$ S(n) = \frac{1}{2}an^2 $$

The distance covered in the $$n$$-th second (between $$t = n-1$$ and $$t = n$$) is:

$$ S_n = S(n) - S(n-1) = \frac{1}{2}an^2 - \frac{1}{2}a(n-1)^2 $$

Expanding $$(n-1)^2 = n^2 - 2n + 1$$:

$$ S_n = \frac{a}{2}\left[n^2 - (n^2 - 2n + 1)\right] = \frac{a}{2}(2n - 1) $$

Similarly, replacing $$n$$ with $$n-1$$ in the formula gives:

$$ S_{n-1} = \frac{a}{2}(2(n-1) - 1) = \frac{a}{2}(2n - 3) $$

Next, we compute the ratio $$\frac{S_{n-1}}{S_n}$$.

$$ \frac{S_{n-1}}{S_n} = \frac{\frac{a}{2}(2n-3)}{\frac{a}{2}(2n-1)} = \frac{2n - 3}{2n - 1} $$

This fraction can be rewritten as:

$$ \frac{2n - 3}{2n - 1} = \frac{(2n-1) - 2}{2n - 1} = 1 - \frac{2}{2n - 1} $$

Since $$\frac{S_{n-1}}{S_n} = 1 - \frac{2}{x}$$, equating this to $$1 - \frac{2}{2n - 1}$$ gives:

$$ \frac{2}{x} = \frac{2}{2n - 1} $$

$$ x = 2n - 1 $$

Finally, substituting $$n = 10$$ gives:

$$ x = 2(10) - 1 = 20 - 1 = 19 $$

The value of $$x$$ is $$\boxed{19}$$.

We need to find the horizontal range of a body thrown horizontally from a tower, given modified initial conditions.

Recall the formula for horizontal range in projectile motion (horizontal throw). When a body is thrown horizontally with velocity $$v$$ from a height $$H$$, the horizontal range $$R$$ is given by $$R = v \times t$$ where $$t$$ is the time to fall height $$H$$. Since vertical motion is free fall (initial vertical velocity = 0), we have $$H = \frac{1}{2}gt^2 \implies t = \sqrt{\frac{2H}{g}}$$.

Substituting this into the expression for $$R$$ gives $$R = v\sqrt{\frac{2H}{g}}$$, showing that the range does not depend on the mass of the body.

Applying the formula to the first case, where the mass is $$M$$, the horizontal velocity is $$v$$, and the height is $$H$$, and given that the range is 100 m, we get $$100 = v\sqrt{\frac{2H}{g}} \quad \ldots (1)$$.

In the second case, the mass becomes $$2M$$ (which is irrelevant since the range is independent of mass), the horizontal velocity is $$\frac{v}{2}$$, and the height is $$4H$$. Hence, $$R_2 = \frac{v}{2}\sqrt{\frac{2 \times 4H}{g}} = \frac{v}{2}\sqrt{\frac{8H}{g}} = \frac{v}{2} \times 2\sqrt{\frac{2H}{g}} = v\sqrt{\frac{2H}{g}}$$.

From equation (1), $$v\sqrt{\frac{2H}{g}} = 100$$ m, so $$R_2 = 100 \text{ m}$$. The halving of velocity is exactly compensated by the doubling of fall time (due to the four times greater height), resulting in the same horizontal range.

The answer is 100 m.

A body falls freely from height $$H$$, hits an inclined plane at height $$h$$, and after a perfectly elastic impact, its velocity becomes horizontal. We need to find $$\frac{H}{h}$$ for which the total time to reach the ground is maximized.

We start by finding the velocity at the inclined plane.

The body falls from height $$H$$ to height $$h$$, so it falls through a distance $$(H - h)$$. Using energy conservation (or kinematics):

$$ v = \sqrt{2g(H - h)} $$

Next, we analyze the motion after the perfectly elastic impact.

After the perfectly elastic impact, the direction of velocity changes to horizontal (magnitude remains the same). From height $$h$$ above the ground, the body undergoes projectile motion with:

- Initial horizontal velocity: $$v = \sqrt{2g(H-h)}$$

- Initial vertical velocity: 0 (horizontal direction)

Then, we calculate the total time.

Time for free fall from $$H$$ to $$h$$ (falling distance $$H - h$$):

$$ t_1 = \sqrt{\frac{2(H-h)}{g}} $$

Time for projectile fall from height $$h$$ to ground (vertical free fall from rest):

$$ t_2 = \sqrt{\frac{2h}{g}} $$

The total time is:

$$ T = t_1 + t_2 = \sqrt{\frac{2(H-h)}{g}} + \sqrt{\frac{2h}{g}} = \sqrt{\frac{2}{g}}\left(\sqrt{H-h} + \sqrt{h}\right) $$

To maximize the total time with respect to $$h$$, we consider:

We need to maximize $$f(h) = \sqrt{H - h} + \sqrt{h}$$ for $$0 \leq h \leq H$$.

Taking the derivative and setting it to zero:

$$ f'(h) = \frac{-1}{2\sqrt{H-h}} + \frac{1}{2\sqrt{h}} = 0 $$

$$ \frac{1}{2\sqrt{h}} = \frac{1}{2\sqrt{H-h}} $$

$$ \sqrt{h} = \sqrt{H - h} $$

$$ h = H - h $$

$$ 2h = H $$

$$ h = \frac{H}{2} $$

Finally, we find the ratio:

$$ \frac{H}{h} = \frac{H}{H/2} = 2 $$

The answer is $$\boxed{2}$$.

Find the acceleration of a particle with velocity $$v = 4\sqrt{x}$$ m/s.

Acceleration can be expressed as $$ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = \frac{dv}{dx} \cdot v $$ because by the chain rule, $$\frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt}$$ and $$\frac{dx}{dt} = v$$.

Since $$v = 4\sqrt{x} = 4x^{1/2}$$, it follows that $$ \frac{dv}{dx} = 4 \times \frac{1}{2}x^{-1/2} = \frac{2}{\sqrt{x}} $$.

Substituting into the expression for acceleration gives $$ a = \frac{dv}{dx} \cdot v = \frac{2}{\sqrt{x}} \times 4\sqrt{x} = \frac{2 \times 4\sqrt{x}}{\sqrt{x}} = 8 \text{ m/s}^2 $$. Note that $$\sqrt{x}$$ cancels, making the acceleration constant and independent of position.

The answer is 8 m/s$$^2$$.

Given $$x = -3t^3 + 18t^2 + 16t$$, we want to find the velocity when the acceleration is zero.

Differentiating the position with respect to time gives the velocity: $$ v = \frac{dx}{dt} = \frac{d}{dt}(-3t^3 + 18t^2 + 16t) = -9t^2 + 36t + 16 $$.

Further differentiating the velocity yields the acceleration: $$ a = \frac{dv}{dt} = \frac{d}{dt}(-9t^2 + 36t + 16) = -18t + 36 $$.

Setting the acceleration to zero gives $$ -18t + 36 = 0 \implies 18t = 36 \implies t = 2 \text{ s} $$.

Substituting $$t = 2$$ s into the velocity expression gives $$ v(2) = -9(2)^2 + 36(2) + 16 = -9(4) + 72 + 16 = -36 + 72 + 16 = 52 \text{ m/s} $$.

The answer is 52 m/s.

Initial velocity: $$u_x = 5$$ m/s, $$u_y = 0$$.

Acceleration: $$a_x = 3$$ m/s$$^2$$, $$a_y = 2$$ m/s$$^2$$.

