JEE Kinematics - 2D Motion Questions

consider forces on the car on a banked road

forces acting:

• weight = mg (vertical downward)
• normal reaction = N (perpendicular to surface)
• friction = f (along surface)

for maximum speed, the car tends to slip up the incline, so friction acts down the slope

resolve forces:

along horizontal (towards center → provides centripetal force)

N sinθ + f cosθ = m v₀² / r

along vertical (no vertical acceleration)

N cosθ − f sinθ = mg

at limiting condition:

f = μN

substitute f = μN

horizontal:

N sinθ + μN cosθ = m v₀² / r

N (sinθ + μ cosθ) = m v₀² / r …(1)

vertical:

N cosθ − μN sinθ = mg

N (cosθ − μ sinθ) = mg …(2)

divide (1) by (2):

(sinθ + μ cosθ) / (cosθ − μ sinθ) = (v₀² / rg)

cross multiply:

sinθ + μ cosθ = (v₀² / rg)(cosθ − μ sinθ)

expand:

sinθ + μ cosθ = (v₀² / rg)cosθ − (v₀² / rg)μ sinθ

collect μ terms:

μ cosθ + (v₀² / rg)μ sinθ = (v₀² / rg)cosθ − sinθ

factor μ:

μ [cosθ + (v₀² / rg) sinθ] = (v₀² / rg)cosθ − sinθ

final expression:

μ = [ (v₀² / rg)cosθ − sinθ ] / [ cosθ + (v₀² / rg) sinθ ]

$$ \mu=\frac{v_0^2-rg\tan\theta}{rg+v_0^2\tan\theta} $$

this is the required coefficient of friction for maximum speed.

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.

We need to find the ratio of maximum heights of two projectiles fired at angles $$(45° - \alpha)$$ and $$(45° + \alpha)$$ with the same initial speed. Since the maximum height of a projectile is given by $$H = \frac{u^2 \sin^2\theta}{2g}$$, we can proceed to express each height in this form.

Accordingly, the maximum heights of the two projectiles become $$H_1 = \frac{u^2 \sin^2(45° - \alpha)}{2g}$$ and $$H_2 = \frac{u^2 \sin^2(45° + \alpha)}{2g}$$.

Forming their ratio gives $$\frac{H_1}{H_2} = \frac{\sin^2(45° - \alpha)}{\sin^2(45° + \alpha)}$$.

Using the identity $$\sin(45° \mp \alpha) = \frac{\cos\alpha \mp \sin\alpha}{\sqrt{2}},$$ one obtains $$\sin^2(45° - \alpha) = \frac{(\cos\alpha - \sin\alpha)^2}{2} = \frac{1 - \sin 2\alpha}{2}$$ since $$(\cos\alpha - \sin\alpha)^2 = \cos^2\alpha - 2\sin\alpha\cos\alpha + \sin^2\alpha = 1 - \sin 2\alpha$$, and similarly $$\sin^2(45° + \alpha) = \frac{(\cos\alpha + \sin\alpha)^2}{2} = \frac{1 + \sin 2\alpha}{2}$$.

Substituting these results back into the ratio yields $$\frac{H_1}{H_2} = \frac{(1 - \sin 2\alpha)/2}{(1 + \sin 2\alpha)/2} = \frac{1 - \sin 2\alpha}{1 + \sin 2\alpha}$$.

The correct answer is Option (2): $$\frac{1 - \sin 2\alpha}{1 + \sin 2\alpha}$$.

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).

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.

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.

Boat speed downstream = 27+9 = 36 km/h = 10 m/s. Ball thrown at 10 m/s upward.

Time of flight = 2u/g = 2×10/10 = 2s. Range = horizontal speed × time = 10×2 = 20m = 2000 cm.

The answer is 2000.

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): $$0°$$.

We need to find the stopping potential when UV light of wavelength 300 nm falls on a metal with work function 2.13 eV.

Key Formula: The photoelectric equation (Einstein’s equation) is:

$$ E = W + eV_0 $$

Here $$E = \frac{hc}{\lambda}$$ is the energy of the incident photon, $$W$$ is the work function of the metal, and $$V_0$$ is the stopping potential. Rearranging this equation yields:

$$ eV_0 = \frac{hc}{\lambda} - W $$

The energy of the incident photon is calculated as:

$$ E = \frac{hc}{\lambda} = \frac{1240 \, \text{eV}\cdot\text{nm}}{300 \, \text{nm}} = 4.133 \, \text{eV} $$

Substituting into the photoelectric equation to determine the stopping potential gives:

$$ eV_0 = E - W = 4.133 - 2.13 = 2.003 \, \text{eV} $$

Since $$eV_0$$ is expressed in electronvolts, the stopping potential $$V_0$$ is approximately 2 V.

The correct answer is Option C: 2 V.

given:
radius of circle = 2 km
start point = P (top of circle)
end point = S (left side of circle)

step 1: what is displacement?

displacement = shortest straight-line distance between start and end
NOT the curved path along the circle

so we just need straight line PS

step 2: visualize positions

center = O

P is directly above O → distance 2
S is directly left of O → distance 2

so OP = 2 km (vertical)
OS = 2 km (horizontal)

and OP ⟂ OS (90° angle)

step 3: triangle OPS

you now have a right triangle:

OP = 2
OS = 2
PS = hypotenuse

apply Pythagoras:

PS² = OP² + OS²
PS² = 2² + 2²
PS² = 4 + 4 = 8

PS = √8 km

step 4: final understanding

even though cyclist moved along a curved path, displacement ignores path
it only depends on start and end points

so straight-line distance = √8 km

A particle of mass $$m$$ is projected with velocity $$u$$ at angle $$30°$$ with the horizontal. We need the angular momentum about the point of projection at maximum height.

Find velocity and position at maximum height: at maximum height, only the horizontal component of velocity remains:

$$v_x = u\cos 30° = \frac{u\sqrt{3}}{2}$$

The maximum height is:

$$h = \frac{u^2 \sin^2 30°}{2g} = \frac{u^2 \cdot \frac{1}{4}}{2g} = \frac{u^2}{8g}$$

Compute angular momentum: the angular momentum formula about the point of projection is $$L = m \cdot v_x \cdot h$$ (since the velocity is purely horizontal and the perpendicular distance from the projection point to the velocity line is $$h$$).

$$ L = m \cdot \frac{u\sqrt{3}}{2} \cdot \frac{u^2}{8g} = \frac{\sqrt{3}}{16} \cdot \frac{mu^3}{g} $$

The correct answer is Option (1): $$\frac{\sqrt{3}}{16} \frac{mu^3}{g}$$.

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).

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.

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

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

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

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

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: $$x = 2$$

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 $$\boxed{2}$$.

We need to find $$\alpha$$ where the time for the first revolution is $$\frac{1}{\alpha}(1 - e^{-2\pi})$$ s.

Considering radius $$R = 0.5$$ m and initial speed $$v_0 = 4$$ m/s, and using $$a_t = a_n$$, we have $$\frac{dv}{dt} = \frac{v^2}{R}\,. $$

Separating variables gives $$\frac{dv}{v^2} = \frac{dt}{R}\,,$$ and integrating yields $$-\frac{1}{v} = \frac{t}{R} + C\,. $$ At $$t = 0$$, $$C = -\frac{1}{v_0} = -\frac{1}{4}$$, so $$v = \frac{4}{1 - 8t}\,. $$

The distance for one revolution is $$s = 2\pi R = \pi$$ and also $$s = \int_0^T v\,dt = \int_0^T \frac{4}{1-8t}\,dt = -\frac{1}{2}\ln(1-8T)\,. $$ Setting this equal to $$\pi$$ gives $$-\frac{1}{2}\ln(1-8T) = \pi\,, $$ so $$1-8T = e^{-2\pi}$$ and $$T = \frac{1 - e^{-2\pi}}{8} = \frac{1}{8}(1 - e^{-2\pi})\,. $$

Comparing this result with the given expression $$\frac{1}{\alpha}(1-e^{-2\pi})$$ shows that $$\alpha = 8\,. $$

Option: 8

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 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 $$3$$.

given
mass = m
projection speed = u
angle = 45°

we need angular momentum about point of projection at highest point

formula:

L = m | r × v |

for 2D motion:

$$L=m(xv_y−yv_x)$$

step 1: velocity at highest point

at highest point
vertical velocity = 0
horizontal velocity =

$$u\cos45°=u/\sqrt{2}$$

so:

v_x = u/√2 , v_y = 0

step 2: time to reach highest point

$$t=(u\sin45°)/g$$

= (u/√2)/g

= u/(√2 g)

step 3: coordinates of highest point

horizontal distance:

$$x=(u\cos45°)\times t$$

= (u/√2) × (u/(√2 g))

= u²/(2g)

vertical height:

$$y=(u\sin45°)^2/(2g)$$

= (u²/2)/(2g)

= u²/(4g)

step 4: angular momentum

$$L=m(xv_y−yv_x)$$

$$\sin ce\ v_y=0:$$

$$L=−myv_x$$

magnitude:

$$L=myv_x$$

$$=m\times(u^2/(4g))\times(u/\sqrt{2})$$

$$=mu^3/(4\sqrt{2}g)$$

Now comparing with the given expression :-

$$\frac{\sqrt{2}mu^3}{Xg}$$

we get it by multiplying and dividing our answer by $$\sqrt{\ 2}$$

so we get $$=\sqrt{\ 2}mu^3/8g$$
so Value of X is 8

Second hand: in 30 min = 30 revolutions. Distance = 30×2π×0.75 = 45π m.

Minute hand: in 30 min = 0.5 revolution. Distance = 0.5×2π×0.60 = 0.6π m.

x = 45π - 0.6π = 44.4π = 44.4×3.14 = 139.4 m.

The correct answer is Option (3): 139.4.

120 rev in 3 min = 40 rev/min = 2/3 rev/s. ω=2π·2/3=4π/3 rad/s.

a=ω²r=(4π/3)²·9=16π²/9·9=16π² m/s².

The answer is Option (4): 16π² m/s².

In uniform circular motion, speed $$v$$ is given by dividing the circumference by the period:
$$v = \frac{2\pi R}{T}$$ $$-(1)$$

The maximum height $$H$$ reached by a projectile with initial speed $$u$$ at angle $$\theta$$ is given by the formula:
Formula: $$H = \frac{u^2 \sin^2 \theta}{2g}$$.

Setting $$u = v$$, we have:
$$H = \frac{v^2 \sin^2 \theta}{2g}$$ $$-(2)$$.

Given that this height equals $$4R$$, substitute (1) into (2):
$$4R = \frac{\Bigl(\frac{2\pi R}{T}\Bigr)^2 \sin^2 \theta}{2g}$$.

Simplify the square:
$$\Bigl(\frac{2\pi R}{T}\Bigr)^2 = \frac{4\pi^2 R^2}{T^2}$$. Thus
$$4R = \frac{4\pi^2 R^2 \sin^2 \theta}{2g T^2} = \frac{2\pi^2 R^2 \sin^2 \theta}{g T^2}$$.

Divide both sides by $$R$$ and rearrange:
$$4 = \frac{2\pi^2 R \sin^2 \theta}{g T^2} \quad\Longrightarrow\quad \sin^2 \theta = \frac{4gT^2}{2\pi^2 R} = \frac{2g T^2}{\pi^2 R}$$ $$-(3)$$.

Taking the positive square root (angle is acute):
$$\sin \theta = \sqrt{\frac{2g T^2}{\pi^2 R}}\quad\Longrightarrow\quad \theta = \sin^{-1}\!\sqrt{\frac{2g T^2}{\pi^2 R}}\!.$$

Therefore, the required angle of projection is
$$\theta = \sin^{-1}\Bigl(\frac{2gT^2}{\pi^2R}\Bigr)^{1/2},$$ which corresponds to Option A.

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 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 have a particle moving with constant speed $$v$$ in a circular path of radius $$r$$. When it turns by $$90°$$, we need to find the ratio of instantaneous velocity to average velocity.

The instantaneous speed at any point is $$v$$. For a $$90°$$ turn, the displacement is the chord subtending $$90°$$ at the centre. If the particle moves from one point to another $$90°$$ apart, the displacement is

$$|\vec{d}| = \sqrt{r^2 + r^2} = r\sqrt{2}$$

Now, the time taken for a quarter circle is

$$t = \frac{\text{arc length}}{v} = \frac{\frac{2\pi r}{4}}{v} = \frac{\pi r}{2v}$$

So the magnitude of average velocity becomes

$$v_{avg} = \frac{|\vec{d}|}{t} = \frac{r\sqrt{2}}{\frac{\pi r}{2v}} = \frac{2v\sqrt{2}}{\pi}$$

The ratio of instantaneous velocity to average velocity is then

$$\frac{v}{v_{avg}} = \frac{v}{\frac{2v\sqrt{2}}{\pi}} = \frac{\pi}{2\sqrt{2}}$$

This gives the ratio as $$\pi : 2\sqrt{2}$$. Comparing with $$\pi : x\sqrt{2}$$, we get $$x = 2$$. So, the answer is $$2$$.

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 object is in uniform circular motion with a radius of $$r=+2$$ meters (since it happens to be at $$x=+2$$ at an instant when the center of the motion is origin). At $$x=+2$$ meters, the velocity $$\vec{v}$$ is $$-4\hat{j}$$ m/s. This indicates that the object is moving clockwise.

