Horizontal Projectile Motion Questions in English

(C) The time taken by the food packet to reach the ground is given by $t = \sqrt{\frac{2h}{g}}$.
The horizontal distance (range) covered by the packet during this time is $R = v \cdot t = v \sqrt{\frac{2h}{g}}$.
When the packet is dropped,the helicopter is at a horizontal distance $R$ and vertical distance $h$ from the man.
The distance $D$ of the helicopter from the man is the hypotenuse of the right-angled triangle formed by $R$ and $h$:
$D = \sqrt{R^{2} + h^{2}}$
Substituting the value of $R$:
$D = \sqrt{\left(v \sqrt{\frac{2h}{g}}\right)^{2} + h^{2}}$
$D = \sqrt{\frac{2v^{2}h}{g} + h^{2}}$

(C) $1$. First,calculate the velocity $v$ of the ball as it leaves the spring gun using the principle of conservation of mechanical energy:
$\frac{1}{2} k x^2 = \frac{1}{2} m v^2$
Given $k = 100\, \text{N/m}$,$x = 0.05\, \text{m}$,and $m = 100\, \text{g} = 0.1\, \text{kg}$.
$100 \times (0.05)^2 = 0.1 \times v^2$
$100 \times 0.0025 = 0.1 \times v^2$
$0.25 = 0.1 \times v^2$
$v^2 = 2.5$
$v = \sqrt{2.5}\, \text{m/s}$.
$2$. Next,calculate the time $t$ taken by the ball to reach the ground from a height $h = 2\, \text{m}$ using the equation of motion for horizontal projection:
$h = \frac{1}{2} g t^2$
$2 = \frac{1}{2} \times 10 \times t^2$
$2 = 5 t^2$
$t^2 = 0.4$
$t = \sqrt{0.4}\, \text{s}$.
$3$. Finally,calculate the horizontal distance $d$:
$d = v \times t$
$d = \sqrt{2.5} \times \sqrt{0.4}$
$d = \sqrt{2.5 \times 0.4}$
$d = \sqrt{1} = 1\, \text{m}$.
Thus,the value of $d$ is $1\, \text{m}$.

(A) The angle of projection with the horizontal is $\theta = 90^{\circ} - 60^{\circ} = 30^{\circ}$.
At the highest point of the trajectory,the vertical component of velocity becomes zero,and only the horizontal component of velocity remains.
The horizontal component of velocity is given by $v_x = u \cos \theta$.
Given $u = 10 \ ms^{-1}$ and $\theta = 30^{\circ}$,we have:
$v_x = 10 \cos 30^{\circ} = 10 \times \frac{\sqrt{3}}{2} = 5 \sqrt{3} \ ms^{-1}$.
Thus,the speed at the highest point is $5 \sqrt{3} \ ms^{-1}$.

(A) Let the initial velocity be $u$. The horizontal and vertical components of velocity at any time $t$ are given by:
$v_x = u \cos 45^{\circ} = \frac{u}{\sqrt{2}}$
$v_y = u \sin 45^{\circ} - gt = \frac{u}{\sqrt{2}} - 10(2) = \frac{u}{\sqrt{2}} - 20$
Given that the resultant velocity at $t = 2 \, s$ is $20 \, m/s$,we have:
$v^2 = v_x^2 + v_y^2$
$20^2 = \left(\frac{u}{\sqrt{2}}\right)^2 + \left(\frac{u}{\sqrt{2}} - 20\right)^2$
$400 = \frac{u^2}{2} + \frac{u^2}{2} - 20\sqrt{2}u + 400$
$0 = u^2 - 20\sqrt{2}u$
Since $u \neq 0$,we get $u = 20\sqrt{2} \, m/s$.
The maximum height $H$ is given by:
$H = \frac{u^2 \sin^2 45^{\circ}}{2g} = \frac{(20\sqrt{2})^2 \times (1/2)}{2 \times 10} = \frac{800 \times 0.5}{20} = \frac{400}{20} = 20 \, m$.

(D) The horizontal range $R$ is given by $R = \frac{V^2 \sin(2\theta)}{g} = \frac{2V^2 \sin\theta \cos\theta}{g}$.
The time taken to reach the highest point (where inclination is zero) is $t = \frac{V \sin\theta}{g}$.
From this,$V \sin\theta = gt$,so $V = \frac{gt}{\sin\theta}$.
Substitute $V$ into the range formula: $R = \frac{2(gt/\sin\theta)^2 \sin\theta \cos\theta}{g} = \frac{2g^2 t^2 \sin\theta \cos\theta}{g \sin^2\theta} = \frac{2gt^2 \cos\theta}{\sin\theta} = \frac{2gt^2}{\tan\theta}$.
Given $g = 10 \, m/s^2$,we have $R = \frac{2(10)t^2}{\tan\theta} = \frac{20t^2}{\tan\theta}$.
Rearranging for $\theta$,we get $\tan\theta = \frac{20t^2}{R}$,which implies $\cot\theta = \frac{R}{20t^2}$.
Therefore,$\theta = \cot^{-1}\left(\frac{R}{20t^2}\right)$.

