Horizontal Projectile Motion Questions in English

(C) In horizontal projectile motion,the horizontal component of velocity $(v_x)$ remains constant throughout the motion.
Given,initial horizontal velocity $u_x = 18 \, ms^{-1}$.
Therefore,at the time of striking the ground,$v_x = 18 \, ms^{-1}$.
The angle of impact with the ground is given as $\theta = 45^o$.
The relationship between the velocity components and the angle is given by $\tan \theta = \frac{v_y}{v_x}$.
Substituting the values: $\tan 45^o = \frac{v_y}{18}$.
Since $\tan 45^o = 1$,we have $1 = \frac{v_y}{18}$.
Thus,$v_y = 18 \, ms^{-1}$.

(B) The horizontal range $R$ of a projectile is given by the formula: $R = \frac{u^2 \sin(2\alpha)}{g}$,where $u$ is the initial velocity,$\alpha$ is the angle of projection,and $g$ is the acceleration due to gravity.
For the first angle of projection $\alpha_1 = (45^\circ - \theta)$:
$R_1 = \frac{u^2 \sin[2(45^\circ - \theta)]}{g} = \frac{u^2 \sin(90^\circ - 2\theta)}{g} = \frac{u^2 \cos(2\theta)}{g}$.
For the second angle of projection $\alpha_2 = (45^\circ + \theta)$:
$R_2 = \frac{u^2 \sin[2(45^\circ + \theta)]}{g} = \frac{u^2 \sin(90^\circ + 2\theta)}{g} = \frac{u^2 \cos(2\theta)}{g}$.
Comparing the two ranges:
$\frac{R_1}{R_2} = \frac{\frac{u^2 \cos(2\theta)}{g}}{\frac{u^2 \cos(2\theta)}{g}} = \frac{1}{1}$.
Thus,the ratio of the horizontal ranges is $1:1$.

(C) The horizontal range $R$ of a projectile is given by $R = \frac{u^2 \sin 2\theta}{g}$.
The maximum height $H$ of a projectile is given by $H = \frac{u^2 \sin^2 \theta}{2g}$.
According to the problem,$R = H$.
Therefore,$\frac{u^2 \sin 2\theta}{g} = \frac{u^2 \sin^2 \theta}{2g}$.
Using the identity $\sin 2\theta = 2 \sin \theta \cos \theta$,we get:
$\frac{2 u^2 \sin \theta \cos \theta}{g} = \frac{u^2 \sin^2 \theta}{2g}$.
Canceling $u^2$,$g$,and $\sin \theta$ (assuming $\sin \theta \neq 0$):
$2 \cos \theta = \frac{\sin \theta}{2}$.
Rearranging the terms,we get $\frac{\sin \theta}{\cos \theta} = 4$,which means $\tan \theta = 4$.
Thus,$\theta = \tan^{-1}(4)$.

(C) In projectile motion,the horizontal component of velocity $(v_x)$ remains constant throughout the motion because there is no acceleration in the horizontal direction.
At point $A$,the velocity is $\vec{v}_A = 2\hat i + 3\hat j \text{ m/s}$.
At point $B$,which is at the same horizontal level as point $A$,the horizontal component remains $2\hat i \text{ m/s}$.
The vertical component of velocity $(v_y)$ changes due to gravity. At the same horizontal level,the magnitude of the vertical component remains the same,but its direction is reversed.
Therefore,the vertical component at point $B$ becomes $-3\hat j \text{ m/s}$.
Thus,the velocity at point $B$ is $\vec{v}_B = 2\hat i - 3\hat j \text{ m/s}$.

(C) For a projectile launched with an initial velocity $u$ at an angle $\theta$,the velocity at any point is given by $v = \sqrt{v_x^2 + v_y^2}$.
At the highest point of the trajectory,the vertical component of velocity $v_y$ becomes $0$,while the horizontal component $v_x = u \cos \theta$ remains constant.
Since the kinetic energy is given by $K = \frac{1}{2}mv^2$,it is proportional to the square of the velocity.
At the highest point,the velocity is minimum $(v = v_x = u \cos \theta)$,therefore the kinetic energy is minimum.
Conversely,the potential energy $U = mgh$ is maximum at the highest point because the height $h$ is maximum.