Thus, at $$x=-2$$ meters, the velocity vector will be equal in magnitude but opposite in direction, and will therefore be $$\vec{v'} = 4\hat{j}$$. 

Furthermore, the magnitude of acceleration at $$x=-2$$ will be given by $$a= \dfrac{v^2}{r} = \dfrac{4^2}{2} = \dfrac{16}{2} = 8 m/s^2$$

And the direction at $$x=-2$$ will be towards the origin (centripetal force), i.e. along $$+\hat{i}$$, thus, we have, $$\vec{a}=8\hat{i}$$.

Option B is the correct answer.

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

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:

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

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.

Two forces of magnitudes $$A$$ and $$\frac{A}{2}$$ are perpendicular to each other.

The magnitude of the resultant when two forces are perpendicular is:

$$ R = \sqrt{F_1^2 + F_2^2} = \sqrt{A^2 + \frac{A^2}{4}} = \sqrt{\frac{5A^2}{4}} = \frac{\sqrt{5}A}{2} $$

The correct answer is $$\dfrac{\sqrt{5}A}{2}$$.

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.

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$$^{-1}$$.

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,

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

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.

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.

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.

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.

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: $$\sqrt{2} : 1$$.

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: $$4$$.

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.

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 the equations of motion: $$x = 4\sin\left(\frac{\pi}{2} - \omega t\right)$$ and $$y = 4\sin(\omega t)$$. Using the identity $$\sin\left(\frac{\pi}{2} - \theta\right) = \cos\theta$$, this simplifies to $$x = 4\cos(\omega t)$$, so we have $$x = 4\cos(\omega t)$$ and $$y = 4\sin(\omega t)$$.

Squaring and adding both equations gives:

$$x^2 + y^2 = 16\cos^2(\omega t) + 16\sin^2(\omega t)$$

$$x^2 + y^2 = 16(\cos^2(\omega t) + \sin^2(\omega t))$$

$$x^2 + y^2 = 16$$

This is the equation of a circle with radius 4 m centered at the origin. Therefore, the path of the particle is circular. The correct answer is Option A.

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 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, $$x = 1$$.

A body is projected at $$45°$$ with the horizontal. After $$2$$ s, its velocity is $$20$$ m/s, and we need to find the maximum height reached, with $$g = 10$$ m/s$$^2$$.

Let the initial speed be $$u$$; since $$\theta = 45°$$, the horizontal and vertical components of the initial velocity are

$$u_x = u\cos 45° = \frac{u}{\sqrt{2}}$$

$$u_y = u\sin 45° = \frac{u}{\sqrt{2}}$$

At time $$t = 2$$ s, the horizontal component remains unchanged while the vertical component becomes

$$v_x = \frac{u}{\sqrt{2}}$$

$$v_y = \frac{u}{\sqrt{2}} - gt = \frac{u}{\sqrt{2}} - 10 \times 2 = \frac{u}{\sqrt{2}} - 20$$

Since the resultant speed at $$t = 2$$ s satisfies

$$v^2 = v_x^2 + v_y^2 = 400$$

$$\left(\frac{u}{\sqrt{2}}\right)^2 + \left(\frac{u}{\sqrt{2}} - 20\right)^2 = 400$$

$$\frac{u^2}{2} + \frac{u^2}{2} - 2 \times 20 \times \frac{u}{\sqrt{2}} + 400 = 400$$

$$u^2 - \frac{40u}{\sqrt{2}} + 400 = 400$$

$$u^2 - 20\sqrt{2} \cdot u = 0$$

$$u(u - 20\sqrt{2}) = 0$$

Since $$u \neq 0$$, it follows that $$u = 20\sqrt{2}$$ m/s.

Now the maximum height is given by

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

$$H = \frac{(20\sqrt{2})^2 \times \sin^2 45°}{2 \times 10}$$

$$H = \frac{800 \times \frac{1}{2}}{20} = \frac{400}{20} = 20 \text{ m}$$

The maximum height reached by the body is $$20$$ m.

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.

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

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 $$\hat{j}$$.

A body of mass 10 kg is projected at $$45^\circ$$ with the horizontal, and its trajectory passes through the point (20, 10). We need to determine the momentum vector at time $$t = \frac{T}{\sqrt{2}}$$, where $$T$$ is the total time of flight.

The trajectory equation for projectile motion is $$y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}.$$ With $$\theta = 45^\circ$$, $$\tan45^\circ = 1$$ and $$\cos^245^\circ = \tfrac12$$, this becomes $$y = x - \frac{gx^2}{u^2}.$$ Substituting the point $$(20,10)$$ and $$g = 10$$ yields $$10 = 20 - \frac{10\times400}{u^2},$$ so $$\tfrac{4000}{u^2} = 10$$ and hence $$u^2 = 400$$, giving $$u = 20\text{ m/s}.$$ The horizontal and vertical components of the initial velocity are $$u_x = u\cos45^\circ = 20\times\frac{1}{\sqrt2} = 10\sqrt2\text{ m/s},\quad u_y = u\sin45^\circ = 10\sqrt2\text{ m/s}.$$ The total time of flight is $$T = \frac{2u\sin\theta}{g} = \frac{2\times20\times\frac{1}{\sqrt2}}{10} = 2\sqrt2\text{ s},$$ so $$t = \frac{T}{\sqrt2} = 2\text{ s}.$$

Because the horizontal velocity remains constant, $$v_x = u_x = 10\sqrt2\text{ m/s},$$ and the vertical velocity at $$t=2$$ s is $$v_y = u_y - gt = 10\sqrt2 - 10\times2 = 10\sqrt2 - 20\text{ m/s}.$$ Thus the momentum vector is $$p = mv = 10(10√2 i + (10√2 - 20) j) = 100√2 i + (100√2 - 200) j N·s$$

Answer: Option D: $$100\sqrt{2}\hat{i} + (100\sqrt{2} - 200)\hat{j}$$ N s

When the girl is standing still, she holds the umbrella at $$45°$$ with the vertical. This means the rain has equal horizontal and vertical velocity components.

Let the velocity of rain be $$\vec{v_r}$$ with horizontal component $$v_h$$ and vertical component $$v_v$$.

Since the umbrella is at $$45°$$ with vertical:

$$\tan 45° = \frac{v_h}{v_v} = 1$$

$$\therefore v_h = v_v$$

When the girl runs with speed $$15\sqrt{2}$$ km/h in the horizontal direction, the rain appears to fall vertically on her head. This means the horizontal component of rain velocity relative to the girl is zero.

$$v_h = 15\sqrt{2} \text{ km/h}$$

$$v_v = v_h = 15\sqrt{2} \text{ km/h}$$

The velocity of rain with respect to the moving girl has only a vertical component (since horizontal component cancels out):

$$v_{rain, girl} = v_v = 15\sqrt{2} \text{ km/h}$$

We can also write this as:

$$v_{rain, girl} = 15\sqrt{2} = \frac{30}{\sqrt{2}} \text{ km/h}$$

The correct answer is Option C.

We have a ball projected with kinetic energy $$E$$ at an angle of $$60^\circ$$ to the horizontal. At the highest point of its flight, the vertical component of velocity becomes zero, and only the horizontal component remains.

The horizontal component of velocity is $$v_x = v\cos 60^\circ = \frac{v}{2}$$, where $$v$$ is the initial speed. The initial kinetic energy is $$E = \frac{1}{2}mv^2$$.

At the highest point, the kinetic energy is $$E' = \frac{1}{2}mv_x^2 = \frac{1}{2}m\left(\frac{v}{2}\right)^2 = \frac{1}{2}m \cdot \frac{v^2}{4} = \frac{1}{4}\left(\frac{1}{2}mv^2\right) = \frac{E}{4}$$.

Hence, the correct answer is Option C.

For a particle in uniform circular motion of radius $$R$$, the position at angle $$\theta$$ from the positive x-axis is:

$$\vec{r} = R\cos\theta\,\hat{i} + R\sin\theta\,\hat{j}$$

In uniform circular motion, the acceleration is centripetal (directed toward the center), i.e., opposite to the position vector from the center:

$$\vec{a} = -\frac{v^2}{R}\hat{r}$$

where $$\hat{r}$$ is the unit vector along the radial direction (outward from center).

The unit radial vector at angle $$\theta$$ is:

$$\hat{r} = \cos\theta\,\hat{i} + \sin\theta\,\hat{j}$$

Therefore:

$$\vec{a} = -\frac{v^2}{R}(\cos\theta\,\hat{i} + \sin\theta\,\hat{j})$$

$$\vec{a} = -\frac{v^2}{R}\cos\theta\,\hat{i} - \frac{v^2}{R}\sin\theta\,\hat{j}$$

Hence, the correct answer is Option C.

Analysis of the Assertion:

The assertion states that for four points $$A,\ B, \ C, \ D$$ on a semi-circular arc with center $$O$$ such that $$|\vec{AB}|=|\vec{BC}|=|\vec{CD}|$$:

$$\vec{AB}+\vec{AC}+\vec{AD} = 4\vec{AO} + \vec{OB} + \vec{OC}$$

We can rewrite the vectors on the left side in terms of the origin (center) $$O$$:

1. $$\vec{AB} = \vec{OB} - \vec{OA}$$
2. $$\vec{AC} = \vec{OC} - \vec{OA}$$
3. $$\vec{AD} = \vec{OD} - \vec{OA}$$

Summing these gives,

$$\vec{AB}+\vec{AC}+\vec{AD} = (\vec{OB}+\vec{OC}+\vec{OD}) - 3\vec{OA}$$

We find that, since $$\vec{OA}$$ and $$\vec{OD}$$ are opposite radii originating from $$O$$, $$\vec{OD} = -\vec{OA} = \vec{AO}$$, since $$\vec{OA} = -\vec{AO}$$ (same length but opposite direction), we therefore get,

$$\vec{AB}+\vec{AC}+\vec{AD} = (\vec{OB}+\vec{OC}+\vec{AO}) + 3\vec{AO}$$

or,

$$\vec{AB}+\vec{AC}+\vec{AD} = 4\vec{AO} + \vec{OB} + \vec{OC}$$

Assertion A is true.

Analysis of the Reason:

By the polygon law of vector addition, $$\vec{AB}+\vec{BC}+\vec{CD} = \vec{AD}$$, and we also see that $$\vec{AD} = 2\vec{AO}$$, since $$\vec{AD}$$ is the diameter, and $$\vec{AO}$$ is the radius.

Thus, the reason R is correct.

But the polygon law of vector addition has not been useful in examining the assertion A while we derived equality in assertion A. Thus, although both the assertion A and the reason R are correct, R is not the correct explanation of A.

Let the magnitude of each vector be $$P$$, and let $$\theta$$ be the angle between $$\vec{P}$$ and $$\vec{Q}$$.

The magnitude of $$\vec{P} + \vec{Q}$$ is $$|\vec{P} + \vec{Q}| = \sqrt{P^2 + Q^2 + 2PQ\cos\theta}$$. Since $$P = Q$$, this becomes $$\sqrt{2P^2 + 2P^2\cos\theta} = P\sqrt{2(1+\cos\theta)}$$.

The magnitude of $$\vec{P} - \vec{Q}$$ is $$|\vec{P} - \vec{Q}| = \sqrt{P^2 + Q^2 - 2PQ\cos\theta} = P\sqrt{2(1-\cos\theta)}$$.

Setting $$|\vec{P} + \vec{Q}| = n|\vec{P} - \vec{Q}|$$ gives $$P\sqrt{2(1+\cos\theta)} = nP\sqrt{2(1-\cos\theta)}$$, so $$1+\cos\theta = n^2(1-\cos\theta)$$.

Solving: $$1 + \cos\theta = n^2 - n^2\cos\theta$$, hence $$\cos\theta(1 + n^2) = n^2 - 1$$, giving $$\cos\theta = \dfrac{n^2 - 1}{n^2 + 1}$$.

Therefore the angle between the two vectors is $$\theta = \cos^{-1}\!\left(\dfrac{n^2-1}{n^2+1}\right)$$.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Therefore,

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

Hence the angle is

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

Comparing with the given options, this matches Option B.

Answer: Option B

The vector projection of $$\vec{A}$$ onto $$\vec{B}$$ is given by the formula $$\text{proj}_{\vec{B}}\vec{A} = \dfrac{\vec{A} \cdot \vec{B}}{|\vec{B}|^2}\,\vec{B}$$.

First, compute the dot product $$\vec{A} \cdot \vec{B} = (1)(1) + (1)(1) + (1)(0) = 2$$.

Next, find the magnitude squared of $$\vec{B}$$: $$|\vec{B}|^2 = 1^2 + 1^2 = 2$$.

Therefore the projection is $$\dfrac{2}{2}(\hat{i} + \hat{j}) = \hat{i} + \hat{j}$$.

The butterfly flies in the north-east direction at $$4\sqrt{2}$$ m s$$^{-1}$$. Resolving this into components: the northward component is $$4\sqrt{2} \cdot \cos 45° = 4$$ m s$$^{-1}$$ and the eastward component is $$4\sqrt{2} \cdot \sin 45° = 4$$ m s$$^{-1}$$.

The wind blows from north to south at 1 m s$$^{-1}$$, adding $$-1$$ m s$$^{-1}$$ to the northward component. The resultant velocity of the butterfly is: eastward = 4 m s$$^{-1}$$, northward = $$4 - 1 = 3$$ m s$$^{-1}$$.

The magnitude of the resultant velocity is $$v = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$$ m s$$^{-1}$$.