(A) For the same range,the angles of projection must satisfy $\theta_{1} + \theta_{2} = 90^{\circ}$,which means $\theta_{2} = 90^{\circ} - \theta_{1}$.
The maximum heights are given by $h_{1} = \frac{u^{2} \sin^{2} \theta_{1}}{2g}$ and $h_{2} = \frac{u^{2} \sin^{2} \theta_{2}}{2g} = \frac{u^{2} \cos^{2} \theta_{1}}{2g}$.
Multiplying these heights,we get $h_{1} h_{2} = \left(\frac{u^{2} \sin^{2} \theta_{1}}{2g}\right) \cdot \left(\frac{u^{2} \cos^{2} \theta_{1}}{2g}\right)$.
This can be rewritten as $h_{1} h_{2} = \frac{u^{4} \sin^{2} \theta_{1} \cos^{2} \theta_{1}}{4g^{2}} = \left(\frac{u^{2} \cdot 2 \sin \theta_{1} \cos \theta_{1}}{4g}\right)^{2} = \left(\frac{u^{2} \sin(2\theta_{1})}{4g}\right)^{2}$.
Since the range $R = \frac{u^{2} \sin(2\theta_{1})}{g}$,we have $h_{1} h_{2} = \left(\frac{R}{4}\right)^{2} = \frac{R^{2}}{16}$.
Therefore,$R^{2} = 16 h_{1} h_{2}$,which implies $R = 4 \sqrt{h_{1} h_{2}}$.
Both Assertion $A$ and Reason $R$ are true,and $R$ provides the correct mathematical derivation for $A$.

(C) For the bullet to hit the fighter jet,the horizontal component of the bullet's velocity must be equal to the horizontal velocity of the jet so that the bullet stays directly under the jet throughout its flight.
Let $v_j = 200 \; m/s$ be the speed of the jet and $v_b = 400 \; m/s$ be the speed of the bullet.
The horizontal component of the bullet's velocity is given by $v_{bx} = v_b \cos \theta$.
Equating the horizontal components: $v_j = v_b \cos \theta$.
Substituting the values: $200 = 400 \cos \theta$.
Therefore,$\cos \theta = \frac{200}{400} = 0.5$.
Thus,$\theta = \cos^{-1}(0.5) = 60^\circ$.

(B) The maximum horizontal range $R_{\max}$ of a projectile is given by the formula $R_{\max} = \frac{u^2}{g}$,where $u$ is the initial velocity and $g$ is the acceleration due to gravity.
Given $R_{\max} = 100 \, m$,we have $\frac{u^2}{g} = 100 \, m$.
The maximum height $H_{\max}$ reached by a projectile when thrown vertically upwards is given by $H_{\max} = \frac{u^2}{2g}$.
Substituting the value of $\frac{u^2}{g}$ into the height formula:
$H_{\max} = \frac{1}{2} \times \left(\frac{u^2}{g}\right) = \frac{1}{2} \times 100 \, m = 50 \, m$.
Therefore,the person can throw the ball to a maximum height of $50 \, m$.

(D) The horizontal range $R$ of a projectile is given by $R = \frac{u^2 \sin(2\theta)}{g} = \frac{2u^2 \sin\theta \cos\theta}{g}$.
The maximum height $H$ of a projectile is given by $H = \frac{u^2 \sin^2\theta}{2g}$.
According to the problem,the range is equal to the maximum height,so $R = H$.
Substituting the formulas: $\frac{2u^2 \sin\theta \cos\theta}{g} = \frac{u^2 \sin^2\theta}{2g}$.
Canceling $u^2/g$ from both sides: $2 \sin\theta \cos\theta = \frac{\sin^2\theta}{2}$.
Dividing both sides by $\sin\theta$ (assuming $\sin\theta \neq 0$): $2 \cos\theta = \frac{\sin\theta}{2}$.
Rearranging to solve for $\tan\theta = \frac{\sin\theta}{\cos\theta}$: $\tan\theta = 4$.

(C) The horizontal range $R$ of a projectile is given by the formula $R = \frac{u^2 \sin(2\theta)}{g}$,where $u$ is the initial velocity,$\theta$ is the angle of projection,and $g$ is the acceleration due to gravity.
Given $u$ is the same for both projectiles,the ratio of their ranges is $\frac{R_1}{R_2} = \frac{\sin(2\theta_1)}{\sin(2\theta_2)}$.
For $\theta_1 = 45^{\circ}$,$2\theta_1 = 90^{\circ}$,so $\sin(90^{\circ}) = 1$.
For $\theta_2 = 30^{\circ}$,$2\theta_2 = 60^{\circ}$,so $\sin(60^{\circ}) = \frac{\sqrt{3}}{2}$.
Therefore,$\frac{R_1}{R_2} = \frac{1}{\sqrt{3}/2} = \frac{2}{\sqrt{3}}$.
Thus,the ratio is $2: \sqrt{3}$.

(C) The time taken to reach the maximum height for a projectile is given by the formula:
$t = \frac{u \sin \theta}{g}$
Given that both projectiles reach the maximum height in the same time,we have:
$t_1 = t_2$
$\frac{u_1 \sin \theta_1}{g} = \frac{u_2 \sin \theta_2}{g}$
Substituting the given angles $\theta_1 = 30^{\circ}$ and $\theta_2 = 45^{\circ}$:
$u_1 \sin 30^{\circ} = u_2 \sin 45^{\circ}$
$u_1 \left( \frac{1}{2} \right) = u_2 \left( \frac{1}{\sqrt{2}} \right)$
Rearranging to find the ratio $\frac{u_1}{u_2}$:
$\frac{u_1}{u_2} = \frac{1/\sqrt{2}}{1/2} = \frac{2}{\sqrt{2}} = \sqrt{2}$
Therefore,the ratio of their initial velocities is $\sqrt{2}: 1$.

(C) The initial horizontal velocity is given as $u_x = 1\,m/s$.
The equation of the trajectory is $y = 5x - 5x^2$.
Differentiating the equation with respect to time $t$,we get the vertical velocity component $v_y = \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$.
First,find $\frac{dy}{dx} = \frac{d}{dx}(5x - 5x^2) = 5 - 10x$.
Since $v_x = \frac{dx}{dt} = u_x = 1$ (assuming constant horizontal velocity),we have $v_y = (5 - 10x) \cdot 1$.
At the initial point,$x = 0$.
Therefore,the initial vertical velocity $u_y = (5 - 10(0)) = 5\,m/s$.
Thus,the $y$-component vector of the initial velocity is $5\,\hat{j}$.