The x-coordinate: $$x = u_x t + \frac{1}{2}a_x t^2 = 5t + \frac{3}{2}t^2$$

Setting $$x = 84$$:

$$5t + \frac{3}{2}t^2 = 84$$

$$3t^2 + 10t - 168 = 0$$

Using the quadratic formula:

$$t = \frac{-10 + \sqrt{100 + 2016}}{6} = \frac{-10 + \sqrt{2116}}{6} = \frac{-10 + 46}{6} = \frac{36}{6} = 6$$ s

Velocity components at $$t = 6$$ s:

$$v_x = u_x + a_x t = 5 + 3(6) = 23$$ m/s

$$v_y = u_y + a_y t = 0 + 2(6) = 12$$ m/s

Speed: $$v = \sqrt{v_x^2 + v_y^2} = \sqrt{529 + 144} = \sqrt{673}$$ m/s

So $$\alpha = 673$$.

The answer is $$\boxed{673}$$.

$$|\vec{A}| = \sqrt{9 + 16} = 5$$. $$\vec{B} = 4\hat{i} + 3\hat{j}$$, $$|\vec{B}| = 5$$.

Unit vector along $$\vec{B}$$: $$\hat{B} = \frac{4\hat{i} + 3\hat{j}}{5}$$.

Required vector = $$5\hat{B} = 4\hat{i} + 3\hat{j}$$.

The x-component is 4 and y-component is 3.

Therefore, $$x = \boxed{4}$$.

$$\vec{A} \cdot (\vec{B} \times \vec{C}) = 0$$ (scalar triple product).

$$\vec{B} \times \vec{C} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 4 & 3 \\ -8 & -1 & 3 \end{vmatrix} = (12+3)\hat{i} - (-3+24)\hat{j} + (1+32)\hat{k} = 15\hat{i} - 21\hat{j} + 33\hat{k}$$.

$$\vec{A} \cdot (\vec{B} \times \vec{C}) = -15x + (-6)(-21) + (-2)(33) = -15x + 126 - 66 = -15x + 60 = 0$$.

$$x = 4$$.

The answer is 4.

The angle between $$\vec{a}$$ and $$\vec{b}$$ is $$\cos^{-1}\left(\frac{5}{9}\right)$$ and $$|\vec{a}+\vec{b}| = \sqrt{2}|\vec{a}-\vec{b}|$$.

Let $$\cos\theta = \frac{5}{9}$$.

From the formulas for the square of the length of the sum and difference of vectors we have:

$$|\vec{a}+\vec{b}|^2 = a^2 + b^2 + 2ab\cos\theta$$

$$|\vec{a}-\vec{b}|^2 = a^2 + b^2 - 2ab\cos\theta$$

Using the given condition we write:

$$a^2 + b^2 + 2ab\cos\theta = 2\bigl(a^2 + b^2 - 2ab\cos\theta\bigr)$$

Expanding the right side gives:

$$a^2 + b^2 + 2ab\cos\theta = 2a^2 + 2b^2 - 4ab\cos\theta$$

Collecting like terms leads to:

$$6ab\cos\theta = a^2 + b^2$$

Substituting $$\cos\theta = \frac{5}{9}$$ into the equation gives:

$$6ab \cdot \frac{5}{9} = a^2 + b^2$$

which simplifies to:

$$\frac{10ab}{3} = a^2 + b^2$$

Assume $$|\vec{a}| = n|\vec{b}|$$ so that $$a = nb$$. Substituting into the last equation yields:

$$\frac{10n b^2}{3} = n^2b^2 + b^2$$

Dividing both sides by $$b^2$$ gives:

$$\frac{10n}{3} = n^2 + 1$$

Rearranging to a standard quadratic form:

$$3n^2 - 10n + 3 = 0$$

Factoring the quadratic leads to:

$$ (3n-1)(n-3) = 0$$

Thus $$n = 3$$ or $$n = \frac{1}{3}$$. Since the integer value is required we choose $$n = 3$$.

The answer is $$\boxed{3}$$.

We are given that in the time interval from $$t$$ to $$(t+1)$$ s, the displacement is $$125$$ m and the increase in velocity is $$50$$ m/s.

Since the increase in velocity in 1 second equals the acceleration:

$$a = 50 \text{ m/s}^2$$

The displacement in the interval from $$t$$ to $$(t+1)$$ s can be written as:

$$s = v_t \cdot (1) + \frac{1}{2}a(1)^2 = v_t + \frac{a}{2}$$

where $$v_t$$ is the velocity at time $$t$$.

$$125 = v_t + \frac{50}{2} = v_t + 25$$

$$v_t = 100 \text{ m/s}$$

The velocity at time $$(t+1)$$:

$$v_{t+1} = v_t + a = 100 + 50 = 150 \text{ m/s}$$

The distance travelled in the $$(t+2)$$-th second (i.e., from $$(t+1)$$ to $$(t+2)$$ s) is:

$$s_{(t+2)\text{th}} = v_{t+1} + \frac{a}{2} = 150 + \frac{50}{2} = 150 + 25 = 175 \text{ m}$$

The correct answer is $$175$$ m.

We are told that the maximum height reached by a projectile is 64 m, and we need to find the new maximum height when the initial velocity is halved. Since the maximum height depends on the vertical component of motion, we recall that for a projectile launched with initial velocity $$u$$ at angle $$\theta$$ with the horizontal, the maximum height is given by $$H = \frac{u^2 \sin^2\theta}{2g}$$. This formula follows from the condition that at maximum height the vertical velocity becomes zero; applying $$v^2 = u_y^2 - 2gH$$ with $$v = 0$$ and $$u_y = u\sin\theta$$ yields $$0 = u^2\sin^2\theta - 2gH \implies H = \frac{u^2\sin^2\theta}{2g}$$.

From this expression, it is clear that with $$\theta$$ and $$g$$ held constant, the maximum height is proportional to the square of the initial velocity, $$H \propto u^2$$. Therefore, if the initial velocity is halved ($$u' = u/2$$), the new maximum height becomes $$H' = \frac{(u/2)^2 \sin^2\theta}{2g} = \frac{u^2 \sin^2\theta}{4 \times 2g} = \frac{H}{4}$$. Substituting the known value $$H = 64$$ m gives $$H' = \frac{64}{4} = 16 \text{ m}$$.

The answer is 16 m.

Let's write the three vectors given in their deconstructed form:

$$\vec{OP} = A\hat{i}$$

$$\vec{OQ} = A\hat{j}$$

$$\vec{OR} = \dfrac{A}{\sqrt{2}}\hat{i} - \dfrac{A}{\sqrt{2}}\hat{j}$$

Thus, adding the vectors, we get,

$$\vec{OP}+\vec{OQ}+\vec{OR} =\vec{R} =  \left(A + \dfrac{A}{\sqrt{2}}\right)\hat{i} + \left(A - \dfrac{A}{\sqrt{2}}\right)\hat{j}$$

And hence, the magnitude of the resultant vector,

$$|\vec{R}| = \sqrt{\left(A + \dfrac{A}{\sqrt{2}}\right)^2 + \left(A - \dfrac{A}{\sqrt{2}}\right)^2}$$

$$|\vec{R}| = \sqrt{2A^2 + 2\dfrac{A^2}{2}} = A\sqrt{3}$$

Therefore, the value of $$x$$ is $$3$$.

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

A body travels 102.5 m in the $$n$$-th second and 115.0 m in the $$(n+2)$$-th second. We need to find the acceleration.