The resultant displacement in 3 seconds is $$d = v \times t = 5 \times 3 = 15$$ m.

The velocity of the mosquito at any time $$t$$ is given by $$\vec{v} = 0.5t^2\hat{i} + 3t\hat{j} + 9\hat{k}$$ m/s.

At $$t = 2$$ s, the velocity components are: $$v_x = 0.5 \times 4 = 2$$ m/s, $$v_y = 3 \times 2 = 6$$ m/s, and $$v_z = 9$$ m/s.

The direction of the mosquito is along its velocity vector $$\vec{v} = 2\hat{i} + 6\hat{j} + 9\hat{k}$$. To find the angle this vector makes with the $$y$$-axis, we compute the components perpendicular and parallel to $$\hat{j}$$.

The component along $$y$$-axis is $$v_y = 6$$. The component perpendicular to the $$y$$-axis lies in the $$xz$$-plane and has magnitude $$\sqrt{v_x^2 + v_z^2} = \sqrt{4 + 81} = \sqrt{85}$$.

The angle from the $$y$$-axis is therefore $$\theta = \tan^{-1}\left(\frac{\sqrt{85}}{6}\right)$$.

Particle P moves along $$\vec{A} = \hat{i} + \hat{j}$$ and then along a direction normal to the plane containing $$\vec{A}$$ and $$\vec{B} = \hat{j} + \hat{k}$$. So the direction of P after the collision is $$\vec{A} \times \vec{B}$$.

$$\vec{A} \times \vec{B} = (\hat{i} + \hat{j}) \times (\hat{j} + \hat{k}) = \hat{i}\times\hat{j} + \hat{i}\times\hat{k} + \hat{j}\times\hat{j} + \hat{j}\times\hat{k}$$ $$= \hat{k} - \hat{j} + 0 + \hat{i} = \hat{i} - \hat{j} + \hat{k}$$

Particle Q moves along $$\vec{A} = \hat{i} + \hat{j}$$ and then along a direction normal to the plane containing $$\vec{A}$$ and $$\vec{C} = -\hat{i} + \hat{j}$$. So the direction of Q after the collision is $$\vec{A} \times \vec{C}$$.

$$\vec{A} \times \vec{C} = (\hat{i} + \hat{j}) \times (-\hat{i} + \hat{j}) = \hat{i}\times(-\hat{i}) + \hat{i}\times\hat{j} + \hat{j}\times(-\hat{i}) + \hat{j}\times\hat{j}$$ $$= 0 + \hat{k} + \hat{k} + 0 = 2\hat{k}$$

The angle $$\theta$$ between these two directions is found using the dot product formula:

$$\cos\theta = \dfrac{(\hat{i} - \hat{j} + \hat{k}) \cdot (2\hat{k})}{|\hat{i} - \hat{j} + \hat{k}| \times |2\hat{k}|}$$

The numerator is $$2$$, and the denominator is $$\sqrt{1^2 + (-1)^2 + 1^2} \times 2 = \sqrt{3} \times 2$$.

$$\cos\theta = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}$$

Comparing with $$\cos^{-1}\!\left(\dfrac{1}{\sqrt{x}}\right)$$, we get $$x = 3$$.

We are given a swimmer with speed $$v_s = 12$$ km/h in still water and a river flowing with speed $$v_r = 6$$ km/h. The swimmer wants to reach the point directly opposite on the other bank, so the net displacement must be purely perpendicular to the river flow.

Let the swimmer swim at an angle $$\theta$$ measured from the upstream direction of the river flow. The swimmer's velocity has two components: along the river, $$v_s \cos\theta$$ (upstream), and perpendicular to the river, $$v_s \sin\theta$$ (across). For the swimmer to reach the point directly opposite, the net velocity along the river must be zero. The river carries the swimmer downstream at $$v_r = 6$$ km/h, and the upstream component of the swimmer's velocity is $$v_s \cos\theta$$. Setting these equal:

$$v_s \cos\theta = v_r$$

$$12 \cos\theta = 6$$

$$\cos\theta = \frac{1}{2}$$

$$\theta = 60°$$

This angle $$\theta = 60°$$ is measured from the upstream direction. But the question asks for the angle with respect to the direction of flow of the river, which is the downstream direction. The angle from the downstream direction to the swimmer's direction (going upstream and across) is $$180° - 60° = 120°$$.

To verify: the perpendicular component of velocity is $$12 \sin 60° = 12 \times \frac{\sqrt{3}}{2} = 6\sqrt{3}$$ km/h (crossing the river), and the along-river component is $$12 \cos 60° = 6$$ km/h upstream, which exactly cancels the 6 km/h downstream river flow. The net velocity is purely perpendicular, confirming the swimmer reaches directly opposite.

The required angle with respect to the direction of flow is $$\boxed{120}$$°.

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

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

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

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

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

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

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

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

So, the answer is $$1$$.

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

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

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

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

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

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

Solving this for $$t$$ we get

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

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

$$x = v t.$$

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

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

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

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

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

We already have $$x^2$$:

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

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

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

This matches Option D.

Hence, the correct answer is Option D.

We start by recalling the standard results for a projectile launched with an initial speed $$u$$ at an angle $$\theta$$ to the horizontal.

Formula for horizontal range: $$R = \dfrac{u^{2}\sin 2\theta}{g}.$$

Formula for maximum height: $$H = \dfrac{u^{2}\sin^{2}\theta}{2g}.$$

Both projectiles are thrown with the same speed $$u$$, so $$u$$ and $$g$$ are common for the two cases. Let us denote the first angle by $$\theta_{1}=42^{\circ}$$ and the second by $$\theta_{2}=48^{\circ}.$$ Their corresponding ranges and heights are $$R_{1},\,H_{1}$$ and $$R_{2},\,H_{2}.$$

Evaluating the ranges

For the first projectile we have

$$R_{1}= \dfrac{u^{2}\sin 2\theta_{1}}{g} = \dfrac{u^{2}\sin(2\times 42^{\circ})}{g} = \dfrac{u^{2}\sin 84^{\circ}}{g}.$$

For the second projectile we obtain

$$R_{2}= \dfrac{u^{2}\sin 2\theta_{2}}{g} = \dfrac{u^{2}\sin(2\times 48^{\circ})}{g} = \dfrac{u^{2}\sin 96^{\circ}}{g}.$$

Now we use the trigonometric identity $$\sin(180^{\circ}-\alpha)=\sin\alpha.$$ Since $$96^{\circ}=180^{\circ}-84^{\circ},$$ we get $$\sin 96^{\circ}=\sin 84^{\circ}.$$ Substituting this equality above,

$$R_{2}= \dfrac{u^{2}\sin 96^{\circ}}{g}= \dfrac{u^{2}\sin 84^{\circ}}{g}=R_{1}.$$

Thus $$R_{1}=R_{2}.$$

Evaluating the heights

For the first projectile:

$$H_{1}=\dfrac{u^{2}\sin^{2}\theta_{1}}{2g} =\dfrac{u^{2}\sin^{2}42^{\circ}}{2g}.$$

For the second projectile:

$$H_{2}=\dfrac{u^{2}\sin^{2}\theta_{2}}{2g} =\dfrac{u^{2}\sin^{2}48^{\circ}}{2g}.$$

Because $$48^{\circ}>42^{\circ}$$ and $$\sin\theta$$ increases from $$0^{\circ}$$ to $$90^{\circ},$$ we have $$\sin 48^{\circ}>\sin 42^{\circ}.$$ Squaring preserves the inequality, so $$\sin^{2}48^{\circ}>\sin^{2}42^{\circ}.$$ Substituting this into the expressions for $$H_{1}$$ and $$H_{2}$$ gives

$$H_{2}>H_{1}.$$

Combining both comparisons, we find

$$R_{1}=R_{2}\quad\text{and}\quad H_{1}<H_{2}.$$

These results match Option B.

Hence, the correct answer is Option B.

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

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

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

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

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

These match Option A.

The trajectory of the projectile is given by $$y = \alpha x - \beta x^2$$.

The angle of projection $$\theta$$ is the angle the velocity makes with the horizontal at the point of projection $$(x = 0)$$. Since $$\tan\theta = \frac{dy}{dx}\bigg|_{x=0}$$, we differentiate: $$\frac{dy}{dx} = \alpha - 2\beta x$$. At $$x = 0$$, $$\frac{dy}{dx} = \alpha$$, so $$\tan\theta = \alpha$$, which gives $$\theta = \tan^{-1}\alpha$$.

The maximum height occurs where $$\frac{dy}{dx} = 0$$, i.e., $$\alpha - 2\beta x = 0$$, giving $$x = \frac{\alpha}{2\beta}$$.

Substituting this into the trajectory equation: $$H = \alpha \cdot \frac{\alpha}{2\beta} - \beta \cdot \left(\frac{\alpha}{2\beta}\right)^2 = \frac{\alpha^2}{2\beta} - \frac{\alpha^2}{4\beta} = \frac{\alpha^2}{4\beta}$$.

Therefore, the angle of projection is $$\tan^{-1}\alpha$$ and the maximum height is $$\frac{\alpha^2}{4\beta}$$.

Let us fix an $$x$$-$$y$$ coordinate system that, at the instant the bomb leaves the aircraft, has its origin at the bomb itself. The $$x$$-axis is chosen horizontally in the direction in which the plane is flying, while the $$y$$-axis is chosen vertically upward (opposite to the direction of gravity). The positive $$y$$ direction is therefore upward and the acceleration due to gravity acts downward, that is, along the negative $$y$$ direction with magnitude $$g$$.

First, we write the equations of motion of the bomb as seen by an observer on the ground (an inertial frame). Because the plane is flying horizontally with constant speed $$u$$, the bomb, at the moment of release, possesses the same horizontal speed $$u$$. Therefore, in the ground frame, the initial conditions are

$$x(0)=0, \qquad y(0)=0,$$

$$\dot x(0)=u, \qquad \dot y(0)=0,$$

and the only acceleration acting on the bomb is gravitational:

$$\ddot x = 0, \qquad \ddot y = -g.$$

Integrating these with respect to time $$t$$, we obtain the bomb’s position in the ground frame:

Horizontal motion:

$$\ddot x = 0 \;\Rightarrow\; \dot x = u \quad(\text{since }\dot x(0)=u),$$

Integrating once more,

$$x(t)=ut.$$

Vertical motion:

$$\ddot y = -g \;\Rightarrow\; \dot y = -gt \quad(\text{since }\dot y(0)=0),$$

Integrating again,

$$y(t) = -\tfrac{1}{2}gt^{2}.$$

Eliminating the time $$t$$ between $$x(t)$$ and $$y(t)$$, we get

$$t = \frac{x}{u}\quad\Longrightarrow\quad y = -\tfrac{1}{2}g\left(\frac{x}{u}\right)^{2},$$

or

$$y = -\frac{g}{2u^{2}}\,x^{2}.$$

This is the standard equation of a parabola opening downward, confirming that in the ground frame the bomb follows a parabolic path.

Now we must determine what the observer inside the plane sees. The plane itself forms a non-inertial (moving) frame that translates with the constant horizontal velocity $$u$$. According to Galilean relativity, to convert coordinates from the ground frame $$(x,y)$$ to the plane’s frame $$(x',y')$$, we subtract the uniform motion of the plane:

$$x' = x - ut, \qquad y' = y.$$

Substituting $$x = ut$$ from the earlier result, we find

$$x' = ut - ut = 0.$$

Thus, at every instant of time,

$$x' = 0.$$

The vertical coordinate, on the other hand, remains exactly what it was in the ground frame because the transformation does not affect $$y$$:

$$y' = y = -\tfrac{1}{2}gt^{2}.$$

Therefore the equations of motion in the plane’s frame reduce to

$$x'(t) = 0,\qquad y'(t) = -\tfrac{1}{2}gt^{2}.$$

Because $$x'$$ is identically zero at all times, the bomb has no horizontal motion relative to the plane. Its entire motion, as perceived by the observer sitting in the aircraft, consists solely of a uniform downward acceleration under gravity.

Consequently, the bomb appears to fall straight down along the vertical line that passes through the point of release. This straight vertical line is precisely Option A in the given list.

Hence, the correct answer is Option A.

We start by noting that the tip of the second’s hand performs uniform circular motion with radius equal to the length of the hand. Thus, for all instants its speed is constant in magnitude and the only acceleration it possesses is the centripetal acceleration directed toward the centre of the clock face.

The length (radius) of the second’s hand is given as $$r = 0.1\;\text{m}.$$

First we need the angular speed. A second’s hand completes one full revolution in $$T = 60\;\text{s}.$$ The standard formula connecting time-period and angular speed is

$$\omega = \frac{2\pi}{T}.$$

Substituting $$T = 60\;\text{s},$$ we obtain

$$\omega = \frac{2\pi}{60} = \frac{\pi}{30}\;\text{rad s}^{-1}.$$

To find the linear speed $$v$$ of the tip we use the relation

$$v = \omega r.$$

Putting $$\omega = \dfrac{\pi}{30}\;\text{rad s}^{-1}$$ and $$r = 0.1\;\text{m},$$ we get

$$v = \left(\frac{\pi}{30}\right)\!(0.1) = \frac{0.1\pi}{30} = \frac{\pi}{300} \;\text{m s}^{-1}.$$

Taking $$\pi \approx 3.14,$$ this yields

$$v \approx \frac{3.14}{300} = 0.01047\;\text{m s}^{-1}.$$

For uniform circular motion the instantaneous (centripetal) acceleration magnitude is

$$a_c = \frac{v^{2}}{r}.$$

Substituting $$v = 0.01047\;\text{m s}^{-1}$$ and $$r = 0.1\;\text{m},$$ we have

$$a_c = \frac{(0.01047)^{2}}{0.1} = \frac{0.0001097}{0.1} = 0.001097\;\text{m s}^{-2}.$$

This value is approximately $$1.1 \times 10^{-3}\;\text{m s}^{-2}.$$ Since the direction of the centripetal acceleration keeps changing but its magnitude stays the same, the average magnitude over any interval is of the same order. Hence, the required order of the average acceleration is $$10^{-3}\;\text{m s}^{-2}.$$

Hence, the correct answer is Option A.