(D) The equation of trajectory is $y = x \tan \theta - \frac{gx^2}{2u^2 \cos^2 \theta}$.
Given $\theta = 45^{\circ}$,$x = 20$,$y = 10$,and $g = 10 \ m/s^2$.
$10 = 20(1) - \frac{10(20)^2}{2u^2(1/2)} \Rightarrow 10 = 20 - \frac{4000}{u^2} \Rightarrow \frac{4000}{u^2} = 10 \Rightarrow u^2 = 400 \Rightarrow u = 20 \ m/s$.
The time of flight $T = \frac{2u \sin \theta}{g} = \frac{2(20)(1/\sqrt{2})}{10} = 2\sqrt{2} \ s$.
At time $t = \frac{T}{\sqrt{2}} = \frac{2\sqrt{2}}{\sqrt{2}} = 2 \ s$.
The velocity components are $v_x = u \cos \theta = 20(1/\sqrt{2}) = 10\sqrt{2} \ m/s$ and $v_y = u \sin \theta - gt = 20(1/\sqrt{2}) - 10(2) = 10\sqrt{2} - 20 \ m/s$.
The momentum vector $\vec{p} = m\vec{v} = 10(10\sqrt{2} \hat{i} + (10\sqrt{2} - 20) \hat{j}) = 100\sqrt{2} \hat{i} + (100\sqrt{2} - 200) \hat{j} \ kg \cdot m/s$.

(C) The initial kinetic energy of the ball is given by $E = \frac{1}{2} mu^2$, where $u$ is the initial velocity.
At the highest point of the trajectory, the vertical component of velocity becomes zero, and the ball only possesses the horizontal component of velocity.
The horizontal component of velocity at the highest point is $v_x = u \cos 60^{\circ} = u \times \frac{1}{2} = \frac{u}{2}$.
Therefore, the kinetic energy at the highest point is $E' = \frac{1}{2} m v_x^2 = \frac{1}{2} m \left(\frac{u}{2}\right)^2$.
$E' = \frac{1}{2} m \frac{u^2}{4} = \frac{1}{4} \left(\frac{1}{2} mu^2\right) = \frac{E}{4}$.

(C) The horizontal range $R$ is given by $R = \frac{u^2 \sin 2\theta}{g}$.
For maximum range,$\theta = 45^{\circ}$,so $R_{\max} = \frac{u^2}{g}$.
For the second object,the range is $R' = \frac{R_{\max}}{2} = \frac{u^2}{2g}$.
Equating the range formula for the second object: $\frac{u^2 \sin 2\theta'}{g} = \frac{u^2}{2g}$.
This simplifies to $\sin 2\theta' = \frac{1}{2}$.
Therefore,$2\theta' = 30^{\circ}$ or $2\theta' = 150^{\circ}$.
Solving for $\theta'$,we get $\theta' = 15^{\circ}$ or $\theta' = 75^{\circ}$.

(D) Given,distance of landing from the building,$s = 12 \, m$.
Height of each floor,$h = 3 \, m$.
Total height of the building,$H = 15 \times 3 = 45 \, m$.
Using the second equation of motion for vertical motion under gravity (with initial vertical velocity $u_y = 0$):
$H = u_y t + \frac{1}{2} g t^2$
$45 = 0 + \frac{1}{2} \times 10 \times t^2$
$45 = 5 t^2$
$t^2 = 9 \Rightarrow t = 3 \, s$.
Now,for horizontal motion with constant speed $v$:
$s = v \times t$
$12 = v \times 3$
$v = 4 \, m/s$.
To convert $m/s$ to $km/h$,multiply by $\frac{18}{5}$:
$v = 4 \times \frac{18}{5} = \frac{72}{5} = 14.4 \, km/h$.
Rounding to the nearest integer,the speed is $15 \, km/h$.

(B) Since the ranges are equal for the same projection velocity,the angles of projection must be complementary.
Let the angles be $\theta_1$ and $\theta_2$. Then $\theta_1 + \theta_2 = 90^{\circ}$.
Given $\theta_1 = 30^{\circ}$,therefore $\theta_2 = 90^{\circ} - 30^{\circ} = 60^{\circ}$.
The formula for the maximum height of a projectile is $H = \frac{u^2 \sin^2 \theta}{2g}$.
For the first ball,$h = \frac{u^2 \sin^2(30^{\circ})}{2g} = \frac{u^2}{2g} \left(\frac{1}{2}\right)^2 = \frac{u^2}{8g}$.
For the second ball,$H_2 = \frac{u^2 \sin^2(60^{\circ})}{2g} = \frac{u^2}{2g} \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{u^2}{2g} \cdot \frac{3}{4} = \frac{3u^2}{8g}$.
Comparing the two,$H_2 = 3 \left(\frac{u^2}{8g}\right) = 3h$.

(A) In time $t$,a particle of mass $m$ falls by a vertical distance $h$ under gravity,given by the equation of motion:
$h = \frac{1}{2} g t^2$
Now,the distance covered horizontally is $d$ and the speed of the photon in the horizontal direction is $c$. The time taken to travel this horizontal distance $d$ is:
$t = \frac{d}{c}$
Substituting the value of $t$ into the equation for vertical fall $h$,we get:
$h = \frac{1}{2} g \left( \frac{d}{c} \right)^2$
$h = \frac{g d^2}{2 c^2}$
Thus,the photon falls through a vertical distance of $\frac{g d^2}{2 c^2}$.

(B) The time taken to reach the maximum height is known as the time of ascent $(t_a = 4 \, s)$.
For a projectile motion under gravity,the time of descent $(t_d)$ is equal to the time of ascent $(t_a)$ if the point of projection and the point of landing are at the same horizontal level.
Therefore,$t_d = 4 \, s$.
The total time of flight $(T)$ is the sum of the time of ascent and the time of descent:
$T = t_a + t_d = 4 \, s + 4 \, s = 8 \, s$.