(D) The initial kinetic energy is $K = \frac{1}{2}mv^2$,where $v$ is the initial velocity.
At the highest point of the trajectory,the vertical component of velocity becomes zero,and the velocity is only the horizontal component,$v_x = v \cos \theta$.
The kinetic energy at the highest point is $K' = \frac{1}{2}m(v \cos \theta)^2$.
Substituting $K = \frac{1}{2}mv^2$,we get $K' = K \cos^2 \theta$.
Given $\theta = 30^{\circ}$,we have $K' = K \cos^2(30^{\circ}) = K \left( \frac{\sqrt{3}}{2} \right)^2 = \frac{3K}{4}$.

(C) The trajectory of a projectile fired at an angle $\theta$ with initial velocity $v_0$ is given by the equation:
$y = x \tan \theta - \frac{gx^2}{2v_0^2 \cos^2 \theta}$
Here,$x$ represents the horizontal distance and $y$ represents the vertical height.
Given that the cliff is at a horizontal distance $D$,we substitute $x = D$ into the equation to find the height $h$ at which the shell strikes the cliff:
$h = D \tan \theta - \frac{gD^2}{2v_0^2 \cos^2 \theta}$
Thus,the correct option is $C$.

(C) The total time of flight $T$ is given by $T = \frac{2u \sin \theta}{g}$.
Substituting the values: $T = \frac{2 \times 50 \times \sin(30^o)}{10} = \frac{100 \times 0.5}{10} = 5 \, s$.
Given that the stone crosses the wall at $t_1 = 3 \, s$,the remaining time in the air after crossing the wall is $t_2 = T - t_1 = 5 - 3 = 2 \, s$.
The horizontal velocity $v_x$ remains constant throughout the motion: $v_x = u \cos \theta = 50 \times \cos(30^o) = 50 \times \frac{\sqrt{3}}{2} = 25\sqrt{3} \, m/s$.
The distance $d$ traveled beyond the wall is $d = v_x \times t_2 = 25\sqrt{3} \times 2 = 50\sqrt{3} \, m$.
Using $\sqrt{3} \approx 1.732$,we get $d = 50 \times 1.732 = 86.6 \, m$.

(B) The velocity vector is given by $\vec{V} = a\hat{i} + (b - ct)\hat{j}$.
Comparing this with $\vec{V} = v_x\hat{i} + v_y\hat{j}$,we get $v_x = a$ and $v_y = b - ct$.
At $t = 0$,the initial velocity components are $u_x = a$ and $u_y = b$.
The acceleration in the $y$-direction is $a_y = \frac{dv_y}{dt} = -c$.
The time of flight $T$ is the time when the particle returns to the horizontal plane,i.e.,displacement in $y$ is $0$.
Using $s_y = u_y T + \frac{1}{2} a_y T^2 = 0$,we get $bT - \frac{1}{2} cT^2 = 0$,which gives $T = \frac{2b}{c}$.
The range $R$ is the displacement in the $x$-direction during the time of flight: $R = u_x T = a \times \left( \frac{2b}{c} \right) = \frac{2ab}{c}$.

(C) Let the ball be thrown with velocity components $v_x$ and $v_y$. The bird moves horizontally with speed $u$. Since the ball touches the bird,the horizontal velocity of the ball must be equal to the speed of the bird,so $v_x = u$.
The vertical displacement of the ball at the point of contact is $h$. Using the equation of motion $h = v_y t - \frac{1}{2}gt^2$ and the condition that the ball 'just touches' the bird (meaning the vertical velocity of the ball at that height is zero relative to the bird's path,or simply considering the peak of the trajectory),we have $h = \frac{v_y^2}{2g}$,which gives $v_y = \sqrt{2gh}$.
The time taken to reach the height $h$ is $t = \frac{v_y}{g} = \sqrt{\frac{2h}{g}}$.
The total time of flight for the projectile is $T = 2t = 2\sqrt{\frac{2h}{g}}$.
The horizontal range $R$ is given by $R = v_x T = u \times 2\sqrt{\frac{2h}{g}} = 2u\sqrt{\frac{2h}{g}}$.