First let us choose a convenient coordinate system. We take the horizontal road surface as the $$x$$-axis and the vertically downward direction (the direction in which raindrops actually fall) as the negative $$y$$-axis. All speeds mentioned are measured with respect to the ground.

When the car is standing still, the driver observes the rain to be falling exactly vertically downward. Because the car is at rest in this situation, the velocity of the rain relative to the ground must itself be vertical. We therefore write the ground-frame velocity of the rain as

$$\vec u=-u\,\hat{\jmath},$$

where $$u>0$$ is the speed of the rain and the minus sign signifies the downward (negative $$y$$) direction.

Now the car starts moving to the right (positive $$x$$-direction) with speed $$v$$. The ground-frame velocity of the car is then

$$\vec v_c = v\,\hat{\imath}.$$

The driver in the car does not see the ground-frame velocity of the rain; instead he sees the relative velocity of the rain with respect to the car. The standard formula for relative velocity is stated first:

$$\vec v_{\text{rain relative to car}}=\vec u-\vec v_c.$$

Substituting the expressions for $$\vec u$$ and $$\vec v_c$$ we obtain

$$\vec v_{rc}=(-u\,\hat{\jmath})-(v\,\hat{\imath})=-v\,\hat{\imath}-u\,\hat{\jmath}.$$

The driver now sees this vector making an angle of $$60^\circ$$ with the horizontal. The horizontal component is $$|-v|=v$$ and the vertical component is $$|-u|=u$$, so from the definition of the tangent of an angle we have

$$\tan 60^\circ=\frac{\text{vertical component}}{\text{horizontal component}}=\frac{u}{v}.$$

Using $$\tan 60^\circ=\sqrt3$$ gives

$$\sqrt3=\frac{u}{v}\quad\Longrightarrow\quad u=v\sqrt3. \quad -(1)$$

Next the car’s speed is increased to $$(1+\beta)v$$. Its new ground-frame velocity becomes

$$\vec v_c'=(1+\beta)v\,\hat{\imath}.$$

The new relative velocity of the rain with respect to the faster car is

$$\vec v_{rc}'=\vec u-\vec v_c'=(-u\,\hat{\jmath})-(1+\beta)v\,\hat{\imath}=-(1+\beta)v\,\hat{\imath}-u\,\hat{\jmath}.$$

This time the driver measures the angle made by this vector with the horizontal to be $$45^\circ$$. Therefore

$$\tan 45^\circ=\frac{u}{(1+\beta)v}.$$

Because $$\tan 45^\circ=1$$, we get

$$1=\frac{u}{(1+\beta)v}\quad\Longrightarrow\quad u=(1+\beta)v. \quad -(2)$$

We now have two expressions, (1) and (2), for the same quantity $$u$$. Equating them gives

$$v\sqrt3=(1+\beta)v.$$

Since $$v\neq0$$, we can cancel $$v$$ from both sides to obtain

$$\sqrt3=1+\beta\quad\Longrightarrow\quad\beta=\sqrt3-1.$$

Using $$\sqrt3\approx1.732$$ we calculate

$$\beta\approx1.732-1=0.732\;(\text{approximately}).$$

The closest option to this value is $$0.73$$.

Hence, the correct answer is Option D.

Let the magnitudes of the two given forces be denoted by $$|\vec P| = P$$ and $$|\vec Q| = Q$$.

Their resultant is $$\vec R = \vec P + \vec Q$$ and we are told that its magnitude equals that of $$\vec P$$, that is

$$|\vec R| = |\vec P|\; \Longrightarrow\; |\vec R| = P.$$

First we express the magnitude of the resultant $$\vec R$$ in terms of $$P$$, $$Q$$ and the angle $$\alpha$$ between $$\vec P$$ and $$\vec Q$$. The law of vector addition (or the triangle law) gives

$$|\vec R|^2 \;=\; |\vec P|^2 + |\vec Q|^2 + 2\,|\vec P|\,|\vec Q|\,\cos\alpha.$$

Substituting $$|\vec R| = P$$ and $$|\vec P|=P,\; |\vec Q|=Q$$ we have

$$P^2 = P^2 + Q^2 + 2PQ\cos\alpha.$$

Subtracting $$P^2$$ from both sides yields

$$0 = Q^2 + 2PQ\cos\alpha.$$

Dividing by the non-zero quantity $$Q$$ we get

$$Q + 2P\cos\alpha = 0.$$

Hence

$$\cos\alpha = -\dfrac{Q}{2P}.$$

Now we form a new resultant by doubling $$\vec P$$ and adding $$\vec Q$$. Let

$$\vec S = 2\vec P + \vec Q.$$

We wish to find the angle $$\theta$$ between $$\vec S$$ and $$\vec Q$$. By the definition of the dot product,

$$\vec S\cdot\vec Q = |\vec S|\,|\vec Q|\,\cos\theta.$$

Compute the dot product on the left:

$$\vec S\cdot\vec Q = (2\vec P + \vec Q)\cdot\vec Q = 2\,\vec P\cdot\vec Q + \vec Q\cdot\vec Q.$$

The individual dot products are

$$\vec P\cdot\vec Q = |\vec P|\,|\vec Q|\,\cos\alpha = P Q \cos\alpha,$$

$$\vec Q\cdot\vec Q = |\vec Q|^2 = Q^2.$$

Therefore

$$\vec S\cdot\vec Q = 2(P Q \cos\alpha) + Q^2.$$

Substituting the previously obtained value $$\cos\alpha = -\dfrac{Q}{2P}$$, we find

$$\vec S\cdot\vec Q = 2\!\left(P Q \left(-\dfrac{Q}{2P}\right)\right) + Q^2 = 2\!\left(-\dfrac{Q^2}{\,2\,}\right) + Q^2 = -Q^2 + Q^2 = 0.$$

Because $$\vec S\cdot\vec Q = 0$$, the cosine of the angle $$\theta$$ is

$$\cos\theta = \dfrac{\vec S\cdot\vec Q}{|\vec S|\,|\vec Q|} = \dfrac{0}{|\vec S|\,|\vec Q|} = 0.$$

An angle whose cosine is zero is a right angle, namely $$\theta = 90^\circ$$.

Hence, the correct answer is Option D (90°).

Let us denote the mass of each body by $$m$$ and the common magnitude of their initial velocities by $$v$$. Thus, at the very beginning each body possesses a momentum vector of magnitude

$$p_0 = m\,v$$

but the two momentum vectors point in different directions within the same plane. Suppose the angle between these two directions is $$\theta$$ (this is precisely the angle we are required to find).

To combine the two individual momentum vectors into a single “net” or “resultant” initial momentum, we treat the vectors exactly the way we add two sides of a parallelogram. For two vectors of equal magnitude $$p_0$$ separated by an angle $$\theta$$, the magnitude of their resultant, by the law of cosines for vectors, is

$$\lvert\vec P_{\text{initial}}\rvert \;=\;\sqrt{p_0^{\,2} + p_0^{\,2} + 2\,p_0\,p_0\,\cos\theta}\;=\;\sqrt{2p_0^{\,2}\,(1+\cos\theta)}$$

Inside the square root we notice the standard trigonometric identity $$1 + \cos\theta = 2\cos^{2}\dfrac{\theta}{2}$$. Substituting this, we obtain

$$\lvert\vec P_{\text{initial}}\rvert \;=\;\sqrt{2p_0^{\,2}\,\bigl(2\cos^{2}\tfrac{\theta}{2}\bigr)} \;=\; \sqrt{4p_0^{\,2}\,\cos^{2}\tfrac{\theta}{2}} \;=\; 2p_0\,\cos\dfrac{\theta}{2}.$$

Replacing $$p_0$$ by $$m\,v$$ gives

$$\lvert\vec P_{\text{initial}}\rvert \;=\;2\,m\,v\,\cos\dfrac{\theta}{2}.$$

Now the collision is stated to be completely inelastic, meaning the two bodies stick together and thereafter behave like a single composite body. The mass of this composite body is simply the sum of the individual masses, namely

$$m_{\text{final}} \;=\; m + m \;=\; 2m.$$

The problem also tells us that the common final speed of this combined body is half of the original speed, that is

$$v_{\text{final}} \;=\; \dfrac{v}{2}.$$

Therefore the magnitude of the final momentum is

$$\lvert\vec P_{\text{final}}\rvert \;=\; m_{\text{final}}\,v_{\text{final}} \;=\; 2m \times \dfrac{v}{2} \;=\; m\,v.$$

According to the principle of conservation of linear momentum, the total momentum of an isolated system remains constant, so we can equate the magnitudes of the initial and final momenta:

$$2\,m\,v\,\cos\dfrac{\theta}{2} \;=\; m\,v.$$

Dividing both sides by the common positive factor $$m\,v$$ leaves us with a very simple trigonometric equation:

$$2\,\cos\dfrac{\theta}{2} \;=\; 1.$$

We now isolate the cosine term:

$$\cos\dfrac{\theta}{2} \;=\; \dfrac{1}{2}.$$

The standard value of the cosine function that gives one-half is $$60^\circ$$, so we can write

$$\dfrac{\theta}{2} \;=\; 60^\circ.$$

Multiplying both sides by 2, we finally obtain

$$\theta \;=\; 120^\circ.$$

So, the answer is $$120^\circ$$.

We have the position vector of the particle as $$\vec r(t)=\cos\omega t\,\hat i+\sin\omega t\,\hat j,$$ where $$\omega$$ is a constant and $$t$$ is time.

By definition, the velocity vector is the time derivative of the position vector, that is $$\vec v(t)=\dfrac{d\vec r}{dt}.$$

Now, differentiating each component separately and using the standard derivatives $$\dfrac{d}{dt}\bigl(\cos\omega t\bigr)=-\omega\sin\omega t$$ and $$\dfrac{d}{dt}\bigl(\sin\omega t\bigr)=\ \ \omega\cos\omega t,$$ we obtain

$$\vec v(t)=\dfrac{d}{dt}\Bigl(\cos\omega t\,\hat i+\sin\omega t\,\hat j\Bigr)=\bigl(-\omega\sin\omega t\bigr)\hat i+\bigl(\ \ \omega\cos\omega t\bigr)\hat j.$$

Next, we verify whether the velocity is perpendicular to the position vector by taking their dot product. Using the formula $$\vec A\cdot\vec B=A_xB_x+A_yB_y$$ for two‐dimensional vectors, we write

$$\vec r(t)\cdot\vec v(t)=\bigl(\cos\omega t\bigr)\bigl(-\omega\sin\omega t\bigr)+\bigl(\sin\omega t\bigr)\bigl(\ \ \omega\cos\omega t\bigr).$$

Simplifying each term, we have

$$\vec r(t)\cdot\vec v(t)=-\omega\cos\omega t\sin\omega t+\omega\sin\omega t\cos\omega t=0.$$

Because the dot product is zero, $$\vec v(t)\perp\vec r(t).$$ So the velocity is indeed perpendicular to the position vector at every instant.

Now, by definition, the acceleration vector is the time derivative of the velocity vector, that is $$\vec a(t)=\dfrac{d\vec v}{dt}.$$

Differentiating $$\vec v(t)=\bigl(-\omega\sin\omega t\bigr)\hat i+\bigl(\omega\cos\omega t\bigr)\hat j$$ term by term and again using the same trigonometric derivatives, we get

$$\vec a(t)=\dfrac{d}{dt}\Bigl(-\omega\sin\omega t\,\hat i+\omega\cos\omega t\,\hat j\Bigr)=\bigl(-\omega^2\cos\omega t\bigr)\hat i+\bigl(-\omega^2\sin\omega t\bigr)\hat j.$$

Notice that every component of $$\vec a(t)$$ is $$-\omega^2$$ times the corresponding component of $$\vec r(t).$$ Writing this compactly, we have

$$\vec a(t)=-\omega^2\Bigl(\cos\omega t\,\hat i+\sin\omega t\,\hat j\Bigr)=-\omega^2\vec r(t).$$

The negative sign means that $$\vec a(t)$$ is opposite in direction to $$\vec r(t).$$ Since $$\vec r(t)$$ points outward from the origin to the particle, $$-\vec r(t)$$ points inward toward the origin. Hence the acceleration is directed toward the origin, i.e., toward the centre of the circular path.

We have thus shown that the velocity is perpendicular to the position vector and the acceleration is directed toward the origin. Among the given options, this description corresponds to Option D.

Hence, the correct answer is Option D.

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

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

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

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

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

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

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

y-component

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

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

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

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

Dividing both sides by 2,

$$t^2 = 16.$$

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

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

x-component

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

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

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

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

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

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

Next the term with acceleration:

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

Adding the two contributions to the displacement,

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

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

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

Hence, the correct answer is Option C.