(C) The formula for the time of flight $T$ of a projectile is given by $T = \frac{2v \sin \theta}{g}$.
Given: Initial speed $v = 20 \, m/s$,angle of projection $\theta = 30^{\circ}$,and acceleration due to gravity $g = 10 \, m/s^2$.
Substituting the values into the formula:
$T = \frac{2 \times 20 \times \sin 30^{\circ}}{10}$
Since $\sin 30^{\circ} = 0.5$ or $\frac{1}{2}$:
$T = \frac{2 \times 20 \times 0.5}{10} = \frac{20}{10} = 2 \, s$.
Therefore,the correct option is $C$.

(C) The formula for the maximum height $H$ of a projectile is given by $H = \frac{u^2 \sin^2 \theta}{2g}$.
Given: $u = 20 \, m/s$,$H = 10 \, m$,and taking $g = 10 \, m/s^2$.
Substituting the values into the formula:
$10 = \frac{(20)^2 \sin^2 \theta}{2 \times 10}$
$10 = \frac{400 \sin^2 \theta}{20}$
$10 = 20 \sin^2 \theta$
$\sin^2 \theta = \frac{10}{20} = \frac{1}{2}$
$\sin \theta = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$
Therefore,$\theta = 45^{\circ}$.

(A) The initial velocity of the body is $\vec{u} = u \cos \theta \hat{i} + u \sin \theta \hat{j}$.
At the end of the flight (when it returns to the ground),the velocity is $\vec{v} = u \cos \theta \hat{i} - u \sin \theta \hat{j}$.
The change in momentum is $\Delta \vec{p} = m\vec{v} - m\vec{u} = m(\vec{v} - \vec{u})$.
$\Delta \vec{p} = m(u \cos \theta \hat{i} - u \sin \theta \hat{j} - (u \cos \theta \hat{i} + u \sin \theta \hat{j}))$.
$\Delta \vec{p} = m(-2u \sin \theta \hat{j})$.
The magnitude of the change in momentum is $|\Delta \vec{p}| = 2mu \sin \theta$.
Given $m = 1 \, kg$,$u = 50 \, m/s$,and $\theta = 30^{\circ}$.
$|\Delta \vec{p}| = 2 \times 1 \times 50 \times \sin 30^{\circ} = 100 \times 0.5 = 50 \, kg \cdot m/s$.

(D) The shell is fired vertically upwards with velocity $v_1$ from a trolley moving horizontally with velocity $v_2$. Therefore,the initial velocity of the shell relative to the ground has a horizontal component $u_x = v_2$ and a vertical component $u_y = v_1$.
In the horizontal direction,there is no acceleration $(a_x = 0)$.
In the vertical direction,the acceleration is due to gravity $(a_y = -g)$.
The time of flight $T$ is determined by the vertical motion. The shell returns to the level of the trolley when the vertical displacement is zero:
$y = u_y T - \frac{1}{2} g T^2 = 0$
$v_1 T - \frac{1}{2} g T^2 = 0$
$T = \frac{2 v_1}{g}$
The horizontal range $R$ is the horizontal distance covered during the time of flight $T$:
$R = u_x \times T$
$R = v_2 \times \frac{2 v_1}{g} = \frac{2 v_1 v_2}{g}$
Thus,the correct option is $(d)$.

(D) The initial velocity components are $u_x = u \cos \theta = 80 \cos 30^{\circ} = 80 \times \frac{\sqrt{3}}{2} = 40 \sqrt{3} \,m/s$ and $u_y = u \sin \theta = 80 \sin 30^{\circ} = 80 \times \frac{1}{2} = 40 \,m/s$.
The position of the particle at any time $t$ is given by $x(t) = u_x t$ and $y(t) = u_y t - \frac{1}{2} g t^2$.
At $t_1 = 2 \,s$:
$x_1 = (40 \sqrt{3}) \times 2 = 80 \sqrt{3} \,m$
$y_1 = 40 \times 2 - \frac{1}{2} \times 10 \times (2)^2 = 80 - 20 = 60 \,m$
At $t_2 = 6 \,s$:
$x_2 = (40 \sqrt{3}) \times 6 = 240 \sqrt{3} \,m$
$y_2 = 40 \times 6 - \frac{1}{2} \times 10 \times (6)^2 = 240 - 180 = 60 \,m$
Since $y_1 = y_2$,the displacement vector is $\Delta \vec{r} = (x_2 - x_1) \hat{i} + (y_2 - y_1) \hat{j} = (240 \sqrt{3} - 80 \sqrt{3}) \hat{i} + 0 \hat{j} = 160 \sqrt{3} \hat{i} \,m$.
The time interval is $\Delta t = 6 - 2 = 4 \,s$.
The average velocity is $\vec{v}_{avg} = \frac{\Delta \vec{r}}{\Delta t} = \frac{160 \sqrt{3}}{4} \hat{i} = 40 \sqrt{3} \hat{i} \,m/s$.
The magnitude of the average velocity is $40 \sqrt{3} \,m/s$.