For a body with initial velocity $$u$$ and constant acceleration $$a$$, the distance covered in the $$n$$-th second (i.e., between $$t = n-1$$ and $$t = n$$) is given by $$ s_n = u + \frac{a}{2}(2n - 1)\,. $$ This formula follows from $$s_n = S(n) - S(n-1) = \left(un + \frac{1}{2}an^2\right) - \left(u(n-1) + \frac{1}{2}a(n-1)^2\right) = u + \frac{a}{2}(2n-1)\,. $$

Using the given data, for the $$n$$-th second we have $$ s_n = u + \frac{a}{2}(2n - 1) = 102.5\,, $$ and for the $$(n+2)$$-th second $$ s_{n+2} = u + \frac{a}{2}(2(n+2) - 1) = u + \frac{a}{2}(2n + 3) = 115.0\,. $$

Subtracting these two equations gives $$ s_{n+2} - s_n = \frac{a}{2}[(2n+3) - (2n-1)]\,, $$ so $$ 115.0 - 102.5 = \frac{a}{2}[2n + 3 - 2n + 1]\,, $$ which simplifies to $$ 12.5 = \frac{a}{2} \times 4 \quad\Longrightarrow\quad 12.5 = 2a\,. $$

Therefore, $$ a = \frac{12.5}{2} = 6.25 \text{ m/s}^2\,. $$ The correct answer is Option (1): 6.25 m/s2.

The time-distance relation is given as $$t = \alpha x^{2} + \beta x$$, where $$\alpha$$ and $$\beta$$ are constants.

First differentiate with respect to $$x$$:
$$\frac{dt}{dx} = 2\alpha x + \beta$$ $$-(1)$$

Velocity is $$v = \frac{dx}{dt} = \frac{1}{\dfrac{dt}{dx}} = \frac{1}{2\alpha x + \beta}$$ $$-(2)$$

Acceleration is $$a = \frac{dv}{dt}$$.
Using the chain rule, $$a = \frac{dv}{dx}\,\frac{dx}{dt} = v \,\frac{dv}{dx}$$ $$-(3)$$

Differentiate $$v$$ from $$(2)$$ with respect to $$x$$:
$$\frac{dv}{dx} = \frac{d}{dx}\left[(2\alpha x + \beta)^{-1}\right] = -1 \cdot (2\alpha)\,(2\alpha x + \beta)^{-2} = -\frac{2\alpha}{(2\alpha x + \beta)^{2}}$$ $$-(4)$$

Substitute $$(2)$$ and $$(4)$$ into $$(3)$$:
$$a = v\left(-\frac{2\alpha}{(2\alpha x + \beta)^{2}}\right) = -\frac{2\alpha\,v}{(2\alpha x + \beta)^{2}}$$ $$-(5)$$

From $$(2)$$, $$v = \dfrac{1}{2\alpha x + \beta} \;\Rightarrow\; 2\alpha x + \beta = \frac{1}{v}$$ $$-(6)$$

Insert $$(6)$$ into $$(5)$$ to eliminate $$x$$:
$$a = -\frac{2\alpha\,v}{\left(\dfrac{1}{v}\right)^{2}} = -2\alpha\,v \, v^{2} = -2\alpha\,v^{3}$$

Thus the required relation is $$a = -2\alpha v^{3}$$.

Option A is correct.

Given: $$x = 2 + 4t$$ and $$y = 3t + 8t^2$$

Velocity components:

$$v_x = \frac{dx}{dt} = 4$$ (constant)

$$v_y = \frac{dy}{dt} = 3 + 16t$$ (varying)

Acceleration components:

$$a_x = 0$$

$$a_y = 16$$ (constant)

Since the acceleration is constant (only in y-direction), the motion is uniformly accelerated.

From $$x = 2 + 4t$$: $$t = \frac{x-2}{4}$$

Substituting in y: $$y = 3\left(\frac{x-2}{4}\right) + 8\left(\frac{x-2}{4}\right)^2 = \frac{3(x-2)}{4} + \frac{(x-2)^2}{2}$$

This is a quadratic in x, so the path is parabolic.

The correct answer is Option 1: uniformly accelerated having motion along a parabolic path.

The body moves in a circle of radius $$R = 10$$ m with period $$T = 4$$ s.

At $$t = 3$$ s, the fraction of revolution completed:

$$ \frac{t}{T} = \frac{3}{4} $$

This means the body has covered $$\frac{3}{4}$$ of a full circle, i.e., $$270°$$.

If the body starts at the top of the circle (say at angle $$0°$$), after $$270°$$ it is at the position that is $$90°$$ short of completing the circle.

The displacement is the straight-line distance from the starting point to the current position. For a $$270°$$ arc (or equivalently $$90°$$ measured the other way), the chord length is:

Using the chord formula $$d = 2R\sin(\theta/2)$$ where $$\theta = 270°$$:

$$ d = 2 \times 10 \times \sin(135°) = 20 \times \frac{\sqrt{2}}{2} = 10\sqrt{2} \text{ m} $$

Alternatively, the starting and ending points are separated by $$90°$$ on the circle (since $$360° - 270° = 90°$$), forming two perpendicular radii. The displacement = $$\sqrt{R^2 + R^2} = R\sqrt{2} = 10\sqrt{2}$$ m.

A particle starts with initial velocity 10.0 m/s and accelerates uniformly at 2.0 m/s$$^2$$. We need to find the time to reach 60.0 m/s.

First, identify the known quantities.

Initial velocity: $$u = 10.0$$ m/s

Acceleration: $$a = 2.0$$ m/s$$^2$$

Final velocity: $$v = 60.0$$ m/s

Time: $$t = ?$$

Next, select the appropriate equation of motion.

Since we know $$u$$, $$v$$, and $$a$$, and need to find $$t$$, we use the first equation of motion:

$$ v = u + at $$

This equation relates velocity to time for uniformly accelerated motion. It follows directly from the definition of acceleration: $$a = \frac{v - u}{t}$$.

Now, substitute and solve for $$t$$.

$$ 60.0 = 10.0 + 2.0 \times t $$

$$ 2.0t = 60.0 - 10.0 = 50.0 $$

$$ t = \frac{50.0}{2.0} = 25\;\text{s} $$

The time taken is 25 s.

The correct answer is Option 1: 25 s.

Average velocity for equal distances: $$v = \frac{2x}{x/v_1 + x/v_2} = \frac{2v_1v_2}{v_1 + v_2}$$

This gives: $$\frac{2}{v} = \frac{v_1 + v_2}{v_1v_2} = \frac{1}{v_1} + \frac{1}{v_2}$$

The correct answer is Option 4.

The vector makes an angle of 30° with the y-axis.

If the angle with y-axis is 30°, then:

$$V_y = V\cos 30° = 2\sqrt{3}$$

$$V = \frac{2\sqrt{3}}{\cos 30°} = \frac{2\sqrt{3}}{\sqrt{3}/2} = 4$$

$$V_x = V\sin 30° = 4 \times \frac{1}{2} = 2$$

The magnitude of the x-component is $$\mathbf{2}$$.

We need to find the length of the tunnel $$L$$ given that train $$A$$ (length $$l$$) takes 35 s less than train $$B$$ (length $$4l$$) to cross the tunnel.