We have particle A with mass $$m_1$$ and initial velocity $$\vec u_1 = \left(\sqrt{3}\,\hat i + \hat j\right)\,\text{m s}^{-1}$$. Particle B has mass $$m_2$$ and is initially at rest, so $$\vec u_2 = \vec 0$$. After the collision the given velocity of A is $$\vec v_1 = \left(\hat i + \sqrt{3}\,\hat j\right)\,\text{m s}^{-1}$$, while the velocity of B is the unknown vector $$\vec v_2 = a\,\hat i + b\,\hat j$$. The masses are related by $$m_1 = 2m_2$$.

Because the collision takes place in isolation, the law of conservation of linear momentum applies:

$$m_1\vec u_1 + m_2\vec u_2 = m_1\vec v_1 + m_2\vec v_2.$$

Substituting the known quantities and remembering that $$\vec u_2 = \vec 0$$ we get

$$m_1\left(\sqrt{3}\,\hat i + \hat j\right) = m_1\left(\hat i + \sqrt{3}\,\hat j\right) + m_2\left(a\,\hat i + b\,\hat j\right).$$

We now compare the $$\hat i$$ and $$\hat j$$ components separately. First, however, it is convenient to express $$m_2$$ in terms of $$m_1$$. Since $$m_1 = 2m_2$$ we have $$m_2 = \dfrac{m_1}{2}$$.

$$\hat i$$-components:

$$m_1\sqrt{3} = m_1\cdot 1 + \dfrac{m_1}{2}\,a.$$

Dividing by $$m_1$$ gives

$$\sqrt{3} = 1 + \dfrac{a}{2}\;\;\Longrightarrow\;\;\dfrac{a}{2} = \sqrt{3}-1\;\;\Longrightarrow\;\;a = 2\!\left(\sqrt{3}-1\right).$$

$$\hat j$$-components:

$$m_1\cdot 1 = m_1\sqrt{3} + \dfrac{m_1}{2}\,b.$$

Again dividing by $$m_1$$ we obtain

$$1 = \sqrt{3} + \dfrac{b}{2}\;\;\Longrightarrow\;\;\dfrac{b}{2} = 1-\sqrt{3} = -\!\left(\sqrt{3}-1\right)\;\;\Longrightarrow\;\;b = -2\!\left(\sqrt{3}-1\right).$$

Thus the velocity of particle B after the collision is

$$\vec v_2 = a\,\hat i + b\,\hat j = 2\!\left(\sqrt{3}-1\right)\hat i - 2\!\left(\sqrt{3}-1\right)\hat j = 2\!\left(\sqrt{3}-1\right)\left(\hat i - \hat j\right)\,\text{m s}^{-1}.$$

Let us call the angle between $$\vec v_1$$ and $$\vec v_2$$ by the symbol $$\theta$$. To find it we use the definition

$$\cos\theta = \dfrac{\vec v_1\!\cdot\!\vec v_2}{\lvert\vec v_1\rvert\,\lvert\vec v_2\rvert}.$$

Dot product:

$$\vec v_1\!\cdot\!\vec v_2 = \bigl(1\,\hat i + \sqrt{3}\,\hat j\bigr)\!\cdot\! \Bigl[2\bigl(\sqrt{3}-1\bigr)\,(\hat i - \hat j)\Bigr]$$ $$= 2\!\left(\sqrt{3}-1\right)\left[1\cdot 1 + \sqrt{3}\,(-1)\right] = 2\!\left(\sqrt{3}-1\right)\left(1-\sqrt{3}\right) = -2\!\left(\sqrt{3}-1\right)^{\!2}.$$

Magnitudes:

$$\lvert\vec v_1\rvert = \sqrt{1^{2} + (\sqrt{3})^{2}} = \sqrt{1+3} = 2,$$ $$\lvert\vec v_2\rvert = 2\!\left(\sqrt{3}-1\right)\sqrt{\,(\hat i - \hat j)\!\cdot\!(\hat i - \hat j)} = 2\!\left(\sqrt{3}-1\right)\sqrt{1+1} = 2\!\left(\sqrt{3}-1\right)\sqrt{2}.$$

Cosine of the angle:

$$\cos\theta = \dfrac{-2\!\left(\sqrt{3}-1\right)^{2}} {2 \times 2\!\left(\sqrt{3}-1\right)\sqrt{2}} = \dfrac{-\left(\sqrt{3}-1\right)}{2\sqrt{2}} \approx \dfrac{-0.732}{2.828} \approx -0.259.$$

The value $$\cos\theta = -0.259$$ corresponds to an angle of approximately $$105^\circ$$ (since $$\cos105^\circ \approx -0.2588$$).

Hence, the correct answer is Option D.

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

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

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

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

x-component

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

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

Simplifying the right-hand side step by step,

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

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

$$5t^2 = 20.$$

Dividing both sides by 5,

$$t^2 = 4.$$

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

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

y-component

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

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

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

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

Calculate each term carefully. The first term is

$$5 \times 2 = 10.$$

The second term involves several steps:

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

and

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

so

$$2 \times 4 = 8.$$

Adding the two contributions gives

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

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

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

These match the first option in the list.

Hence, the correct answer is Option A.

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

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

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

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

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

Therefore

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

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

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

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

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

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

Adding the momenta on the left gives

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Writing it in the same form as the options,

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

Hence, the correct answer is Option A.

We have to find the position of the particle after $$t = 2\ \text{s}$$ and then its distance from the origin.

The standard kinematic relation for constant acceleration in vector form is stated first:

$$\vec r(t) \;=\; \vec r_0 \;+\; \vec v_0\,t \;+\; \tfrac12\,\vec a\,t^{\,2}.$$

Here

$$\vec r_0 = 2.0\,\hat i + 4.0\,\hat j\ \text{m},\qquad \vec v_0 = 5.0\,\hat i + 4.0\,\hat j\ \text{m s}^{-1},\qquad \vec a = 4.0\,\hat i + 4.0\,\hat j\ \text{m s}^{-2}.$$

Now we substitute $$t = 2\ \text{s}$$ step by step.

First, the displacement due to initial velocity:

$$\vec v_0\,t \;=\; (5.0\,\hat i + 4.0\,\hat j)(2) \;=\; 10\,\hat i + 8\,\hat j\ \text{m}.$$

Next, the displacement due to acceleration:

$$\tfrac12\,\vec a\,t^{\,2} \;=\; \tfrac12\,(4.0\,\hat i + 4.0\,\hat j)\,(2)^{2} \;=\; \tfrac12\,(4.0\,\hat i + 4.0\,\hat j)\,(4) \;=\; 8\,\hat i + 8\,\hat j\ \text{m}.$$

Now we add all three contributions:

$$\vec r(2) \;=\; \vec r_0 + \vec v_0\,t + \tfrac12\,\vec a\,t^{\,2}$$

$$\phantom{\vec r(2)} \;=\; (2\,\hat i + 4\,\hat j) + (10\,\hat i + 8\,\hat j) + (8\,\hat i + 8\,\hat j).$$

Combining the $$\hat i$$ components: $$2 + 10 + 8 = 20,$$ so we get $$20\,\hat i.$$

Combining the $$\hat j$$ components: $$4 + 8 + 8 = 20,$$ so we get $$20\,\hat j.$$

Thus the position vector at $$t = 2\ \text{s}$$ is

$$\vec r(2) = 20\,\hat i + 20\,\hat j\ \text{m}.$$

The distance from the origin is the magnitude of this vector:

$$|\vec r(2)| \;=\; \sqrt{(20)^{2} + (20)^{2}} \;=\; \sqrt{400 + 400} \;=\; \sqrt{800} \;=\; 20\sqrt{2}\ \text{m}.$$

Hence, the correct answer is Option B.

We are told that the magnitudes of the two vectors are $$|\vec A_1| = 3$$ and $$|\vec A_2| = 5$$, while the magnitude of their sum is $$|\vec A_1 + \vec A_2| = 5$$.

First, we need the dot product $$\vec A_1 \cdot \vec A_2$$. We use the standard magnitude formula for the sum of two vectors:

$$|\vec A_1 + \vec A_2|^2 \;=\; |\vec A_1|^2 + |\vec A_2|^2 + 2\,\vec A_1 \cdot \vec A_2.$$

Substituting the given magnitudes, we have

$$5^2 \;=\; 3^2 + 5^2 + 2\,\vec A_1 \cdot \vec A_2.$$

Writing the squares explicitly,

$$25 \;=\; 9 + 25 + 2\,\vec A_1 \cdot \vec A_2.$$

Bringing the numerical terms on the right to the left side,

$$25 - 34 \;=\; 2\,\vec A_1 \cdot \vec A_2.$$

So

$$-9 \;=\; 2\,\vec A_1 \cdot \vec A_2,$$

and therefore

$$\vec A_1 \cdot \vec A_2 \;=\; -\frac{9}{2} = -4.5.$$

We also note the squared magnitudes that will be needed shortly:

$$|\vec A_1|^2 = 3^2 = 9, \qquad |\vec A_2|^2 = 5^2 = 25.$$

Now we must evaluate the required scalar product

$$\bigl(2\vec A_1 + 3\vec A_2\bigr)\,\cdot\,\bigl(3\vec A_1 - 2\vec A_2\bigr).$$

We expand this dot product term by term, using linearity and the commutativity of the dot product ($$\vec A_2 \cdot \vec A_1 = \vec A_1 \cdot \vec A_2$$):

$$\begin{aligned} (2\vec A_1 + 3\vec A_2)\cdot(3\vec A_1 - 2\vec A_2) &= (2\vec A_1)\cdot(3\vec A_1) \;+\; (2\vec A_1)\cdot(-2\vec A_2) \\ &\quad +\; (3\vec A_2)\cdot(3\vec A_1) \;+\; (3\vec A_2)\cdot(-2\vec A_2). \end{aligned}$$

Evaluating each term separately:

$$\begin{aligned} (2\vec A_1)\cdot(3\vec A_1) &= 6\,(\vec A_1\cdot\vec A_1) = 6\,|\vec A_1|^2, \\ (2\vec A_1)\cdot(-2\vec A_2) &= -4\,(\vec A_1\cdot\vec A_2), \\ (3\vec A_2)\cdot(3\vec A_1) &= 9\,(\vec A_2\cdot\vec A_1) = 9\,(\vec A_1\cdot\vec A_2), \\ (3\vec A_2)\cdot(-2\vec A_2) &= -6\,(\vec A_2\cdot\vec A_2) = -6\,|\vec A_2|^2. \end{aligned}$$

Combining these, we obtain

$$\begin{aligned} (2\vec A_1 + 3\vec A_2)\cdot(3\vec A_1 - 2\vec A_2) &= 6|\vec A_1|^2 \;+\; \bigl(-4 + 9\bigr)\,(\vec A_1\cdot\vec A_2) \;-\; 6|\vec A_2|^2 \\ &= 6|\vec A_1|^2 \;+\; 5\,(\vec A_1\cdot\vec A_2) \;-\; 6|\vec A_2|^2. \end{aligned}$$

Now we substitute the numerical values we have found:

$$\begin{aligned} 6|\vec A_1|^2 &= 6 \times 9 = 54, \\ 5\,(\vec A_1\cdot\vec A_2) &= 5 \times (-4.5) = -22.5, \\ -6|\vec A_2|^2 &= -6 \times 25 = -150. \end{aligned}$$

Adding these three results together:

$$ 54 \;+\; (-22.5) \;+\; (-150) \;=\; 54 - 22.5 - 150 = 31.5 - 150 = -118.5. $$

Thus the required scalar product is $$-118.5$$.

Hence, the correct answer is Option B.

Let us first describe the data in vector language. We adopt east as the positive $$x$$-direction and north as the positive $$y$$-direction, so the unit vectors are $$\hat{i}$$ to the east and $$\hat{j}$$ to the north.

Ship A is moving with velocity

$$\vec v_A = 30\hat i + 50\hat j\ \text{km h}^{-1}.$$

Ship B is initially 80 km east and 150 km north of A. Measured from A, its initial position vector is therefore

$$\vec r_0 = 80\hat i + 150\hat j\ \text{km}.$$

Ship B is sailing due west at 10 km h$$^{-1}$$, hence

$$\vec v_B = -10\hat i\ \text{km h}^{-1}.$$

We are interested in the separation between the two ships, so we examine the relative motion of B with respect to A. The relative velocity is given by the formula

$$\vec v_{\text{rel}} = \vec v_B - \vec v_A.$$

Substituting the individual velocities, we have

$$\vec v_{\text{rel}} = (-10\hat i) - (30\hat i + 50\hat j)$$ $$= -10\hat i - 30\hat i - 50\hat j$$ $$= -40\hat i - 50\hat j\ \text{km h}^{-1}.$$

At any time $$t$$ hours after the instant chosen as $$t=0$$, the relative position vector of B with respect to A is

$$\vec r(t) = \vec r_0 + \vec v_{\text{rel}}\,t.$$

Substituting $$\vec r_0$$ and $$\vec v_{\text{rel}}$$, we obtain

$$\vec r(t) = (80\hat i + 150\hat j) + (-40\hat i - 50\hat j)t$$ $$= (80 - 40t)\hat i + (150 - 50t)\hat j.$$

The actual separation between the ships is the magnitude of $$\vec r(t)$$. To find the time of minimum distance we minimise the square of that magnitude, because the square is easier to differentiate and has its minimum at the same instant. Using the formula $$|\vec r|^2 = x^2 + y^2$$, we write

$$D^2(t) = (80 - 40t)^2 + (150 - 50t)^2.$$

Now we differentiate $$D^2(t)$$ with respect to $$t$$ and set the derivative equal to zero. This ensures the distance is extremal (and here it will be a minimum):

$$\frac{d}{dt}D^2(t) = 2(80 - 40t)(-40) + 2(150 - 50t)(-50) = 0.$$

We can cancel the factor 2 to simplify:

$$(80 - 40t)(-40) + (150 - 50t)(-50) = 0.$$

Expanding each product, we get

$$-40\cdot80 + 40\cdot40t - 50\cdot150 + 50\cdot50t = 0,$$ $$-3200 + 1600t - 7500 + 2500t = 0.$$

Collecting like terms,

$$(-3200 - 7500) + (1600 + 2500)t = 0,$$ $$-10700 + 4100t = 0.$$

Isolating $$t$$ yields

$$4100t = 10700,$$ $$t = \frac{10700}{4100}.$$

We divide numerator and denominator by 100 to simplify:

$$t = \frac{107}{41} \ \text{h}.$$

Performing the division,

$$t \approx 2.609756\ \text{h}.$$

Rounded to one decimal place this is 2.6 h, which matches Option C.