(D) The maximum vertical height $H$ attained by a projectile is given by the formula $H = \frac{u^2 \sin^2 \theta}{2g}$.
Given that both objects attain the same vertical height,we have $H_1 = H_2$.
Substituting the formula,we get $\frac{u_1^2 \sin^2 \theta_1}{2g} = \frac{u_2^2 \sin^2 \theta_2}{2g}$.
Canceling $2g$ from both sides,we get $u_1^2 \sin^2 \theta_1 = u_2^2 \sin^2 \theta_2$.
Given $\theta_1 = 45^{\circ}$ and $\theta_2 = 60^{\circ}$,we have $u_1^2 \sin^2 45^{\circ} = u_2^2 \sin^2 60^{\circ}$.
Substituting the values $\sin 45^{\circ} = \frac{1}{\sqrt{2}}$ and $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$,we get $u_1^2 \left(\frac{1}{\sqrt{2}}\right)^2 = u_2^2 \left(\frac{\sqrt{3}}{2}\right)^2$.
This simplifies to $u_1^2 \left(\frac{1}{2}\right) = u_2^2 \left(\frac{3}{4}\right)$.
Rearranging for the ratio $\frac{u_1^2}{u_2^2}$,we get $\frac{u_1^2}{u_2^2} = \frac{3/4}{1/2} = \frac{3}{4} \times 2 = \frac{3}{2}$.
Taking the square root on both sides,the ratio of the velocities is $\frac{u_1}{u_2} = \sqrt{\frac{3}{2}}$.

(D) Given that the horizontal range $R$ is twice the maximum height $H$,so $R = 2H$.
We know the formulas for horizontal range and maximum height are $R = \frac{u^2 \sin(2\theta)}{g}$ and $H = \frac{u^2 \sin^2 \theta}{2g}$.
Substituting these into the given condition $R = 2H$:
$\frac{u^2 (2 \sin \theta \cos \theta)}{g} = 2 \left( \frac{u^2 \sin^2 \theta}{2g} \right)$
$2 \sin \theta \cos \theta = \sin^2 \theta$
Dividing both sides by $\sin \theta$ (assuming $\sin \theta \neq 0$):
$2 \cos \theta = \sin \theta \Rightarrow \tan \theta = 2$.
From the triangle with $\tan \theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{2}{1}$,the hypotenuse is $\sqrt{2^2 + 1^2} = \sqrt{5}$.
Thus,$\sin \theta = \frac{2}{\sqrt{5}}$ and $\cos \theta = \frac{1}{\sqrt{5}}$.
Now,calculate the range $R$:
$R = \frac{2 u^2 \sin \theta \cos \theta}{g} = \frac{2 u^2}{g} \left( \frac{2}{\sqrt{5}} \right) \left( \frac{1}{\sqrt{5}} \right) = \frac{4 u^2}{5g}$.

(A) At the maximum height,the vertical component of velocity is zero,so the velocity of the projectile is equal to its horizontal component,$v = u \cos \theta$.
Given that the velocity at the maximum height is $\frac{\sqrt{3}}{2}$ times the initial velocity $u$,we have:
$u \cos \theta = \frac{\sqrt{3}}{2} u$
$\cos \theta = \frac{\sqrt{3}}{2}$
$\theta = 30^{\circ}$
The horizontal range $R$ of a projectile is given by the formula:
$R = \frac{u^2 \sin 2\theta}{g}$
Substituting $\theta = 30^{\circ}$ into the formula:
$R = \frac{u^2 \sin(2 \times 30^{\circ})}{g} = \frac{u^2 \sin 60^{\circ}}{g}$
Since $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$,we get:
$R = \frac{u^2 \times \frac{\sqrt{3}}{2}}{g} = \frac{\sqrt{3} u^2}{2 g}$
Thus,the correct option is $A$.

(B) For a projectile,the maximum horizontal range $R_{\max}$ is given by $R_{\max} = \frac{u^2}{g}$.
Given $R_{\max} = 400 \, m$.
The velocity of a projectile is minimum at its highest point.
At the highest point,the horizontal distance covered is $x = \frac{R}{2} = \frac{400}{2} = 200 \, m$.
The maximum height $H$ is given by $H = \frac{u^2 \sin^2 \theta}{2g}$. For maximum range,$\theta = 45^\circ$,so $H = \frac{u^2 \sin^2 45^\circ}{2g} = \frac{u^2 (1/2)}{2g} = \frac{u^2}{4g} = \frac{R_{\max}}{4}$.
$H = \frac{400}{4} = 100 \, m$.
Thus,the coordinates of the highest point are $(x, H) = (200, 100) \, m$.

(A) The time of flight $T$ for a projectile is given by $T = \frac{2 u \sin \theta}{g}$,where $u$ is the initial velocity and $\theta$ is the angle of projection.
From this,we can express the initial velocity as $u = \frac{g T}{2 \sin \theta}$.
The horizontal range $R$ is given by $R = \frac{u^2 \sin(2 \theta)}{g} = \frac{2 u^2 \sin \theta \cos \theta}{g}$.
Substituting $u$ into the range formula: $R = \frac{2}{g} \left(\frac{g T}{2 \sin \theta}\right)^2 \sin \theta \cos \theta$.
Simplifying the expression: $R = \frac{2}{g} \cdot \frac{g^2 T^2}{4 \sin^2 \theta} \cdot \sin \theta \cos \theta$.
$R = \frac{g T^2}{2} \cdot \frac{\cos \theta}{\sin \theta} = \frac{g T^2}{2 \tan \theta}$.
Rearranging for $\tan \theta$: $\tan \theta = \frac{g T^2}{2 R}$.
Therefore,the angle of projection is $\theta = \tan^{-1}\left(\frac{g T^2}{2 R}\right)$.

(C) The graph shows a constant value over time $t$.
In projectile motion, the horizontal component of velocity $(v_x)$ remains constant throughout the motion because there is no acceleration in the horizontal direction $(a_x = 0)$.
Therefore, $v_x = u_x = \text{constant}$.
Other quantities like kinetic energy, momentum, and vertical velocity change with time.
Thus, the quantity plotted along the $y$-axis is the horizontal velocity.