(B) The displacement vector $\vec{s}$ from the starting point $(0,0)$ to the highest point $(\frac{R}{2}, H)$ is given by $\vec{s} = \frac{R}{2} \hat{i} + H \hat{j}$.
The magnitude of displacement is $|\vec{s}| = \sqrt{(\frac{R}{2})^2 + H^2}$.
Given $R = \frac{u^2 \sin 2\theta}{g} = \frac{2u^2 \sin \theta \cos \theta}{g}$ and $H = \frac{u^2 \sin^2 \theta}{2g}$.
So,$\frac{R}{2} = \frac{u^2 \sin \theta \cos \theta}{g}$.
$|\vec{s}| = \sqrt{(\frac{u^2 \sin \theta \cos \theta}{g})^2 + (\frac{u^2 \sin^2 \theta}{2g})^2} = \frac{u^2 \sin \theta}{g} \sqrt{\cos^2 \theta + \frac{\sin^2 \theta}{4}} = \frac{u^2 \sin \theta}{2g} \sqrt{4 \cos^2 \theta + \sin^2 \theta} = \frac{u^2 \sin \theta}{2g} \sqrt{3 \cos^2 \theta + 1}$.
The time taken to reach the highest point is $t = \frac{u \sin \theta}{g}$.
Average velocity $V_{avg} = \frac{|\vec{s}|}{t} = \frac{\frac{u^2 \sin \theta}{2g} \sqrt{3 \cos^2 \theta + 1}}{\frac{u \sin \theta}{g}} = \frac{u}{2} \sqrt{1 + 3 \cos^2 \theta}$.

(C) In projectile motion,the horizontal component of velocity remains constant throughout the flight.
$v_x = u \cos \theta$.
Let the speed of the projectile at an angle $\alpha$ be $v$. The horizontal component of this velocity is $v_x = v \cos \alpha$.
Since the horizontal component is constant,we have $v \cos \alpha = u \cos \theta$.
Solving for $v$,we get $v = \frac{u \cos \theta}{\cos \alpha} = u \cos \theta \sec \alpha$.

(C) Let the maximum height of the projectile be $H$ and the horizontal range be $R$. The shell clears the summit,which is at the peak of the trajectory.
The angular elevation $\beta$ of the summit from the base is given by $\tan \beta = \frac{H}{R/2} = \frac{2H}{R}$.
Substituting the standard formulas for maximum height $H = \frac{u^2 \sin^2 \alpha}{2g}$ and horizontal range $R = \frac{u^2 \sin(2\alpha)}{g} = \frac{2u^2 \sin \alpha \cos \alpha}{g}$:
$\tan \beta = \frac{2 \left( \frac{u^2 \sin^2 \alpha}{2g} \right)}{\left( \frac{2u^2 \sin \alpha \cos \alpha}{g} \right)}$
$\tan \beta = \frac{u^2 \sin^2 \alpha / g}{2u^2 \sin \alpha \cos \alpha / g} = \frac{\sin^2 \alpha}{2 \sin \alpha \cos \alpha} = \frac{\tan \alpha}{2}$.
Therefore,$\beta = \tan^{-1}\left(\frac{\tan \alpha}{2}\right)$.