Hence, the correct answer is Option C.

We are told that the instantaneous position of the particle is described, as a function of time $$t,$$ by the three co-ordinate equations

$$x = a \cos \omega t,$$

$$y = a \sin \omega t,$$

and

$$z = a \, \omega t.$$

To obtain the speed we must first find the velocity vector. By definition, the velocity components are the time derivatives of the corresponding position components. In symbols, if $$x(t),\,y(t),\,z(t)$$ are known, then

$$v_x = \dfrac{dx}{dt}, \qquad v_y = \dfrac{dy}{dt}, \qquad v_z = \dfrac{dz}{dt}.$$

Carrying out these differentiations one by one, we have

$$v_x = \dfrac{d}{dt}\!\bigl(a \cos \omega t\bigr).$$

Because the derivative of $$\cos (\omega t)$$ with respect to $$t$$ is $$-\omega \sin (\omega t),$$ we obtain

$$v_x = a\;(-\omega \sin \omega t) = -\,a \omega \sin \omega t.$$

Next, for the $$y$$ component,

$$v_y = \dfrac{d}{dt}\!\bigl(a \sin \omega t\bigr) = a \,\omega \cos \omega t,$$

since the derivative of $$\sin (\omega t)$$ is $$\omega \cos (\omega t).$$

Finally, for the $$z$$ component,

$$v_z = \dfrac{d}{dt}\!\bigl(a \omega t\bigr) = a \omega,$$

because the derivative of a linear function $$\omega t$$ is simply $$\omega.$$

Thus the full velocity vector is

$$\vec v = \bigl(-a\omega \sin \omega t\,,\; a\omega \cos \omega t\,,\; a\omega\bigr).$$

The speed $$v$$ is the magnitude of this vector. The magnitude (or Euclidean norm) of a vector $$\!(v_x,v_y,v_z)$$ is given by the formula

$$v = \sqrt{\,v_x^2 + v_y^2 + v_z^2\,}.$$

Now we substitute the three components just calculated:

$$v = \sqrt{\bigl(-a\omega \sin \omega t\bigr)^2 + \bigl(a\omega \cos \omega t\bigr)^2 + \bigl(a\omega\bigr)^2}.$$

Expanding each squared term inside the radical gives

$$v = \sqrt{a^2\omega^2 \sin^2 \omega t + a^2\omega^2 \cos^2 \omega t + a^2\omega^2}.$$

We notice that the first two terms share the common factor $$a^2\omega^2,$$ so we group them:

$$v = \sqrt{a^2\omega^2 \bigl(\sin^2 \omega t + \cos^2 \omega t\bigr) + a^2\omega^2}.$$

From trigonometry we know the fundamental identity $$\sin^2 \theta + \cos^2 \theta = 1.$$ Applying this identity (with $$\theta = \omega t$$) converts the grouped bracket to $$1$$, giving

$$v = \sqrt{a^2\omega^2 \cdot 1 + a^2\omega^2}.$$

So we simply have two identical terms inside the square root:

$$v = \sqrt{a^2\omega^2 + a^2\omega^2} = \sqrt{2\,a^2\omega^2}.$$

Finally, we take the square root of the product. Because $$a$$ and $$\omega$$ are positive constants (length and angular frequency respectively), their squares inside the root come out as absolute values, which are just themselves:

$$v = \sqrt{2}\; a \omega.$$

This expression is a constant, showing that the particle moves with uniform speed of magnitude $$\sqrt{2}\,a\omega.$$

Hence, the correct answer is Option A.

We start by recalling that in uniform circular motion the speed is constant, but the velocity vector is continuously changing its direction. At any point on the circle, the velocity is tangent to the path and perpendicular to the radius drawn to that point. Let the constant speed be denoted by $$v = 10\ \text{m s}^{-1}$$.

Suppose the particle is at point $$P_{1}$$ on the circle. Its velocity vector at this point is $$\vec v_{1}$$, tangent to the circle. After the particle has moved through a central angle of $$\theta = 60^{\circ}$$ (which is $$\theta = \dfrac{\pi}{3}\ \text{radians}$$) it reaches point $$P_{2}$$. The speed is still $$v$$, and the velocity vector there is $$\vec v_{2}$$, also tangent to the circle.

The two velocity vectors $$\vec v_{1}$$ and $$\vec v_{2}$$ have the same magnitude $$v$$ but they are directed along tangents that make an angle $$\theta$$ with each other, because each tangent is perpendicular to its corresponding radius and the radii themselves are separated by the same central angle $$\theta$$.

We want the magnitude of the change in velocity:

$$|\Delta \vec v| = |\vec v_{2} - \vec v_{1}|$$.

For two vectors of equal magnitude $$v$$ with an included angle $$\theta$$ between them, we can use the law of cosines (vector form) to find the magnitude of their difference. The formula is stated as:

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

Here $$a = b = v$$ and $$\theta = 60^{\circ}$$, so

$$|\Delta \vec v| = \sqrt{v^{2} + v^{2} - 2v\,v\cos\theta}$$.

Substituting $$v = 10\ \text{m s}^{-1}$$ and $$\cos 60^{\circ} = \dfrac{1}{2}$$, we get

$$|\Delta \vec v| = \sqrt{10^{2} + 10^{2} - 2 \cdot 10 \cdot 10 \cdot \dfrac{1}{2}}$$ $$= \sqrt{100 + 100 - 100}$$ $$= \sqrt{100}$$ $$= 10\ \text{m s}^{-1}.$$

Many students also memorize a simpler equivalent relation for uniform circular motion: $$|\Delta \vec v| = 2v\sin\left(\dfrac{\theta}{2}\right)$$. Stating and using that formula gives the same result. We have $$\theta/2 = 30^{\circ}$$ and $$\sin 30^{\circ} = \dfrac{1}{2}$$, so $$|\Delta \vec v| = 2 \times 10 \times \dfrac{1}{2} = 10\ \text{m s}^{-1}$$.

Either way, the magnitude of the change in velocity is $$10\ \text{m s}^{-1}$$. Hence, the correct answer is Option D.

For the motion of a projectile close to the surface of the earth (neglecting air resistance) the standard Cartesian equation of the trajectory is first written. We have the well-known result

$$y = x\tan\theta_0 - \frac{g\,x^2}{2\,v_0^{\;2}\cos^{2}\theta_0}.$$

In this formula $$\theta_0$$ is the angle of projection with the horizontal, $$v_0$$ is the initial speed and $$g$$ is the magnitude of the acceleration due to gravity.

The question gives the actual path followed by the projectile as

$$y = 2x - 9x^{2}.$$

Because two different expressions describe the same curve they must be identical term by term in powers of $$x$$. Therefore we compare the coefficient of $$x$$ first:

$$\tan\theta_0 = 2.$$

Now we convert this tangent value into the corresponding sine and cosine, because they will be needed later. Starting with

$$\tan\theta_0 = \frac{\sin\theta_0}{\cos\theta_0} = 2,$$

write $$\sin\theta_0 = 2\cos\theta_0$$ and impose the identity $$\sin^{2}\theta_0 + \cos^{2}\theta_0 = 1.$$ Substituting gives

$$4\cos^{2}\theta_0 + \cos^{2}\theta_0 = 1 \;\Longrightarrow\; 5\cos^{2}\theta_0 = 1 \;\Longrightarrow\; \cos^{2}\theta_0 = \frac{1}{5}.$$

Taking the positive square-root (because the angle of projection is acute in projectile motion) we get

$$\cos\theta_0 = \frac{1}{\sqrt{5}}.$$

From $$\sin\theta_0 = 2\cos\theta_0$$ we immediately have

$$\sin\theta_0 = \frac{2}{\sqrt{5}}.$$

Hence one may write the angle either as $$\theta_0 = \cos^{-1}\!\left(\dfrac{1}{\sqrt{5}}\right)$$ or as $$\theta_0 = \sin^{-1}\!\left(\dfrac{2}{\sqrt{5}}\right).$$ Both representations describe the same direction because they come from the same sine and cosine pair.

Next compare the coefficient of $$x^{2}$$ in the two trajectory equations. From the standard formula the coefficient of $$x^{2}$$ is

$$-\frac{g}{2\,v_0^{\;2}\cos^{2}\theta_0},$$

whereas in the given curve it is $$-9$$. Setting them equal produces

$$-\frac{g}{2\,v_0^{\;2}\cos^{2}\theta_0} = -9 \;\Longrightarrow\; \frac{g}{2\,v_0^{\;2}\cos^{2}\theta_0} = 9.$$

Substituting $$g = 10\text{ m s}^{-2}$$ and $$\cos^{2}\theta_0 = \dfrac{1}{5}$$, we get

$$\frac{10}{2\,v_0^{\;2}\left(\dfrac{1}{5}\right)} = 9 \;\Longrightarrow\; \frac{10}{\dfrac{2v_0^{\;2}}{5}} = 9 \;\Longrightarrow\; \frac{10 \times 5}{2v_0^{\;2}} = 9 \;\Longrightarrow\; \frac{50}{2v_0^{\;2}} = 9.$$

Now multiply both sides by $$2v_0^{\;2}$$ to isolate $$v_0$$:

$$50 = 18v_0^{\;2} \;\Longrightarrow\; v_0^{\;2} = \frac{50}{18} = \frac{25}{9}.$$

Taking the positive square-root gives the initial speed

$$v_0 = \frac{5}{3}\text{ m s}^{-1}.$$

We have therefore found

$$\theta_0 = \cos^{-1}\!\left(\frac{1}{\sqrt{5}}\right), \qquad v_0 = \frac{5}{3}\text{ m s}^{-1}.$$

These values coincide exactly with those specified in Option A.

Hence, the correct answer is Option A.

Let the two particles be projected with the common speed $$u$$, one making an angle $$\theta$$ with the horizontal, and the other making the complementary angle $$90^\circ-\theta$$. Since the angles are complementary, the two projectiles automatically have the same horizontal range.

First, we recall the standard formula for the range of a projectile launched with speed $$u$$ at an angle $$\alpha$$:

$$R=\frac{u^{2}\sin 2\alpha}{g}\,.$$

Applying this to the angle $$\theta$$, we get

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

For the complementary angle $$90^\circ-\theta$$ we have

$$R=\frac{u^{2}\sin 2(90^\circ-\theta)}{g} =\frac{u^{2}\sin(180^\circ-2\theta)}{g} =\frac{u^{2}\sin 2\theta}{g}\,,$$

which is clearly the same value, confirming that both projectiles indeed possess the same range $$R$$.

Next, we need the individual maximum heights. The well-known formula for the maximum height of a projectile launched with speed $$u$$ at angle $$\alpha$$ is

$$h=\frac{u^{2}\sin^{2}\alpha}{2g}\,.$$

Hence, for the first projectile launched at $$\theta$$, the maximum height is

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

For the second projectile launched at the angle $$90^\circ-\theta$$, we use $$\sin(90^\circ-\theta)=\cos\theta$$, giving

$$h_2=\frac{u^{2}\sin^{2}(90^\circ-\theta)}{2g} =\frac{u^{2}\cos^{2}\theta}{2g}\,.$$

We now form the product of the two maximum heights:

$$h_1h_2=\left(\frac{u^{2}\sin^{2}\theta}{2g}\right) \left(\frac{u^{2}\cos^{2}\theta}{2g}\right) =\frac{u^{4}\sin^{2}\theta\cos^{2}\theta}{4g^{2}}\,.$$

Let us keep this result in mind and move to the square of the common range. From the range expression we have already written,

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

Squaring this gives

$$R^{2} =\left(\frac{2u^{2}\sin\theta\cos\theta}{g}\right)^{2} =\frac{4u^{4}\sin^{2}\theta\cos^{2}\theta}{g^{2}}\,.$$

We now compare $$R^{2}$$ with $$h_1h_2$$. Observe that

$$h_1h_2=\frac{u^{4}\sin^{2}\theta\cos^{2}\theta}{4g^{2}}\,,$$
$$R^{2}=\frac{4u^{4}\sin^{2}\theta\cos^{2}\theta}{g^{2}} =16\left(\frac{u^{4}\sin^{2}\theta\cos^{2}\theta}{4g^{2}}\right) =16\,h_1h_2\,.$$

We have therefore arrived at the simple relation

$$R^{2}=16\,h_1h_2\,.$$

Hence, the correct answer is Option D.