(A) The equation of the trajectory of a projectile is given by $y = ax - bx^2$.
When the projectile lands on the horizontal plane,its vertical displacement $y$ becomes $0$.
At this point,the horizontal distance $x$ is equal to the horizontal range $R$.
Substituting $y = 0$ and $x = R$ into the equation,we get:
$0 = aR - bR^2$
$aR = bR^2$
Since $R \neq 0$ (as it is the range),we can divide both sides by $R$:
$a = bR$
$R = \frac{a}{b}$
Thus,the horizontal range is $\frac{a}{b}$.

(C) The maximum height of a projectile is given by $H = \frac{v^2 \sin^2 \theta}{2g}$.
We are looking for the vertical component of velocity $v_y$ at a height $h = \frac{H}{2} = \frac{v^2 \sin^2 \theta}{4g}$.
Using the kinematic equation $v_y^2 = u_y^2 - 2gh$,where $u_y = v \sin \theta$ is the initial vertical velocity:
$v_y^2 = (v \sin \theta)^2 - 2g \left( \frac{v^2 \sin^2 \theta}{4g} \right)$
$v_y^2 = v^2 \sin^2 \theta - \frac{v^2 \sin^2 \theta}{2}$
$v_y^2 = \frac{v^2 \sin^2 \theta}{2}$
Taking the square root of both sides,we get:
$v_y = \frac{v \sin \theta}{\sqrt{2}}$
Thus,the correct option is $(C)$.

(B) The equation of the trajectory of a projectile is given by:
$y = x \tan \theta \left(1 - \frac{x}{R}\right)$
where $R$ is the horizontal range.
From the figure,the total range $R = x_1 + x_2$.
At point $P$,the horizontal distance is $x = x_1$.
Substituting these values into the trajectory equation:
$y = x_1 \tan \theta \left(1 - \frac{x_1}{x_1 + x_2}\right)$
$y = x_1 \tan \theta \left(\frac{x_1 + x_2 - x_1}{x_1 + x_2}\right)$
$y = x_1 \tan \theta \left(\frac{x_2}{x_1 + x_2}\right)$
$y = \left[\frac{x_1 x_2}{x_1 + x_2}\right] \tan \theta$
Thus,the correct option is $B$.

(D) Given:
Horizontal distance $x = 100 \,m$
Vertical displacement $y = 10 \,cm = 0.1 \,m$
Initial vertical velocity $u_y = 0$
Acceleration due to gravity $g = 10 \,m/s^2$
For horizontal projectile motion,the vertical displacement $y$ is given by:
$y = \frac{1}{2} g t^2$
$0.1 = \frac{1}{2} \times 10 \times t^2$
$0.1 = 5 t^2$
$t^2 = 0.02$
$t = \sqrt{0.02} \,s$
The horizontal distance $x$ is covered with constant horizontal velocity $u$:
$x = u \times t$
$100 = u \times \sqrt{0.02}$
$u = \frac{100}{\sqrt{0.02}} = \frac{100}{\sqrt{2 \times 10^{-2}}} = \frac{100}{0.1414} \approx 707 \,m/s$
Thus,the velocity of the bullet at $A$ is nearly $700 \,m/s$.

(B) At the maximum height,the vertical component of velocity is zero,and the horizontal component of velocity is $v = u \cos \theta$.
The acceleration acting on the object at this point is the acceleration due to gravity,$g$,which acts vertically downwards.
The centripetal acceleration $a_c$ is given by $a_c = \frac{v^2}{R}$,where $R$ is the radius of curvature.
At the maximum height,the acceleration $g$ is perpendicular to the velocity $v$. Thus,$g$ acts as the centripetal acceleration.
Therefore,$g = \frac{(u \cos \theta)^2}{R}$.
Rearranging for $R$,we get $R = \frac{u^2 \cos^2 \theta}{g}$.

(A) The slope $(m)$ of the trajectory at any time $(t)$ is given by the tangent of the angle $\phi$ that the velocity vector makes with the horizontal.
The velocity components at time $(t)$ are $v_x = u \cos \theta$ and $v_y = u \sin \theta - g t$.
The slope $(m)$ is given by:
$m = \tan \phi = \frac{v_y}{v_x} = \frac{u \sin \theta - g t}{u \cos \theta}$
Simplifying this expression:
$m = \frac{u \sin \theta}{u \cos \theta} - \frac{g t}{u \cos \theta}$
$m = \tan \theta - \left( \frac{g}{u \cos \theta} \right) t$
This equation is of the form $m = c - k t$,where $c = \tan \theta$ (a constant) and $k = \frac{g}{u \cos \theta}$ (a positive constant).
This represents a straight line with a positive $y$-intercept $(\tan \theta)$ and a negative slope $(-k)$. Therefore,the graph of $m$ versus $t$ is a straight line with a negative slope.

(B) The maximum height $H$ is given by $H = \frac{u^2 \sin^2 \theta}{2g}$,which implies $\sin \theta = \sqrt{\frac{2gH}{u^2}}$.
The horizontal range $R$ is given by $R = \frac{2u^2 \sin \theta \cos \theta}{g}$.
Substituting $\sin \theta$ and $\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \frac{2gH}{u^2}}$ into the range formula:
$R = \frac{2u^2}{g} \cdot \sqrt{\frac{2gH}{u^2}} \cdot \sqrt{1 - \frac{2gH}{u^2}} = \frac{2u^2}{g} \cdot \frac{\sqrt{2gH}}{u} \cdot \sqrt{\frac{u^2 - 2gH}{u^2}}$.
Simplifying this,we get $R = \frac{2}{g} \cdot \sqrt{2gH} \cdot \sqrt{u^2 - 2gH}$.
Squaring both sides:
$R^2 = \frac{4}{g^2} \cdot 2gH \cdot (u^2 - 2gH) = \frac{8H}{g} (u^2 - 2gH)$.
Rearranging for $u^2$:
$u^2 - 2gH = \frac{R^2 g}{8H} \Rightarrow u^2 = 2gH + \frac{R^2 g}{8H}$.
Thus,the speed of projection is $u = \sqrt{2gH + \frac{R^2 g}{8H}}$.