Convert velocities to m/s:

$$v_A = 108 \text{ km/h} = 108 \times \frac{5}{18} = 30 \text{ m/s}$$

$$v_B = 72 \text{ km/h} = 72 \times \frac{5}{18} = 20 \text{ m/s}$$

Time for a train to cross a tunnel = (length of train + length of tunnel) / speed of train.

$$t_A = \frac{L + l}{30}$$

$$t_B = \frac{L + 4l}{20}$$

Given $$t_B - t_A = 35$$ s:

$$\frac{L + 4l}{20} - \frac{L + l}{30} = 35$$

Taking LCM of 20 and 30 = 60:

$$\frac{3(L + 4l) - 2(L + l)}{60} = 35$$

$$\frac{3L + 12l - 2L - 2l}{60} = 35$$

$$\frac{L + 10l}{60} = 35$$

$$L + 10l = 2100$$

Given $$L = 60l$$, so $$l = \frac{L}{60}$$:

$$L + 10 \times \frac{L}{60} = 2100$$

$$L + \frac{L}{6} = 2100$$

$$\frac{7L}{6} = 2100$$

$$L = \frac{2100 \times 6}{7} = 1800 \text{ m}$$

The length of the tunnel is $$\mathbf{1800}$$ m.

A car travels a distance $$x$$ with speed $$v_1$$ and then the same distance $$x$$ with speed $$v_2$$ in the same direction. We need to find the average speed.

Calculate the total distance.

Total distance = $$x + x = 2x$$

Calculate the total time.

Time for first part: $$t_1 = \frac{x}{v_1}$$

Time for second part: $$t_2 = \frac{x}{v_2}$$

Total time: $$t_1 + t_2 = \frac{x}{v_1} + \frac{x}{v_2} = x\left(\frac{v_1 + v_2}{v_1 v_2}\right)$$

Calculate the average speed.

$$ v_{avg} = \frac{\text{Total distance}}{\text{Total time}} = \frac{2x}{x\left(\frac{v_1 + v_2}{v_1 v_2}\right)} = \frac{2v_1 v_2}{v_1 + v_2} $$

This is the harmonic mean of the two speeds, which is the correct formula when equal distances are covered at different speeds.

The correct answer is Option D: $$\frac{2v_1 v_2}{v_1 + v_2}$$.

Find the ratio of kinetic energy at the point of projection to the kinetic energy at the highest point for a stone projected at $$30°$$ to the horizontal.

The kinetic energy at the point of projection is $$KE_1 = \frac{1}{2}mv^2$$. At the highest point only the horizontal component of velocity remains, namely $$v_x = v\cos 30° = \frac{v\sqrt{3}}{2}$$, so the kinetic energy there is $$KE_2 = \frac{1}{2}mv_x^2 = \frac{1}{2}m \cdot \frac{3v^2}{4} = \frac{3}{4} \cdot \frac{1}{2}mv^2$$.

Therefore, the ratio of the kinetic energies is $$\frac{KE_1}{KE_2} = \frac{\frac{1}{2}mv^2}{\frac{3}{8}mv^2} = \frac{4}{3}$$, giving a ratio of $$4 : 3$$ which corresponds to Option D.

The vehicle travels two equal distances of 4 km each at different speeds. For equal distances, we can find the average speed by computing total distance over total time.

The time for the first 4 km at 3 km/h is $$t_1 = \frac{4}{3}$$ h, and the time for the next 4 km at 5 km/h is $$t_2 = \frac{4}{5}$$ h.

So the total distance is $$4 + 4 = 8$$ km, and the total time is

$$t_1 + t_2 = \frac{4}{3} + \frac{4}{5} = \frac{20 + 12}{15} = \frac{32}{15} \text{ h}$$

Now the average speed is

$$v_{avg} = \frac{\text{Total distance}}{\text{Total time}} = \frac{8}{\frac{32}{15}} = \frac{8 \times 15}{32} = \frac{120}{32} = 3.75 \text{ km h}^{-1}$$

So, the answer is $$3.75$$ km h$$^{-1}$$.

$$v_{avg} = \frac{\text{Total Distance}}{\text{Total Time}}$$

$$AB = BC = x$$

$$AD = 3AB = 3x$$

$$CD = AD - (AB + BC) = 3x - 2x = x$$

$$t_1 = \frac{x}{v_1}, \quad t_2 = \frac{x}{v_2}, \quad t_3 = \frac{x}{v_3}$$

$$v_{avg} = \frac{x + x + x}{t_1 + t_2 + t_3}$$

$$v_{avg} = \frac{3x}{\frac{x}{v_1} + \frac{x}{v_2} + \frac{x}{v_3}}$$

$$v_{avg} = \frac{3}{\frac{1}{v_1} + \frac{1}{v_2} + \frac{1}{v_3}}$$

$$v_{avg} = \frac{3v_1v_2v_3}{v_1v_2 + v_2v_3 + v_3v_1}$$

given
speed = π m/s (constant)
motion is along a circle
angle from A to B = 120°

we need average velocity = displacement / time

step 1: understand what displacement means

even though motion is along a curve, displacement is always the straight line from A to B

so here displacement = chord AB

step 2: find displacement (chord AB)

draw the triangle formed by center O, A, and B

OA = OB = R
angle AOB = 120°

chord formula:

AB = 2R sin(θ/2)

θ = 120° → θ/2 = 60°

AB = 2R sin60°
= 2R × (√3/2)
= R√3

so displacement = R√3

step 3: find time taken

the particle moves along the arc from A to B

arc length = Rθ

convert 120° into radians:

120° = 2π/3

so arc length = R × (2π/3)

speed = π

time = distance / speed

t = [R(2π/3)] / π

cancel π:

t = 2R/3

step 4: compute average velocity

average velocity = displacement / time

= (R√3) / (2R/3)

now simplify:

= R√3 × (3 / 2R)

R cancels:

= (3√3)/2

= 1.5√3 m/s

final answer: 1.5√3 m/s

The shaded region under a $$v\!-\!t$$ graph always gives displacement when the sign (above / below the time-axis) is taken into account and gives distance when only the magnitude of the area is added.

From the given graph it is seen that:

• For the first $$20\ \text{s}$$ the velocity is constant and positive (the horizontal line at height $$v_0$$).
   Area (rectangle) $$A_1 = v_0 \times 20$$

• For the next $$5\ \text{s}$$ (from $$t = 20\ \text{s}$$ to $$t = 25\ \text{s}$$) the velocity is the same in magnitude but negative (the horizontal line at depth $$-v_0$$).
   Area (rectangle) $$A_2 = (-\,v_0) \times 5$$

Distance travelled in 25 s
Add magnitudes of both areas:$$|A_1| + |A_2| = v_0(20) + v_0(5) = 25\,v_0$$

Displacement in 25 s
Add the signed areas:$$A_1 + A_2 = v_0(20) + (-v_0)(5) = v_0(20 - 5) = 15\,v_0$$

Required ratio
$$\text{distance} \; : \; \text{displacement} = \frac{25\,v_0}{15\,v_0} = \frac{5}{3}$$

Hence the ratio of distance to displacement in the first $$25\ \text{s}$$ of motion is $$\frac{5}{3}$$.
Option C is correct.

We need to analyze each statement carefully.

For Statement I, it says the area under the velocity-time graph gives the distance travelled. However, the area under the velocity-time graph actually gives the displacement, not the distance. Distance is the total path length and equals the area only when velocity does not change sign. If velocity becomes negative, the area below the time axis is negative, giving displacement. To get distance, we would need the area under the $$|v|$$ vs $$t$$ graph. So Statement I is incorrect in general.