We are told that the two vectors $$\vec A$$ and $$\vec B$$ have the same magnitude. Let us denote this common magnitude by $$a$$, so that we can write $$|\vec A| = |\vec B| = a$$.

The angle between the two vectors will be denoted by $$\theta$$. Our objective is to find an expression for this angle in terms of the given constant $$n$$.

First, we recall the standard vector identity for the magnitude of the sum of two vectors:

$$|\vec A + \vec B|^2 = |\vec A|^2 + |\vec B|^2 + 2|\vec A|\,|\vec B|\cos\theta.$$

Substituting $$|\vec A| = |\vec B| = a$$ into this formula gives

$$|\vec A + \vec B|^2 \;=\; a^2 + a^2 + 2a\cdot a\cos\theta \;=\; 2a^2(1+\cos\theta).$$

In exactly the same way we use the identity for the magnitude of the difference of two vectors:

$$|\vec A - \vec B|^2 = |\vec A|^2 + |\vec B|^2 - 2|\vec A|\,|\vec B|\cos\theta.$$

Again substituting $$a$$ for the magnitudes, we obtain

$$|\vec A - \vec B|^2 \;=\; a^2 + a^2 - 2a\cdot a\cos\theta \;=\; 2a^2(1-\cos\theta).$$

The statement of the problem says that the magnitude of $$\vec A + \vec B$$ is $$n$$ times the magnitude of $$\vec A - \vec B$$, that is,

$$|\vec A + \vec B| = n\,|\vec A - \vec B|.$$

We square both sides to eliminate the square roots:

$$|\vec A + \vec B|^2 = n^2\,|\vec A - \vec B|^2.$$

Now we substitute the expressions we have already derived:

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

Because the factor $$2a^2$$ is present on both sides, we can divide it out immediately, giving

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

Let us bring all the terms involving $$\cos\theta$$ to one side so that we can solve for $$\cos\theta$$ explicitly:

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

$$\cos\theta + n^2\cos\theta = n^2 - 1,$$

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

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

Thus the angle $$\theta$$ between the two vectors is given by the inverse cosine of the right-hand side:

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

This expression exactly matches Option A.

Hence, the correct answer is Option A.

We have a projectile launched at time $$t = 0$$ with initial speed $$u = 10 \text{ ms}^{-1}$$ making an angle $$\theta = 60^{\circ}$$ with the horizontal. Neglecting air resistance, the horizontal and vertical coordinates of the particle at any instant $$t$$ are obtained from elementary kinematics:

$$x = u\cos\theta \; t, \qquad y = u\sin\theta \; t \;-\; \dfrac12\,g\,t^{2}.$$

For the radius of curvature of a plane curve $$y(x)$$ we use the standard formula

$$R \;=\; \dfrac{\bigl(1 + (dy/dx)^{2}\bigr)^{3/2}}{\left|\,d^{2}y/dx^{2}\right|}.$$

Hence we first need the first and second derivatives of $$y$$ with respect to $$x$$. As derivatives with respect to $$t$$ are easier to obtain, we start there and then convert.

The time-derivatives of the coordinates are the velocity components:

$$\frac{dx}{dt} \;=\; u\cos\theta, \qquad \frac{dy}{dt} \;=\; u\sin\theta - g\,t.$$

Therefore the slope of the trajectory is

$$\frac{dy}{dx} \;=\; \frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}} \;=\; \frac{u\sin\theta - g\,t}{u\cos\theta} \;=\; \tan\theta \;-\;\frac{g\,t}{u\cos\theta}.$$

Now the second derivative is obtained by differentiating this result with respect to $$x$$. Using the chain rule $$\dfrac{d}{dx} = \dfrac{1}{dx/dt}\dfrac{d}{dt}$$, we write

$$\frac{d^{2}y}{dx^{2}} \;=\; \frac{1}{dx/dt}\, \frac{d}{dt}\!\Bigl(\frac{dy}{dx}\Bigr) \;=\; \frac{1}{u\cos\theta}\; \frac{d}{dt}\!\Bigl(\tan\theta - \frac{g\,t}{u\cos\theta}\Bigr).$$

The term $$\tan\theta$$ is a constant, so its derivative is zero, and we get

$$\frac{d^{2}y}{dx^{2}} \;=\; \frac{1}{u\cos\theta}\; \Bigl(-\frac{g}{u\cos\theta}\Bigr) \;=\; -\,\frac{g}{u^{2}\cos^{2}\theta}.$$

Observe that this quantity is a constant; it does not depend on time. We may now substitute the numerical values $$u = 10 \text{ ms}^{-1}, \quad g = 10 \text{ ms}^{-2}, \quad \cos\theta = \cos60^{\circ} = \tfrac12.$$ This gives

$$\frac{d^{2}y}{dx^{2}} \;=\; -\frac{10}{(10)^{2}(1/2)^{2}} \;=\; -\frac{10}{100 \times 1/4} \;=\; -\frac{10}{25} \;=\; -0.4.$$

Next we evaluate the first derivative at the required instant $$t = 1 \text{ s}$$:

$$\biggl(\frac{dy}{dx}\biggr)_{t=1} \;=\; \tan60^{\circ} - \frac{g\,(1)}{u\cos60^{\circ}} \;=\; \sqrt{3} - \frac{10}{10 \times \tfrac12} \;=\; \sqrt{3} - 2.$$

Because $$\sqrt{3} \approx 1.732$$, this numerically equals

$$\frac{dy}{dx}\;\approx\; 1.732 - 2 = -0.268.$$

Now we assemble the parts of the curvature formula. First,

$$1 + \bigl(dy/dx\bigr)^{2} = 1 + (-0.268)^{2} = 1 + 0.0717 = 1.0717.$$

Taking the power $$3/2$$:

$$\bigl(1 + (dy/dx)^{2}\bigr)^{3/2} = (1.0717)^{3/2} = \sqrt{1.0717}\;\times 1.0717 \approx 1.0355 \times 1.0717 \approx 1.110.$$

The denominator in the curvature expression is the magnitude of the second derivative, i.e. $$|d^{2}y/dx^{2}| = 0.4.$$

Thus the radius of curvature at $$t = 1 \text{ s}$$ becomes

$$R = \frac{1.110}{0.4} = 2.775 \text{ m} \;\approx\; 2.8 \text{ m}.$$

Hence, the correct answer is Option B.

We consider the usual equations for the motion of a projectile launched from ground level.

For an initial speed $$u$$ and an angle of projection $$\theta$$ with the horizontal, the time of flight is given by the well-known formula

$$t \;=\; \frac{2u\sin\theta}{g},$$

because the vertical component of velocity is $$u\sin\theta$$ and the projectile returns to the same vertical level after rising for $$u\sin\theta/g$$ seconds and falling for the same duration.

The horizontal range reached in that flight is, by the standard range formula,

$$R \;=\; \frac{u^{2}\sin2\theta}{g}.$$

We are told that the shell actually lands at the given distance $$R$$ from the gun, and that two different angles of projection are possible. Let those angles be $$\theta_{1}$$ and $$\theta_{2}$$, with the corresponding times of flight $$t_{1}$$ and $$t_{2}$$.

Since both angles produce the same range, we must have

$$\sin2\theta_{1} \;=\; \sin2\theta_{2} \;=\; \frac{gR}{u^{2}}.$$

Because $$\sin(180^{\circ}-\alpha)=\sin\alpha$$, the two values of $$2\theta$$ that give the same sine add up to $$180^{\circ}$$, so

$$2\theta_{1} + 2\theta_{2} = 180^{\circ} \;\;\Longrightarrow\;\; \theta_{1} + \theta_{2} = 90^{\circ}.$$

Now their times of flight, from the first formula, are

$$t_{1} = \frac{2u\sin\theta_{1}}{g}, \quad t_{2} = \frac{2u\sin\theta_{2}}{g}.$$

We multiply these two expressions to obtain the required product:

$$ t_{1}t_{2} = \left(\frac{2u\sin\theta_{1}}{g}\right)\!\!\left(\frac{2u\sin\theta_{2}}{g}\right) = \frac{4u^{2}}{g^{2}}\,\sin\theta_{1}\sin\theta_{2}. $$

Because $$\theta_{1}+\theta_{2}=90^{\circ}$$, we may write $$\sin\theta_{2}=\cos\theta_{1}$$, giving

$$\sin\theta_{1}\sin\theta_{2} = \sin\theta_{1}\cos\theta_{1} = \tfrac12\sin2\theta_{1}.$$

But from the range equality stated earlier,

$$\sin2\theta_{1} = \frac{gR}{u^{2}}.$$

Substituting this into the product of sines gives

$$\sin\theta_{1}\sin\theta_{2} = \tfrac12 \left(\frac{gR}{u^{2}}\right) = \frac{gR}{2u^{2}}.$$

Putting this back into the expression for $$t_{1}t_{2}$$, we have

$$ t_{1}t_{2} = \frac{4u^{2}}{g^{2}} \times \frac{gR}{2u^{2}} = \frac{4}{g^{2}}\times\frac{gR}{2} = \frac{2R}{g}. $$

Thus

$$t_{1}t_{2} = \frac{2R}{g}.$$

Hence, the correct answer is Option C.

The position of the particle is given as $$\vec r(t)=15t^{2}\hat i+(4-20t^{2})\hat j$$.

To obtain the acceleration we must first find the velocity by differentiating the position with respect to time. We state the basic relation $$\vec v(t)=\dfrac{d\vec r}{dt}.$$

Differentiating each component separately, we get

$$\vec v(t)=\dfrac{d}{dt}\bigl(15t^{2}\bigr)\hat i+\dfrac{d}{dt}\bigl(4-20t^{2}\bigr)\hat j.$$

Performing the derivatives term-by-term, we have

$$\dfrac{d}{dt}(15t^{2}) = 30t,$$

$$\dfrac{d}{dt}(4) = 0,$$

$$\dfrac{d}{dt}(-20t^{2}) = -40t.$$

Substituting these results back, the velocity becomes

$$\vec v(t)=\bigl(30t\bigr)\hat i+\bigl(-40t\bigr)\hat j=30t\hat i-40t\hat j.$$

Acceleration is the time derivative of velocity, expressed by the formula $$\vec a(t)=\dfrac{d\vec v}{dt}.$$

Again differentiating component-wise, we write

$$\vec a(t)=\dfrac{d}{dt}\bigl(30t\bigr)\hat i+\dfrac{d}{dt}\bigl(-40t\bigr)\hat j.$$

The derivatives are straightforward:

$$\dfrac{d}{dt}(30t)=30,$$

$$\dfrac{d}{dt}(-40t)=-40.$$

Hence the acceleration vector is

$$\vec a(t)=30\hat i-40\hat j.$$

We now evaluate this vector at the required instant, $$t=1$$. Since the vector is already independent of $$t$$, we simply have

$$\vec a(1)=30\hat i-40\hat j.$$

The magnitude of a vector $$\vec a=a_{x}\hat i+a_{y}\hat j$$ is obtained using the Pythagorean formula $$|\vec a|=\sqrt{a_{x}^{2}+a_{y}^{2}}.$$ Substituting $$a_{x}=30$$ and $$a_{y}=-40$$ we get

$$|\vec a(1)|=\sqrt{(30)^{2}+(-40)^{2}}=\sqrt{900+1600}=\sqrt{2500}=50.$$

So the magnitude of the acceleration at $$t=1$$ is $$50$$.

Hence, the correct answer is Option D.

We have a river whose water is flowing in a straight line with a velocity $$\vec v_r = 2\;{\rm km\,h^{-1}}$$. We may imagine this flow to be along the positive $$x$$-axis.

The swimmer is able to propel himself through still water with a speed $$\lvert \vec v_s\rvert = 4\;{\rm km\,h^{-1}}$$. He can choose any direction; let him make an angle $$\theta$$ with the direction of flow of the river. Thus, measured from the positive $$x$$-axis (down-stream direction), his velocity relative to the water can be written in component form as

$$ \vec v_s = 4\cos\theta\,\hat i + 4\sin\theta\,\hat j, $$

where $$\hat i$$ is the unit vector along the flow (east), and $$\hat j$$ is the unit vector pointing straight across the river towards the opposite bank (north).

The actual or ground velocity of the swimmer, which we shall denote by $$\vec v_{g}$$, is obtained by vector addition of the river velocity and the swimmer’s velocity relative to water:

$$ \vec v_{g} \;=\; \vec v_r + \vec v_s \;=\; (2\,\hat i) + \bigl(4\cos\theta\,\hat i + 4\sin\theta\,\hat j\bigr) \;=\; (2 + 4\cos\theta)\,\hat i + 4\sin\theta\,\hat j. $$

To cross the river straight—that is, to reach the point on the opposite bank directly opposite to his starting point—the swimmer’s resultant velocity must point purely across the river. In mathematical terms, the $$x$$-component of $$\vec v_{g}$$ must vanish:

$$ 2 + 4\cos\theta \;=\; 0. $$

Solving for $$\cos\theta$$ gives

$$ \cos\theta \;=\; -\,\frac{2}{4} \;=\; -\frac12. $$

We recall the standard trigonometric result that $$\cos120^\circ = -\dfrac12$$. Hence,

$$ \theta \;=\; 120^\circ. $$

This angle is measured from the direction of the river’s flow and points upstream (because the cosine is negative), exactly the orientation needed to cancel the down-stream drift.