(B) Let the initial velocity be $u$ and the angle of projection be $\theta$.
The horizontal component of velocity is $v_x = u \cos \theta$ (constant).
The vertical component of velocity at time $t$ is $v_y = u \sin \theta - gt$.
Given that at $t = 1 \, s$,the angle with the horizontal is $45^{\circ}$:
$\tan 45^{\circ} = \frac{v_y}{v_x} = \frac{u \sin \theta - g(1)}{u \cos \theta} = 1$
$u \sin \theta - g = u \cos \theta \implies u \sin \theta - u \cos \theta = g \quad \dots (1)$
Given that the speed is minimum at $t = 2 \, s$. The speed of a projectile is minimum at the highest point where the vertical component of velocity is zero $(v_y = 0)$:
$u \sin \theta - g(2) = 0 \implies u \sin \theta = 2g \quad \dots (2)$
Substitute $(2)$ into $(1)$:
$2g - u \cos \theta = g \implies u \cos \theta = g \quad \dots (3)$
Dividing $(2)$ by $(3)$:
$\frac{u \sin \theta}{u \cos \theta} = \frac{2g}{g} \implies \tan \theta = 2$
$\theta = \tan^{-1}(2)$.

(D) Let the height of the pole be $h$. The ball is projected at an angle $\theta = 45^{\circ}$.
Let $\alpha$ be the angle of elevation of the top of the pole from the point of projection,and $\beta$ be the angle of depression of the point where the ball lands from the top of the pole.
From the geometry of the trajectory,we have $\tan \alpha = \frac{h}{d_1}$ and $\tan \beta = \frac{h}{d_2}$.
For a projectile passing through a point $(x, y)$,the equation of trajectory is $y = x \tan \theta \left(1 - \frac{x}{R}\right)$,where $R$ is the range.
Alternatively,using the property of projectile motion,$\tan \theta = \tan \alpha + \tan \beta$.
Substituting the values: $\tan 45^{\circ} = \frac{h}{d_1} + \frac{h}{d_2}$.
Since $\tan 45^{\circ} = 1$,we have $1 = h \left(\frac{1}{d_1} + \frac{1}{d_2}\right) = h \left(\frac{d_1 + d_2}{d_1 d_2}\right)$.
Therefore,the height of the pole is $h = \frac{d_1 d_2}{d_1 + d_2}$.

(C) The vertical displacement of the projectile between times $t_1$ and $t_2$ is zero because the particle is at the same height at both times.
Average velocity is defined as the total displacement divided by the total time interval.
Average velocity $\vec{v}_{avg} = \frac{\Delta \vec{r}}{\Delta t} = \frac{\Delta x \hat{i} + \Delta y \hat{j}}{t_2 - t_1}$.
Since the height is the same,$\Delta y = 0$.
The horizontal displacement $\Delta x$ is given by $v_x \times (t_2 - t_1)$,where $v_x = u \cos \theta$.
Thus,$\Delta x = (u \cos \theta)(t_2 - t_1)$.
Substituting this into the average velocity formula: $\vec{v}_{avg} = \frac{(u \cos \theta)(t_2 - t_1) \hat{i} + 0 \hat{j}}{t_2 - t_1} = u \cos \theta \hat{i}$.
Therefore,the magnitude of the average velocity is $u \cos \theta$.

(A) The time of flight $T$ is given by $T = \frac{2u \sin \theta}{g}$.
Substituting the values: $T = \frac{2 \times 20 \times \sin(30^{\circ})}{10} = \frac{2 \times 20 \times 0.5}{10} = 2 \, s$.
The time at which we need to find the centripetal acceleration is $t = 1 \, s$.
Since the total time of flight is $2 \, s$,the object reaches its highest point at $t = \frac{T}{2} = 1 \, s$.
At the highest point of a projectile's trajectory,the velocity is purely horizontal,and the acceleration is purely vertical (equal to $g$).
The centripetal acceleration $a_c$ is defined as $a_c = \frac{v^2}{R}$. At the highest point,the radius of curvature $R$ is given by $R = \frac{v_x^2}{g}$.
Substituting $R$ into the formula for $a_c$,we get $a_c = \frac{v_x^2}{(v_x^2/g)} = g$.
Therefore,the centripetal acceleration at the highest point is $g = 10 \, m/s^2$.

(B) When a constant force acts on a moving body in a direction that is not parallel or antiparallel to its initial velocity,the body experiences a constant acceleration in that direction.
Let the initial velocity be $\vec{u}$ and the constant acceleration be $\vec{a}$.
The position of the body at any time $t$ is given by $\vec{r}(t) = \vec{u}t + \frac{1}{2}\vec{a}t^2$.
This equation represents a quadratic relationship between displacement and time,which is the characteristic equation of a parabola.
Therefore,the path followed by the body is parabolic.

(B) If the range is the same for two projectiles thrown with the same speed,the angles of projection must be complementary,i.e.,$\theta$ and $(90^\circ - \theta)$.
The maximum height attained by a projectile is given by the formula $h = \frac{u^2 \sin^2 \theta}{2g}$.
For the first stone,$h_1 = \frac{u^2 \sin^2 \theta}{2g}$.
For the second stone,$h_2 = \frac{u^2 \sin^2(90^\circ - \theta)}{2g} = \frac{u^2 \cos^2 \theta}{2g}$.
Adding the two heights:
$h_1 + h_2 = \frac{u^2 \sin^2 \theta}{2g} + \frac{u^2 \cos^2 \theta}{2g}$
$h_1 + h_2 = \frac{u^2}{2g} (\sin^2 \theta + \cos^2 \theta)$
Since $\sin^2 \theta + \cos^2 \theta = 1$,we get:
$h_1 + h_2 = \frac{u^2}{2g}$.