Now consider Statement II, which says the area under the acceleration-time graph equals the change in velocity. We know that $$a = \frac{dv}{dt}$$, so

$$\int_{t_1}^{t_2} a \, dt = v(t_2) - v(t_1) = \Delta v$$

The area under the acceleration-time graph indeed gives the change in velocity (since integrating acceleration over time recovers the velocity change). So Statement II is true.

Hence, Statement I is incorrect but Statement II is true. Hence, the correct answer is Option 4.

Key idea:

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

Graph A:-

• Slope increases with time → velocity increases with time
• That means accelerated motion

Matches II (v increases linearly with time)

So:
A → II

Graph B:-

• Initially slope is large (negative direction), then becomes less steep
• So velocity starts negative and approaches zero
• Velocity increasing towards zero (but still negative initially)

Matches IV (velocity increasing from negative towards zero)

So:
B → IV

Graph C:-

• First slope positive → velocity positive
• Then slope negative → velocity negative
• Sudden change → velocity jumps from +v to −v

Matches III (step from +v to −v)

So:
C → III

Graph D:-

• Constant slope → constant velocity

Matches I (constant velocity)

So:
D → I

Final answer:

A→II,B→IV,C→III,D→I

Given: $$x = 4t^2$$

Velocity: $$v = \frac{dx}{dt} = 8t$$

At $$t = 5$$ s:

$$v = 8 \times 5 = 40$$ m/s

The correct answer is Option 1: 40 m s⁻¹.

Given: $$s = 2.5t^2$$

Instantaneous speed is the derivative of distance with respect to time:

$$v = \frac{ds}{dt} = 2 \times 2.5t = 5t$$

At $$t = 5$$ s:

$$v = 5 \times 5 = 25 \text{ m/s}$$

The instantaneous speed is $$\mathbf{25}$$ m/s.

At the highest point of projectile motion, the vertical component of velocity is zero, so the speed equals the horizontal component:

$$ v_{\text{top}} = u\cos\theta $$

Given that $$v_{\text{top}} = \dfrac{\sqrt{3}}{2}u$$:

$$ u\cos\theta = \dfrac{\sqrt{3}}{2}u $$

$$ \cos\theta = \dfrac{\sqrt{3}}{2} $$

$$ \theta = 30° $$

The time of flight of a projectile is:

$$ T = \dfrac{2u\sin\theta}{g} = \dfrac{2u\sin 30°}{g} = \dfrac{2u \times \frac{1}{2}}{g} = \dfrac{u}{g} $$

We know that the maximum vertical height a man can throw a ball is 136 m, and our goal is to determine the maximum horizontal distance.

First, by finding the initial speed from the maximum height, we note that the height is reached when the ball is thrown vertically upward ($$\theta = 90°$$). Using the standard relation for maximum height, we have:

$$ h_{max} = \frac{u^2}{2g} $$

Solving for $$u^2$$ gives $$u^2 = 2g \times h_{max} = 2 \times 10 \times 136 = 2720\;\text{m}^2/\text{s}^2$$, so the square of the initial speed is 2720.

Next, the maximum horizontal range on flat ground occurs at a launch angle of $$\theta = 45°$$, yielding:

$$ R_{max} = \frac{u^2 \sin(2 \times 45°)}{g} = \frac{u^2 \sin 90°}{g} = \frac{u^2}{g} $$

Substituting $$u^2 = 2720$$ and $$g = 10\;\text{m}/\text{s}^2$$ gives:

$$ R_{max} = \frac{2720}{10} = 272\;\text{m} $$

We also note the useful general result that $$R_{max} = 2h_{max}$$. Therefore, the maximum horizontal distance is 272 m, which corresponds to Option 3: 272 m.

Let the maximum speed with which the man can release the ball be $$u$$.

Case 1: Ball thrown vertically upward.
Maximum height reached is given by the kinematic formula

$$h_{\max}= \frac{u^{2}}{2g}$$

The question states that $$h_{\max}=136\; \text{m}$$. Hence

$$\frac{u^{2}}{2g}=136$$

$$\Rightarrow u^{2}=2g \times 136 = 272\,g \qquad -(1)$$

Case 2: Ball thrown for maximum horizontal range.
For a given speed $$u$$, the horizontal range $$R$$ of a projectile is

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

The range becomes maximum when $$\theta = 45^{\circ}$$ because $$\sin 90^{\circ}=1$$. Therefore, the maximum range is

$$R_{\max}= \frac{u^{2}}{g} \qquad -(2)$$

Substitute $$u^{2}$$ from equation $$(1)$$ into $$(2)$$:

$$R_{\max}= \frac{272\,g}{g}=272\;\text{m}$$

Thus the farthest horizontal distance the man can throw the ball is $$272\;\text{m}$$.

Hence, Option C is correct.

look at the graph slowly and interpret each thing correctly

horizontal axis → time
vertical axis → position (distance from school)

both students start from school (origin)

now key ideas
slope of line = speed
final height = how far their home is
final time coordinate = time taken

1. time taken

both lines end at the same time coordinate on x-axis
so both students reach home at the same time

so A does NOT take lesser time
(C) is wrong

2.who is faster

compare slopes

line B is steeper than A
steeper line → greater slope → higher speed

so B is faster
(E) is correct
(D) is wrong

3. who lives closer

compare final heights (end points on y-axis)

B’s endpoint is higher → B has travelled more distance
so B lives farther

A’s endpoint is lower → A lives closer

so (A) is correct
(B) is wrong

final answer: A and E

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

$$\frac{H_{30}}{H_{60}} = \frac{\sin^2 30°}{\sin^2 60°} = \frac{1/4}{3/4} = \frac{1}{3}$$

The correct answer is Option 4: 1:3.

The velocity field is $$\vec v = \alpha\bigl(y\,\hat x + 2x\,\hat y\bigr)$$.
Hence

$$v_x = \alpha y , \qquad v_y = 2\alpha x$$

Acceleration is the material (total) derivative of velocity:

$$\vec a = \frac{d\vec v}{dt} = v_x\frac{\partial\vec v}{\partial x} + v_y\frac{\partial\vec v}{\partial y}$$

Component $$a_x$$:

$$a_x = v_x\frac{\partial v_x}{\partial x} + v_y\frac{\partial v_x}{\partial y}$$

Since $$v_x = \alpha y$$, we have $$\frac{\partial v_x}{\partial x}=0$$ and $$\frac{\partial v_x}{\partial y}=\alpha$$.

Therefore $$a_x = v_x(0) + v_y(\alpha) = (2\alpha x)(\alpha) = 2\alpha^{2}x$$.

Component $$a_y$$:

$$a_y = v_x\frac{\partial v_y}{\partial x} + v_y\frac{\partial v_y}{\partial y}$$

Because $$v_y = 2\alpha x$$, we have $$\frac{\partial v_y}{\partial x}=2\alpha$$ and $$\frac{\partial v_y}{\partial y}=0$$.

Thus $$a_y = v_x(2\alpha) + v_y(0) = (\alpha y)(2\alpha) = 2\alpha^{2}y$$.

Combining the two components:

$$\vec a = 2\alpha^{2}\bigl(x\,\hat x + y\,\hat y\bigr)$$

Using Newton’s second law $$\vec F = m\vec a$$:

$$\vec F = 2m\alpha^{2}\bigl(x\,\hat x + y\,\hat y\bigr)$$

Therefore, the correct option is:
Option A which is: $$\vec{F} = 2m\alpha^{2}(x\hat{x} + y\hat{y})$$.