So the swimmer must head at an angle of $$120^\circ$$ with respect to the flow of the river.

Hence, the correct answer is Option C.

Let us consider one of the guns first. A bullet is projected from ground level with an initial speed $$v$$ and makes an angle $$\theta$$ with the horizontal.

For a projectile that starts and finishes at the same height, the horizontal range is given by the well-known kinematic result

$$R \;=\; \frac{v^{2}\sin 2\theta}{g},$$

where $$g$$ denotes the magnitude of gravitational acceleration.

We are asked for the maximum horizontal distance the bullet can reach. The factor $$\sin 2\theta$$ attains its greatest value, namely $$1$$, when $$2\theta = 90^{\circ}$$, i.e. $$\theta = 45^{\circ}$$. Hence the greatest possible range for speed $$v$$ is

$$R_{\max} \;=\; \frac{v^{2}}{g}.$$

The bullet can be fired in any azimuthal direction. Therefore every direction in the horizontal plane is available, and the set of all landing points forms a full circle centred on the firing point with radius $$R_{\max}$$.

The area of this circle is thus

$$A \;=\; \pi R_{\max}^{2} \;=\; \pi \left( \frac{v^{2}}{g} \right)^{2} \;=\; \pi \frac{v^{4}}{g^{2}}.$$

Observe that $$\pi$$ and $$g$$ are the same for both guns, so the area is proportional to the fourth power of the muzzle speed:

$$A \;\propto\; v^{4}.$$

Now compare the two specific guns.

For gun A we have $$v_{A}=1\ \text{km/s},$$ so $$A_{A}\propto (1)^{4}=1.$$

For gun B we have $$v_{B}=2\ \text{km/s},$$ so $$A_{B}\propto (2)^{4}=16.$$

Taking the ratio of the two areas,

$$\frac{A_{A}}{A_{B}} = \frac{1}{16}.$$

Therefore the required ratio is $$1 : 16.$$

Hence, the correct answer is Option A.

Let the distance $$RM$$ be $$x$$. And let the distance $$QM$$ be $$L$$. We have, $$PR= \sqrt{x^2+d^2}$$

The total time taken to travel $$PR$$ will be $$\dfrac{\sqrt{x^2+d^2}}{v\div 2} = \dfrac{2\sqrt{x^2+d^2}}{v}$$

Time taken to cover the remaining distance $$L-x$$ from $$Q$$ to $$R$$ will be $$\dfrac{L-x}{v}$$

Thus, the total time,

$$T= \dfrac{L-x}{v} + \dfrac{2\sqrt{x^2+d^2}}{v}$$

$$\Rightarrow T = \dfrac{L-x+2\sqrt{x^2+d^2}}{v}$$

To minimise time, we will take the derivative of $$T$$ with respect to $$x$$, and equate it to $$0$$, we get,

$$\dfrac{dT}{dx} = \dfrac{d}{dx}\left[\dfrac{L-x + 2{(x^2+d^2)}^{1/2}}{v}\right]$$

$$\dfrac{dT}{dx} = 0 - \dfrac{1}{v} + \dfrac{2}{v}\cdot \dfrac{2x}{2}(x^2+d^2)^{-1/2}$$

$$\Rightarrow \dfrac{dT}{dx} = -\dfrac{1}{v}+ \dfrac{2x}{v}\cdot (x^2+d^2)^{-1/2}$$

Equating $$\dfrac{dT}{dx}$$ to zero to get the minimum value, we have,

$$\dfrac{2x}{v}\cdot (x^2+d^2)^{-1/2} = \dfrac{1}{v}$$

$$\dfrac{2x}{\sqrt{x^2+d^2}} = 1$$

$$\Rightarrow {(2x)}^2 = x^2+d^2$$

Which gives $$3x^2 = d^2$$  or  $$x=\dfrac{d}{\sqrt{3}}$$

A particle of mass $$ M $$ is moving in a circle of fixed radius $$ R $$ with centripetal acceleration given by $$ n^2 R t^2 $$, where $$ n $$ is a constant. We need to find the power delivered to the particle by the force acting on it.

First, recall that centripetal acceleration $$ a_c $$ is related to the tangential velocity $$ v $$ by the formula:

$$ a_c = \frac{v^2}{R} $$

Given that $$ a_c = n^2 R t^2 $$, we substitute:

$$ \frac{v^2}{R} = n^2 R t^2 $$

Solving for $$ v^2 $$, multiply both sides by $$ R $$:

$$ v^2 = n^2 R t^2 \times R $$

$$ v^2 = n^2 R^2 t^2 $$

Taking the square root of both sides (and considering speed as positive, so we take the positive root):

$$ v = n R t $$

Now, centripetal acceleration changes with time, which means the speed is changing. Therefore, there must be a tangential acceleration $$ a_t $$ responsible for changing the speed. Tangential acceleration is the derivative of velocity with respect to time:

$$ a_t = \frac{dv}{dt} $$

Substitute $$ v = n R t $$:

$$ a_t = \frac{d}{dt} (n R t) $$

Since $$ n $$ and $$ R $$ are constants:

$$ a_t = n R \frac{d}{dt}(t) = n R \times 1 = n R $$

The tangential force $$ F_t $$ is given by Newton's second law:

$$ F_t = M a_t = M \times n R = M n R $$

The centripetal force $$ F_c $$ is perpendicular to the velocity and does no work, so it delivers no power. Power is delivered only by the tangential force component, which is parallel to the velocity. Power $$ P $$ is the dot product of force and velocity:

$$ P = \vec{F} \cdot \vec{v} = F_t \times v \quad (\text{since they are parallel}) $$

Substitute $$ F_t = M n R $$ and $$ v = n R t $$:

$$ P = (M n R) \times (n R t) $$

$$ P = M n R \times n R t $$

$$ P = M n^2 R^2 t $$

Comparing with the options:

A. $$ \frac{1}{2} M n^2 R^2 t^2 $$

B. $$ M n^2 R^2 t $$

C. $$ M n R^2 t^2 $$

D. $$ M n R^2 t $$

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

We begin by noting that the new vector $$\vec B$$ is obtained from the old vector $$\vec A$$ purely by rotation through a very small angle $$\Delta\theta$$ (measured in radians). Rotation does not change magnitude, so we immediately have

$$|\vec A| = |\vec B|.$$

Our task is to find the magnitude of the difference vector $$\vec B-\vec A$$. The square of this magnitude can be expressed with the dot-product formula

$$|\vec B-\vec A|^{2} = (\vec B-\vec A)\cdot(\vec B-\vec A).$$

Expanding the right-hand side using distributivity of the dot product, we obtain

$$$ |\vec B-\vec A|^{2}= \vec B\cdot\vec B + \vec A\cdot\vec A - 2\,\vec A\cdot\vec B. $$$

Recognising that $$\vec B\cdot\vec B = |\vec B|^{2}$$ and $$\vec A\cdot\vec A = |\vec A|^{2}$$, and recalling that $$\vec A\cdot\vec B = |\vec A|\,|\vec B|\cos\Delta\theta,$$ we rewrite the expression as

$$$ |\vec B-\vec A|^{2}= |\vec B|^{2}+|\vec A|^{2}-2|\vec A|\,|\vec B|\cos\Delta\theta. $$$

Because $$|\vec A|=|\vec B|,$$ let us denote this common magnitude simply by $$|\vec A|$$. Substituting, we have

$$$ |\vec B-\vec A|^{2}= |\vec A|^{2}+|\vec A|^{2}-2|\vec A|^{2}\cos\Delta\theta = 2|\vec A|^{2}\bigl(1-\cos\Delta\theta\bigr). $$$

Now we employ the small-angle approximation for cosine. For very small $$\Delta\theta$$ (in radians), the Taylor series gives

$$$ \cos\Delta\theta \approx 1-\frac{(\Delta\theta)^{2}}{2}. $$$

Substituting this approximation into the previous expression, we find

$$$ |\vec B-\vec A|^{2} \approx 2|\vec A|^{2}\left[1-\left(1-\frac{(\Delta\theta)^{2}}{2}\right)\right] = 2|\vec A|^{2}\left[\frac{(\Delta\theta)^{2}}{2}\right] = |\vec A|^{2}(\Delta\theta)^{2}. $$$

Taking the square root of both sides (and keeping only the positive root because a magnitude is always positive) gives

$$$ |\vec B-\vec A| \approx |\vec A|\,\Delta\theta. $$$

This expression is exactly Option C in the list provided. None of the other options match the derived result.

Hence, the correct answer is Option C.

The formula for centripetal acceleration is $$a = \frac{v^2}{r}$$

Since $$v^2$$ is constant, the relationship becomes $$a \propto \frac{1}{r}$$

This represents a rectangular hyperbola.

Option (D) represents this correctly.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Rearranging,

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

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

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

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

$$t_1 + t_2 = n\,t_1

\;\;\Longrightarrow\;\;

t_2 = (n-1)t_1.$$

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

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

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

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

Squaring both sides to eliminate the square root,

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

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

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

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

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

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

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

Expanding and simplifying the bracket,

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

so that

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

Factoring $$n$$ out of the bracket,

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

Finally writing the result in a compact symmetric form,

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

This is exactly the relation given in Option C.

Hence, the correct answer is Option C.

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

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

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

Simplifying the fraction:

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

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

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

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

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

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

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

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

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

Now, multiply:

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

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

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

Simplify by dividing numerator and denominator by 2:

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

Now, compute the numerical value:

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

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

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

The value 6.05 m is closest to 6.1 m.

Hence, the correct answer is Option C.

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

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

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

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

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

$$ 40 = u_x \cdot 2 $$

Solving for $$ u_x $$:

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

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

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

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

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

So the equation becomes:

$$ 50 = 2u_y - 20 $$

Add 20 to both sides:

$$ 50 + 20 = 2u_y $$

$$ 70 = 2u_y $$

Solving for $$ u_y $$:

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

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

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

Substituting the values:

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

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

Comparing with the options:

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

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

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

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

Option C matches our result.

Hence, the correct answer is Option C.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Alternatively, using exact fractions:

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

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

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

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

Comparing with the options:

A. 1.73 kg m² s⁻¹

B. 3.46 kg m² s⁻¹

C. 3.0 kg m² s⁻¹

D. 6.0 kg m² s⁻¹

Hence, the correct answer is Option C.

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

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

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

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

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

Given R = 10 m:

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

Solving for u²:

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

The trajectory equation for a projectile is:

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

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

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

Simplifying:

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

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

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

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

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

The g cancels out:

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

Simplifying the fraction:

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

Thus:

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

Therefore, the height of the wall is 2.4 m.

Comparing with the options:

A. 4.4 m

B. 2.4 m

C. 3.6 m

D. 1.6 m

Hence, the correct answer is Option B.

We are told that the projectile is launched from the origin with the velocity vector

$$\vec u \;=\; 1\,\hat i \;+\; 2\,\hat j \quad\text{m s}^{-1}.$$

Here $$\hat i$$ is the horizontal (ground) unit vector and $$\hat j$$ is the vertical upward unit vector. Hence the horizontal and vertical components of the initial velocity are

$$u_x = 1\ \text{m s}^{-1}, \qquad u_y = 2\ \text{m s}^{-1}.$$

For a projectile that starts from the origin, the general equation of the trajectory (the path in the $$x\!-\!y$$ plane) is obtained from kinematics. First we write the two independent component equations of motion:

Horizontal motion (no acceleration horizontally):

$$x = u_x\,t.$$

Vertical motion (constant downward acceleration $$g$$):

$$y = u_y\,t - \dfrac{1}{2}g\,t^2.$$

To eliminate the time $$t$$, we first solve the horizontal equation for $$t$$:

$$t = \dfrac{x}{u_x}.$$

Now we substitute this value of $$t$$ into the vertical equation:

$$$ \begin{aligned} y &= u_y\left(\dfrac{x}{u_x}\right) \;-\; \dfrac{1}{2}g\left(\dfrac{x}{u_x}\right)^2 \\[6pt] &= \dfrac{u_y}{u_x}\,x \;-\; \dfrac{g}{2u_x^{\,2}}\,x^2. \end{aligned} $$$

This is the standard textbook form

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

but expressed directly through the components $$u_x, u_y$$ it reads

$$y = \dfrac{u_y}{u_x}\,x \;-\; \dfrac{g}{2u_x^{\,2}}\,x^2.$$

Now we plug in the numerical values $$u_x = 1\ \text{m s}^{-1},\; u_y = 2\ \text{m s}^{-1},\; g = 10\ \text{m s}^{-2}:$$

$$$ \begin{aligned} y &= \dfrac{2}{1}\,x \;-\; \dfrac{10}{2\,(1)^2}\,x^2 \\[6pt] &= 2x \;-\; \dfrac{10}{2}\,x^2 \\[6pt] &= 2x \;-\; 5x^2. \end{aligned} $$$

Thus the explicit equation of the trajectory is

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

Comparing this with the given options, we see it matches option D.

Hence, the correct answer is Option D.

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

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

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

$$ u^2 = 400 $$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Differentiate with respect to $$ \theta $$:

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

Set the derivative to zero for maximum:

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

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

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

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

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

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

Solve the quadratic equation:

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

So:

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

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

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

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

Hence, the correct answer is Option B.

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

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

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

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

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

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

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

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

This simplifies to:

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

Equating components:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

From the original range equation:

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

So:

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

Substituting this into the expression for d:

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

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

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

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

So, the answer is Option C.