(A) When the bomb is released from an aeroplane flying at a constant horizontal velocity,the bomb possesses the same horizontal velocity as the aeroplane at the moment of release.
Since there is no air resistance (assuming ideal conditions) and no horizontal force acting on the bomb,its horizontal velocity remains constant throughout its flight.
Because both the aeroplane and the bomb maintain the same horizontal velocity,the horizontal displacement of both objects at any time $t$ will be identical.
Therefore,the bomb will always remain vertically below the aeroplane during its descent.

(B) The time of flight $T$ for a projectile is given by the formula $T = \frac{2 u \sin \theta}{g}$.
We also know that the maximum height $H$ reached by a projectile is $H = \frac{u^2 \sin^2 \theta}{2g}$.
From the expression for maximum height,we can write $\sqrt{H} = \frac{u \sin \theta}{\sqrt{2g}}$,which implies $u \sin \theta = \sqrt{2gH}$.
Substituting this into the time of flight formula: $T = \frac{2 \sqrt{2gH}}{g} = 2 \sqrt{\frac{2H}{g}}$.
Since $g$ is constant,$T \propto \sqrt{H}$.
Therefore,the baseball that reaches the maximum height will remain in the air for the longest time.

(B) The projectile crosses the two walls at $t_1 = 2 \ s$ and $t_2 = 6 \ s$.
Due to symmetry,the time taken to reach the maximum height is $T_{max} = \frac{t_1 + t_2}{2} = \frac{2 + 6}{2} = 4 \ s$.
Using the equation of motion for vertical displacement,$y = u_y t - \frac{1}{2} g t^2$.
At $t = 4 \ s$,the vertical velocity $v_y = 0$.
Using $v_y = u_y - gt$,we get $0 = u_y - 10(4)$,so $u_y = 40 \ m/s$.
The maximum height $H_{max}$ is given by $H_{max} = \frac{u_y^2}{2g} = \frac{40^2}{2 \times 10} = \frac{1600}{20} = 80 \ m$.

(C) Let the projectile be thrown with velocity $v_0$ at an angle $\theta$ with the horizontal.
The projectile crosses the two walls at times $t_1 = 2 \ s$ and $t_2 = 6 \ s$.
Since the motion is symmetric,the time taken to reach the maximum height is $t_{max} = \frac{t_1 + t_2}{2} = \frac{2 + 6}{2} = 4 \ s$.
At maximum height,the vertical component of velocity is zero,so $v_y = v_0 \sin \theta - gt = 0$.
Using $g = 10 \ ms^{-2}$,we get $v_0 \sin \theta = 10 \times 4 = 40 \ ms^{-1}$.
The horizontal distance between the two walls is $d = 120 \ m$,and the time taken to travel this distance is $\Delta t = t_2 - t_1 = 6 - 2 = 4 \ s$.
The horizontal component of velocity is constant: $v_x = v_0 \cos \theta = \frac{d}{\Delta t} = \frac{120}{4} = 30 \ ms^{-1}$.
The magnitude of the projection velocity is $v_0 = \sqrt{(v_x)^2 + (v_y \sin \theta)^2} = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50 \ ms^{-1}$.

(D) Given: Initial velocity $u = 50 \, ms^{-1}$,angle $\theta = 53^o$.
Horizontal component of velocity: $u_x = u \cos 53^o = 50 \times (3/5) = 30 \, ms^{-1}$.
Vertical component of velocity: $u_y = u \sin 53^o = 50 \times (4/5) = 40 \, ms^{-1}$.
Statement $A$: The vertical component of velocity at the start is $40 \, ms^{-1}$. This is correct.
Statement $B$: The horizontal component of velocity is $30 \, ms^{-1}$. This is correct.
Statement $C$: The minimum velocity of a projectile occurs at the highest point,which is equal to the horizontal component $u_x = 30 \, ms^{-1}$. This is correct.
Since all statements $A, B,$ and $C$ are correct,the incorrect statement is none of these.