When the ball is dropped from a height of $$9.8$$ m, its velocity just before hitting the floor is found using $$v^2 = 2gh$$: $$v = \sqrt{2 \times 10 \times 9.8} = \sqrt{196} = 14$$ m/s (downward).

After rebounding to a height of $$5.0$$ m, the velocity just after leaving the floor is $$v' = \sqrt{2 \times 10 \times 5} = \sqrt{100} = 10$$ m/s (upward).

Taking upward as positive, the velocity changes from $$-14$$ m/s (downward, just before contact) to $$+10$$ m/s (upward, just after contact). The average acceleration during the $$0.2$$ s contact time is $$a = \dfrac{v' - v}{\Delta t} = \dfrac{10 - (-14)}{0.2} = \dfrac{24}{0.2} = \boxed{120}$$ m s$$^{-2}$$.

A body is dropped from height $$h_1$$ and rebounds to height $$h_2$$. The ratio of velocities just before and after hitting the ground is $$4$$.

The velocity just before hitting the ground is $$v_1 = \sqrt{2gh_1}$$ and the velocity just after rebounding is $$v_2 = \sqrt{2gh_2}$$, satisfying $$\dfrac{v_1}{v_2} = 4$$.

Squaring gives $$\dfrac{v_1^2}{v_2^2} = 16$$, so the ratio of kinetic energies after and before impact is $$\dfrac{KE_{\text{after}}}{KE_{\text{before}}} = \dfrac{v_2^2}{v_1^2} = \dfrac{1}{16}$$.

The percentage loss in kinetic energy is $$\% \text{ loss} = \left(1 - \dfrac{v_2^2}{v_1^2}\right)\times 100 = \left(1 - \dfrac{1}{16}\right)\times 100 = \dfrac{15}{16}\times 100 = 93.75\%$$.

Since $$\dfrac{x}{4} = 93.75$$, it follows that $$x = 93.75 \times 4 = 375$$.

The value of $$x$$ is $$\boxed{375}$$.

Solution :

Given :

Radius of circular path,

$$r = 600\text{ m}$$

Initial speed,

$$u = 54\text{ km h}^{-1}$$

$$= \frac{54 \times 1000}{3600}$$

$$= 15\text{ m s}^{-1}$$

Given tangential acceleration equals centripetal acceleration :

$$a_t = a_c$$

$$\frac{dv}{dt} = \frac{v^2}{r}$$

Separating variables :

$$\frac{dv}{v^2} = \frac{dt}{r}$$

Integrating :

$$\int_{u}^{v}\frac{dv}{v^2} = \int_{0}^{t}\frac{dt}{r}$$

$$\left[-\frac{1}{v}\right]_{u}^{v} = \frac{t}{r}$$

$$\frac{1}{u} - \frac{1}{v} = \frac{t}{r}$$

Hence,

$$v = \frac{ur}{r-ut}$$

Angular speed :

$$\omega = \frac{v}{r}$$

Angular displacement for quarter revolution :

$$\theta = \frac{\pi}{2}$$

Therefore,

$$\int_0^t \omega dt = \frac{\pi}{2}$$

$$\int_0^t \frac{u}{r-ut}dt = \frac{\pi}{2}$$

$$\left[-\ln(r-ut)\right]_0^t = \frac{\pi}{2}$$

$$\ln\left(\frac{r}{r-ut}\right) = \frac{\pi}{2}$$

$$\frac{r}{r-ut} = e^{\pi/2}$$

$$r-ut = re^{-\pi/2}$$

$$ut = r(1-e^{-\pi/2})$$

$$t = \frac{r}{u}(1-e^{-\pi/2})$$

Substituting values :

$$t = \frac{600}{15}(1-e^{-\pi/2})$$

$$t = 40(1-e^{-\pi/2})$$

Comparing with given form :

$$t(1-e^{-\pi/2})$$

Hence,

$$t = 40$$

Final Answer :

$$40$$

A horse rider covers half the distance at 5 m/s, then the remaining half distance with speeds 10 and 15 m/s for equal time intervals.

Set up variables.

Let the total distance = $$2d$$.

Time for first half.

$$t_1 = \frac{d}{5}$$

Time for second half.

Let the time for each speed be $$t_2$$. Distance in second half:

$$d = 10t_2 + 15t_2 = 25t_2 \implies t_2 = \frac{d}{25}$$

Total time for second half = $$2t_2 = \frac{2d}{25}$$

Calculate mean speed.

Total time = $$\frac{d}{5} + \frac{2d}{25} = \frac{5d + 2d}{25} = \frac{7d}{25}$$

Mean speed = $$\frac{2d}{\frac{7d}{25}} = \frac{50}{7}$$ m/s

Therefore, $$x = \boxed{50}$$.

For a projectile, if it is at the same height at two different times $$t_1$$ and $$t_2$$, then by symmetry of the trajectory, these times are symmetric about the time of maximum height.

Time of maximum height = $$\frac{t_1 + t_2}{2} = \frac{3 + 5}{2} = 4$$ s

The total time of flight is:

$$T = 2 \times 4 = 8$$ s

Using the time of flight formula:

$$T = \frac{2u\sin\theta}{g}$$

$$8 = \frac{2u\sin 30°}{10} = \frac{2u \times 0.5}{10} = \frac{u}{10}$$

$$u = 80 \text{ m/s}$$

The speed of projection is $$80$$ m s$$^{-1}$$.

Scenario 1: Full stopping distance

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

Final velocity $$v_1 = 0 \text{ m/s}$$

Displacement $$s_1 = 500 \text{ m}$$

$$v_1^2 - u^2 = 2as_1$$

$$0^2 - (20)^2 = 2a(500)$$

$$-400 = 1000a$$

$$a = -0.4 \text{ m/s}^2$$

Scenario 2: Half the stopping distance

The brakes are now applied at half the distance, meaning the train travels $$s_2 = 250 \text{ m}$$ while decelerating. We want to find its speed $$v_2$$ as it crosses the station.

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

Acceleration $$a = -0.4 \text{ m/s}^2$$

Displacement $$s_2 = 250 \text{ m}$$

$$v_2^2 - u^2 = 2as_2$$

$$v_2^2 - (20)^2 = 2(-0.4)(250)$$

$$v_2^2 - 400 = -0.8(250)$$

$$v_2^2 - 400 = -200$$

$$v_2^2 = 200$$

$$v_2 = \sqrt{200} \text{ m/s}$$

We are given $$\vec{P} = 3\hat{i} + \sqrt{3}\hat{j} + 2\hat{k}$$ and $$\vec{Q} = 4\hat{i} + \sqrt{3}\hat{j} + 2.5\hat{k}$$.

Compute $$\vec{P} \times \vec{Q}$$.

$$ \vec{P} \times \vec{Q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & \sqrt{3} & 2 \\ 4 & \sqrt{3} & 2.5 \end{vmatrix} $$

$$ = \hat{i}(\sqrt{3} \cdot 2.5 - 2 \cdot \sqrt{3}) - \hat{j}(3 \cdot 2.5 - 2 \cdot 4) + \hat{k}(3\sqrt{3} - 4\sqrt{3}) $$

$$ = \hat{i}(2.5\sqrt{3} - 2\sqrt{3}) - \hat{j}(7.5 - 8) + \hat{k}(-\sqrt{3}) $$

$$ = \frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j} - \sqrt{3}\hat{k} $$

Find the magnitude.

$$ |\vec{P} \times \vec{Q}| = \sqrt{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2 + (-\sqrt{3})^2} = \sqrt{\frac{3}{4} + \frac{1}{4} + 3} = \sqrt{4} = 2 $$

Find the unit vector.

$$ \hat{n} = \frac{\vec{P} \times \vec{Q}}{|\vec{P} \times \vec{Q}|} = \frac{1}{2}\left(\frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j} - \sqrt{3}\hat{k}\right) = \frac{\sqrt{3}}{4}\hat{i} + \frac{1}{4}\hat{j} - \frac{\sqrt{3}}{2}\hat{k} $$

Compare with the given form.