(A) The projectile is projected with speed $u$ at an angle $\theta$ with the horizontal.
Let the projectile cross the same level at times $t_1$ and $t_2$.
The horizontal component of velocity is $v_x = u \cos \theta$,which remains constant throughout the motion.
The vertical component of velocity is $v_y = u \sin \theta - gt$.
At the same level,the vertical displacement is zero,so the vertical velocity components at $t_1$ and $t_2$ are equal in magnitude but opposite in direction.
The horizontal displacement between these two instants is $\Delta x = (u \cos \theta) \times \Delta t$,where $\Delta t = t_2 - t_1$.
The average velocity is defined as $\vec{v}_{avg} = \frac{\Delta \vec{r}}{\Delta t}$.
Since the vertical displacement is zero,the average velocity is purely horizontal: $\vec{v}_{avg} = \frac{\Delta x}{\Delta t} = \frac{(u \cos \theta) \Delta t}{\Delta t} = u \cos \theta$.

(A) The horizontal range $R$ of a projectile is given by $R = \frac{u^2 \sin(2\theta)}{g} = \frac{2u^2 \sin \theta \cos \theta}{g}$.
The maximum height $H$ of a projectile is given by $H = \frac{u^2 \sin^2 \theta}{2g}$.
According to the problem,the horizontal range is equal to the maximum height,so $R = H$.
Substituting the formulas: $\frac{2u^2 \sin \theta \cos \theta}{g} = \frac{u^2 \sin^2 \theta}{2g}$.
Canceling common terms $(u^2/g)$: $2 \sin \theta \cos \theta = \frac{\sin^2 \theta}{2}$.
Rearranging the terms: $4 \sin \theta \cos \theta = \sin^2 \theta$.
Dividing both sides by $\sin \theta \cos \theta$ (assuming $\theta \neq 0$): $4 = \frac{\sin \theta}{\cos \theta}$.
Therefore,$\tan \theta = 4$.

(C) The vertical displacement $y$ of a projectile at any time $t$ is given by $y = (u \sin \theta)t - \frac{1}{2}gt^2$.
Since the particle is at the same height at times $t_1$ and $t_2$,we have:
$(u \sin \theta)t_1 - \frac{1}{2}gt_1^2 = (u \sin \theta)t_2 - \frac{1}{2}gt_2^2$
$(u \sin \theta)(t_2 - t_1) = \frac{1}{2}g(t_2^2 - t_1^2)$
$u \sin \theta = \frac{g(t_1 + t_2)}{2}$
Given $t_1 = 1 \, s$,$t_2 = 3 \, s$,$\theta = 30^{\circ}$,and $g = 10 \, m/s^2$:
$u \sin 30^{\circ} = \frac{10(1 + 3)}{2}$
$u \times 0.5 = \frac{10 \times 4}{2}$
$0.5u = 20$
$u = 40 \, m/s$.

(B) The horizontal range $R$ of a projectile is given by the formula $R = \frac{u^2 \sin 2\theta}{g}$.
From this expression,it is clear that $R \propto u^2$,meaning the range is proportional to the square of the speed of projection,not the square root.
Therefore,statement $(B)$ is incorrect.
Statement $(A)$ is true because time of flight $T = \frac{2u \sin \theta}{g}$,so $T \propto u$.
Statement $(C)$ is true because $\sin 2\theta$ is maximum at $2\theta = 90^{\circ}$,so $\theta = 45^{\circ}$.
Statement $(D)$ is true because at maximum height,velocity is purely horizontal,while acceleration due to gravity is purely vertical,making them perpendicular.

(B) The vertical displacement of a projectile is given by $y = (u \sin \theta)t - \frac{1}{2}gt^2$.
Since the projectile is at the same height at $t_1 = 1 \, s$ and $t_2 = 3 \, s$,the sum of these times is equal to the time taken to reach the maximum height multiplied by $2$,which is the total time of flight for that specific height level.
Specifically,$t_1 + t_2 = \frac{2u \sin \theta}{g}$.
Given $u = 40 \, m/s$,$g = 10 \, m/s^2$,$t_1 = 1 \, s$,and $t_2 = 3 \, s$:
$1 + 3 = \frac{2 \times 40 \times \sin \theta}{10}$.
$4 = 8 \sin \theta$.
$\sin \theta = \frac{4}{8} = \frac{1}{2}$.
Therefore,$\theta = 30^{\circ}$.
The angle of projection is $\theta = \tan^{-1}(\tan 30^{\circ}) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right)$.

(A) The power delivered by a force is given by the formula $P = \vec{F} \cdot \vec{v} = Fv cos \theta$,where $F$ is the force,$v$ is the velocity,and $ \theta$ is the angle between the force and velocity vectors.
In this case,the force is gravity,which acts vertically downwards $(F = mg)$.
During the downward motion of a projectile,the velocity vector $\vec{v}$ is directed downwards,making an angle $ \theta$ with the vertical force vector $\vec{F}$.
As the body moves downwards,its speed $v$ increases due to the acceleration of gravity.
Since the force of gravity is constant and the angle $ \theta$ between the downward force and the downward velocity is $0^{\circ}$ (or decreasing towards $0^{\circ}$),the product $Fv cos \theta$ increases.
Therefore,the power delivered by gravity increases during the downward motion.

(D) The motion of the centre of mass of a system is determined only by external forces acting on the system.
In this case,the explosion is caused by internal forces,which do not affect the trajectory of the centre of mass.
Therefore,the centre of mass continues to follow the original parabolic path it would have taken if no explosion had occurred.
The horizontal range of a projectile is given by the formula $R = \frac{v^2 \sin 2\theta}{g}$.
Thus,the horizontal range of the centre of mass remains $\frac{v^2 \sin 2\theta}{g}$.