(D) The vertical component of velocity is $u_y = u \sin \theta = 50 \sin 53^o = 50 \times (4/5) = 40 \, m/s$.
The vertical displacement at any time $t$ is given by $y = u_y t - \frac{1}{2} g t^2$.
For a given height $h$,the equation is $\frac{1}{2} g t^2 - u_y t + h = 0$.
This is a quadratic equation in $t$,which has two roots $t_1$ and $t_2$ representing the two times the projectile reaches height $h$.
The sum of the roots is $t_1 + t_2 = \frac{-(-u_y)}{\frac{1}{2} g} = \frac{2 u_y}{g} = \frac{2 \times 40}{10} = 8 \, s$.
Checking the options:
For $A$: $1 + 7 = 8 \, s$.
For $B$: $3 + 5 = 8 \, s$.
For $C$: $2 + 6 = 8 \, s$.
Since all pairs satisfy the condition $t_1 + t_2 = 8 \, s$,the correct option is $D$.

(A) The equation of trajectory for a projectile is given by $y = x \tan \theta - \frac{gx^2}{2u^2 \cos^2 \theta}$.
Given: $u = 50 \, ms^{-1}$,$\theta = 53^o$,$g = 10 \, ms^{-2}$.
We know $\sin 53^o = 0.8$ and $\cos 53^o = 0.6$,so $\tan 53^o = \frac{0.8}{0.6} = \frac{4}{3}$.
Substituting the values:
$y = x \left( \frac{4}{3} \right) - \frac{10 x^2}{2(50)^2 (0.6)^2}$
$y = \frac{4x}{3} - \frac{10 x^2}{2(2500)(0.36)}$
$y = \frac{4x}{3} - \frac{10 x^2}{1800}$
$y = \frac{4x}{3} - \frac{x^2}{180}$
Multiply the entire equation by $180$:
$180y = 180 \left( \frac{4x}{3} \right) - x^2$
$180y = 240x - x^2$.

(D) Given that the particle has acceleration only in the negative $y$-direction,the acceleration in the $x$-direction is $a_x = 0$. Since $a_x = 0$,the $x$-component of velocity $v_x$ is constant.
From the equation $y = ax - bx^2$,differentiate with respect to time $t$: $\frac{dy}{dt} = a\frac{dx}{dt} - 2bx\frac{dx}{dt}$,which gives $v_y = v_x(a - 2bx)$.
Differentiating again with respect to time: $\frac{d^2y}{dt^2} = a\frac{d^2x}{dt^2} - 2b\left[\left(\frac{dx}{dt}\right)^2 + x\frac{d^2x}{dt^2}\right]$.
Since $a_x = 0$ and $a_y = -g$,we get $-g = -2bv_x^2$,which implies $v_x = \sqrt{\frac{g}{2b}}$.
At the origin $(x = 0)$,$v_y = a v_x = a\sqrt{\frac{g}{2b}}$.
The angle $\theta$ with the $x$-axis is given by $\tan \theta = \frac{v_y}{v_x} = \frac{a v_x}{v_x} = a$,so $\theta = \tan^{-1}(a)$.
Thus,all statements are correct.

(B) The time of flight of a projectile depends only on the vertical component of the initial velocity and the acceleration due to gravity.
The initial velocity $V$ is given at an angle $\theta$ with the vertical. Therefore,the vertical component of the velocity is $V_y = V \cos \theta$.
In an elastic collision with a vertical wall,the horizontal component of the velocity reverses direction,but the vertical component of the velocity remains unchanged.
Since the vertical motion is unaffected by the collision with the vertical wall,the total time of flight is the same as that of a projectile moving without any collision.
The formula for the time of flight is $T = \frac{2V_y}{g}$.
Substituting $V_y = V \cos \theta$,we get $T = \frac{2V \cos \theta}{g}$.