The unit vector is given as $$\frac{1}{x}(\sqrt{3}\hat{i} + \hat{j} - 2\sqrt{3}\hat{k})$$.

$$ \frac{1}{x}(\sqrt{3}\hat{i} + \hat{j} - 2\sqrt{3}\hat{k}) = \frac{\sqrt{3}}{4}\hat{i} + \frac{1}{4}\hat{j} - \frac{\sqrt{3}}{2}\hat{k} $$

Comparing the $$\hat{j}$$ components: $$\frac{1}{x} = \frac{1}{4}$$, so $$x = 4$$.

Verification with $$\hat{i}$$ component: $$\frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{4}$$

Verification with $$\hat{k}$$ component: $$\frac{-2\sqrt{3}}{4} = -\frac{\sqrt{3}}{2}$$

The value of $$x$$ is $$\boxed{4}$$.

The swimmer swims perpendicular to the river flow with speed $$v_s = 4$$ km/h. The river has width $$d = 1$$ km.

To begin, time to cross the river,

$$ t = \frac{d}{v_s} = \frac{1}{4} \text{ h} $$

During this time, the river carries the swimmer downstream by 750 m = 0.75 km.

Next, speed of river,

$$ v_r = \frac{\text{drift}}{t} = \frac{0.75}{1/4} = 0.75 \times 4 = 3 \text{ km h}^{-1} $$

Two projectiles launched with the same speed $$u = 40$$ m/s have the same range. One is at $$60°$$.

For same range with same speed, the angles are complementary: $$\theta_2 = 90° - 60° = 30°$$.

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

For $$\theta_1 = 60°$$:

$$H_1 = \frac{40^2 \times \sin^2 60°}{2 \times 10} = \frac{1600 \times \frac{3}{4}}{20} = \frac{1200}{20} = 60 \text{ m}$$

For $$\theta_2 = 30°$$:

$$H_2 = \frac{40^2 \times \sin^2 30°}{2 \times 10} = \frac{1600 \times \frac{1}{4}}{20} = \frac{400}{20} = 20 \text{ m}$$

$$H_1 + H_2 = 60 + 20 = 80 \text{ m}$$

The answer is $$\boxed{80}$$ m.

A ball is dropped from height $$h = 20$$ m. Coefficient of restitution $$e = 0.5$$.

Velocity just before hitting the floor.

Using $$v^2 = u^2 + 2gh$$ with $$u = 0$$:

$$

v_{\text{before}} = \sqrt{2gh} = \sqrt{2 \times 10 \times 20} = 20 \text{ m/s}

$$

Velocity just after rebound.

By definition of coefficient of restitution:

$$

v_{\text{after}} = e \times v_{\text{before}} = 0.5 \times 20 = 10 \text{ m/s}

$$

Maximum rebound height.

Using $$v^2 = u^2 - 2gH$$ with $$v = 0$$:

$$

H = \frac{v_{\text{after}}^2}{2g} = \frac{100}{20} = 5 \text{ m}

$$

Alternatively: $$H = e^2 \times h = (0.5)^2 \times 20 = 0.25 \times 20 = 5$$ m.

The answer is $$\boxed{5}$$ m.

We have a particle of mass $$m = 10$$ g $$= 10^{-2}$$ kg moving in a straight line with retardation $$2x$$, where $$x$$ is the displacement in metres. Since it is retardation, the acceleration is $$a = -2x$$ m/s$$^2$$.

The retarding force is $$F = ma = -2mx$$. Using the work-energy theorem, the work done by this force over a displacement from 0 to $$x$$ is:

$$W = \int_0^x (-2mx) \, dx = -2m \left[\frac{x^2}{2}\right]_0^x = -mx^2$$

Since $$W = KE_f - KE_i$$, the loss of kinetic energy is:

$$\text{Loss of KE} = -W = mx^2 = 10^{-2} \times x^2 \text{ J}$$

We can verify this using kinematics. Since $$a = v\dfrac{dv}{dx} = -2x$$, integrating both sides:

$$\int_{v_0}^{v} v \, dv = \int_0^x (-2x) \, dx$$

$$\frac{v^2 - v_0^2}{2} = -x^2$$

or $$v_0^2 - v^2 = 2x^2$$

So the loss of kinetic energy is $$\frac{1}{2}m(v_0^2 - v^2) = \frac{1}{2} \times 10^{-2} \times 2x^2 = 10^{-2} \times x^2$$ J, which confirms our result.

Comparing with $$10^{-n} \cdot x^2$$, we get $$n = 2$$.

Hence, the answer is $$2$$.

A ball is thrown vertically upward with an initial speed of 150 m/s, and we are asked to determine the value of x such that the ratio of its velocities at t = 3 s and t = 5 s equals $$\frac{x+1}{x}$$.

Using the relation $$v = u - gt$$ with g = 10 m/s2 and upward taken as positive, we find the velocity at t = 3 s and t = 5 s as follows:

$$

v_3 = 150 - 10(3) = 150 - 30 = 120\;\text{m/s}

$$

$$

v_5 = 150 - 10(5) = 150 - 50 = 100\;\text{m/s}

$$

Since the required ratio is $$\frac{v_3}{v_5} = \frac{120}{100} = \frac{6}{5}$$, setting this equal to $$\frac{x+1}{x}$$ leads to

$$

\frac{x+1}{x} = \frac{6}{5}

$$

$$

5(x+1) = 6x \implies 5x + 5 = 6x \implies x = 5

$$

Hence, the correct answer is Option 4: 5.

We have a stone thrown horizontally from a cliff of height $$h = 10$$ m with initial horizontal speed $$v_x = 5$$ m/s.

The horizontal velocity remains constant throughout the motion, so $$v_x = 5$$ m/s. For the vertical motion (starting from rest), we use the kinematic equation

$$v_y^2 = u_y^2 + 2gh = 0 + 2 \times 10 \times 10 = 200$$

Now the speed with which the stone hits the ground is

$$v = \sqrt{v_x^2 + v_y^2} = \sqrt{25 + 200} = \sqrt{225} = 15 \text{ m/s}$$

Hence, the correct answer is 15 m/s.

Train A moves at 90 km/h and train B moves at 54 km/h in the opposite direction.

Relative velocity (since moving in opposite directions):

$$v_{rel} = 90 + 54 = 144 \text{ km/h} = 144 \times \frac{5}{18} = 40 \text{ m/s}$$

The passenger in train A observes train B passing for 8 seconds. The length of train B equals:

$$L_B = v_{rel} \times t = 40 \times 8 = 320 \text{ m}$$

The length of train B is $$\mathbf{320}$$ m.

Given a projection angle of 30° and an initial velocity of 40 m/s, after 2 s the horizontal component of velocity remains constant, so $$v_x = u\cos 30° = 40 \times \frac{\sqrt{3}}{2} = 20\sqrt{3} \text{ m/s}$$.

Next, the vertical component of velocity at t = 2 s is $$v_y = u\sin 30° - gt = 40 \times \frac{1}{2} - 10 \times 2 = 20 - 20 = 0 \text{ m/s}$$.

Substituting these into the expression for resultant velocity gives $$v = \sqrt{v_x^2 + v_y^2} = \sqrt{(20\sqrt{3})^2 + 0^2} = 20\sqrt{3} \text{ m/s}$$. At t = 2 s, the projectile is at its maximum height (vertical velocity = 0), so the resultant velocity equals the horizontal component.

The range of a projectile is given by $$R = \frac{u^2 \sin 2\theta}{g}$$. For maximum range, we need $$\sin 2\theta$$ to be maximum, i.e., $$\sin 2\theta = 1$$, which gives $$2\theta = 90°$$ or $$\theta = 45°$$. Thus Assertion A is true: at $$\theta = 45°$$, the range is indeed maximum.

Reason R states that for maximum range, $$\sin 2\theta$$ should equal one, which is exactly the condition derived above. This is also true, and it directly explains why $$\theta = 45°$$ gives maximum range.

Since both A and R are correct and R is the correct explanation of A, the answer is Option C.

$$x = 5t^2 - 4t + 5$$

Velocity: $$v = \frac{dx}{dt} = 10t - 4$$

At $$t = 2$$ s: $$v = 10(2) - 4 = 20 - 4 = 16$$ m/s

The magnitude of velocity is $$\mathbf{16}$$ m/s.

The range of a projectile is given by: $$R = \frac{u^2 \sin 2\theta}{g}$$

At $$\theta = 15°$$: $$R_1 = \frac{u^2 \sin 30°}{g} = \frac{u^2}{2g} = 50$$ m

At $$\theta = 45°$$: $$R_2 = \frac{u^2 \sin 90°}{g} = \frac{u^2}{g}$$

From the first equation: $$\frac{u^2}{g} = 100$$ m

Therefore $$R_2 = 100$$ m.

The correct answer is Option 1: 100 m.

We have the trajectory of the projectile given by

$$y = x - \frac{x^2}{20}$$

To find the maximum height, we set $$\frac{dy}{dx} = 0$$:

$$\frac{dy}{dx} = 1 - \frac{2x}{20} = 1 - \frac{x}{10} = 0$$

Solving gives $$x = 10$$ m. Now substituting $$x = 10$$ back into the trajectory equation:

$$y_{max} = 10 - \frac{(10)^2}{20} = 10 - \frac{100}{20} = 10 - 5 = 5 \text{ m}$$

So the maximum height attained by the projectile is 5 m. Hence, the correct answer is Option 3.

look at the v-t graph as areas of rectangles

area above time axis → positive displacement
area below time axis → negative displacement

now take each interval one by one

from 0 to 2 s
velocity = +8 m/s (constant)
so displacement = area = 8 × 2 = 16

from 2 to 4 s
velocity = −4 m/s
so displacement = −4 × 2 = −8

from 4 to 8 s
velocity = +4 m/s
so displacement = 4 × 4 = 16

from 8 to 10 s
velocity = −4 m/s
so displacement = −4 × 2 = −8

now combine properly

net displacement = 16 − 8 + 16 − 8 = 16

distance is different → here we don’t cancel signs
we take total path length

distance = 16 + 8 + 16 + 8 = 48

so finally

ratio = displacement / distance = 16 / 48 = 1/3

idea to remember
displacement = signed area
distance = total area (ignore sign)

The horizontal range of a projectile is:

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

For the first object (angle $$\alpha$$): $$R_1 = \frac{u^2 \sin 2\alpha}{g}$$

For the second object (angle $$\beta$$): $$R_2 = \frac{u^2 \sin 2\beta}{g}$$

Since $$\alpha + \beta = 90°$$, we have $$\beta = 90° - \alpha$$.

$$\sin 2\beta = \sin 2(90° - \alpha) = \sin(180° - 2\alpha) = \sin 2\alpha$$

Therefore $$R_1 = R_2$$, and the ratio is $$1 : 1$$.

The correct answer is Option 4: 1 : 1.

The range of a projectile is: $$R = \frac{u^2 \sin 2\theta}{g}$$

For projectile A: $$u_A = 40$$ m/s, $$\theta_A = 30°$$

$$ R_A = \frac{40^2 \sin 60°}{10} = \frac{1600 \times \frac{\sqrt{3}}{2}}{10} = 80\sqrt{3}\;\text{m} $$

For projectile B: $$u_B = 60$$ m/s, $$\theta_B = 60°$$

$$ R_B = \frac{60^2 \sin 120°}{10} = \frac{3600 \times \frac{\sqrt{3}}{2}}{10} = 180\sqrt{3}\;\text{m} $$

The ratio of ranges:

$$ \frac{R_A}{R_B} = \frac{80\sqrt{3}}{180\sqrt{3}} = \frac{80}{180} = \frac{4}{9} $$

The correct answer is 4 : 9.

We need to identify the correct statement about a body projected at an angle with the horizontal from the ground.

Analyzing each option:

Option A: Gravitational potential energy is maximum at the highest point.

At the highest point, the body reaches its maximum height. Since gravitational potential energy $$U = mgh$$, and $$h$$ is maximum at the highest point, the GPE is indeed maximum there. This statement is correct.

Option B: The horizontal component of velocity is zero at the highest point.

In projectile motion, the horizontal component of velocity remains constant throughout the flight ($$v_x = u\cos\theta$$). It is never zero (unless the projection angle is 90°, which is not a general case). This statement is incorrect.

Option C: The vertical component of momentum is maximum at the highest point.

At the highest point, the vertical component of velocity is zero, so the vertical component of momentum $$p_y = mv_y = 0$$, which is its minimum magnitude, not maximum. This statement is incorrect.

Option D: The kinetic energy is zero at the highest point.

At the highest point, the body still has horizontal velocity $$v_x = u\cos\theta$$, so $$KE = \frac{1}{2}m(u\cos\theta)^2 \neq 0$$. This statement is incorrect.

The answer is Option A: Gravitational potential energy is maximum at the highest point.

A toy starts from rest under constant acceleration and travels 10 m in $$t$$ seconds. We need to find the distance in the next $$t$$ seconds.

Using $$s = ut + \frac{1}{2}at^2$$ with $$u = 0$$:

$$10 = \frac{1}{2}at^2 \quad \ldots (1)$$

$$s_{2t} = \frac{1}{2}a(2t)^2 = \frac{1}{2}a \cdot 4t^2 = 4 \times \frac{1}{2}at^2 = 4 \times 10 = 40 \text{ m}$$

$$s_{\text{next}} = s_{2t} - s_t = 40 - 10 = 30 \text{ m}$$

Hence, the correct answer is Option C: 30 m.

We are given the positions of two buses:

$$x_P(t) = \alpha t + \beta t^2$$

$$x_Q(t) = ft - t^2$$

The velocity of each bus is the derivative of its position with respect to time.

Velocity of bus P:

$$v_P = \frac{dx_P}{dt} = \alpha + 2\beta t$$

Velocity of bus Q:

$$v_Q = \frac{dx_Q}{dt} = f - 2t$$

For same velocity, we set $$v_P = v_Q$$:

$$\alpha + 2\beta t = f - 2t$$

$$2\beta t + 2t = f - \alpha$$

$$t(2\beta + 2) = f - \alpha$$

$$t = \frac{f - \alpha}{2(\beta + 1)} = \frac{f - \alpha}{2(1 + \beta)}$$

Hence, the correct answer is Option D.

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 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 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: $$\vec{A} \times \vec{A} = 0$$.

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.

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 $$\boxed{100}$$.

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 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.