(D) The torque $\vec{\tau}$ about the point of projection is given by $\vec{\tau} = \vec{r} \times \vec{F}$.
Here,the force $\vec{F}$ is the weight $m\vec{g}$ acting downwards.
At maximum height,the horizontal distance (range) from the point of projection is $x = R/2$,where $R$ is the horizontal range.
The formula for horizontal range is $R = \frac{u^2 \sin(2\theta)}{g}$.
Given $\theta = 45^o$,$R = \frac{u^2 \sin(90^o)}{g} = \frac{u^2}{g}$.
At maximum height,the horizontal position is $x = \frac{R}{2} = \frac{u^2}{2g}$.
The vertical force is $F = mg$ acting downwards.
The torque magnitude is $\tau = F \cdot x = (mg) \cdot \left( \frac{u^2}{2g} \right) = \frac{1}{2} m u^2$.

(D) Let $u$ be the initial velocity with which the particle is thrown and $m$ be the mass of the particle. The initial kinetic energy is given by $K = \frac{1}{2}mu^2$.
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 $v_x = u \cos(60^\circ)$.
Therefore, the kinetic energy at the highest point $K'$ is:
$K' = \frac{1}{2}m(v_x)^2 = \frac{1}{2}m(u \cos(60^\circ))^2$
$K' = \frac{1}{2}mu^2 \cos^2(60^\circ)$
Since $\cos(60^\circ) = \frac{1}{2}$, we have $\cos^2(60^\circ) = \frac{1}{4}$.
Substituting $K = \frac{1}{2}mu^2$, we get:
$K' = K \times \frac{1}{4} = \frac{K}{4}$.

(D) The water fountain acts as a projectile source spraying water in all directions with speed $v$. The maximum horizontal range $R_{\max}$ of the water is achieved when the angle of projection is $45^\circ$.
Using the formula for horizontal range: $R = \frac{v^2 \sin(2\theta)}{g}$.
For maximum range,$\sin(2\theta) = 1$,so $R_{\max} = \frac{v^2}{g}$.
The water covers a circular area around the fountain with radius $R_{\max}$.
Therefore,the total area $A$ is given by: $A = \pi R_{\max}^2 = \pi \left( \frac{v^2}{g} \right)^2 = \frac{\pi v^4}{g^2}$.

(A) The maximum height $H_{max}$ attained by a projectile is given by $H_{max} = \frac{u^2}{2g}$.
Given $H_{max} = 10 \ m$,we have $10 = \frac{u^2}{2g}$,which implies $u^2 = 20g$.
The maximum horizontal range $R_{max}$ is achieved when the angle of projection is $45^\circ$,given by $R_{max} = \frac{u^2}{g}$.
Substituting the value of $u^2$,we get $R_{max} = \frac{20g}{g} = 20 \ m$.

(B) The initial velocity vector is given by $\vec{u} = (1 \hat{i} + 2 \hat{j}) \ ms^{-1}$.
Comparing this with $\vec{u} = u_x \hat{i} + u_y \hat{j}$,we get the horizontal component $u_x = 1 \ ms^{-1}$ and the vertical component $u_y = 2 \ ms^{-1}$.
The equation of the trajectory of a projectile is given by $y = x \tan \theta - \frac{gx^2}{2u_x^2}$.
Since $\tan \theta = \frac{u_y}{u_x} = \frac{2}{1} = 2$,we substitute the values into the equation:
$y = x(2) - \frac{10 \cdot x^2}{2(1)^2}$
$y = 2x - \frac{10x^2}{2}$
$y = 2x - 5x^2$.

(D) Let the initial velocity of the particle be $u$. The initial kinetic energy is $K = \frac{1}{2}mu^2$.
At the maximum height,the vertical component of velocity becomes $0$,but the horizontal component remains constant.
The horizontal component of velocity is $v_x = u \cos \theta$.
Given $\theta = 60^o$,so $v_x = u \cos 60^o = u/2$.
The kinetic energy at the maximum height is $K' = \frac{1}{2}m(v_x)^2$.
Substituting $v_x = u/2$,we get $K' = \frac{1}{2}m(u/2)^2 = \frac{1}{2}m(u^2/4) = \frac{1}{4} (\frac{1}{2}mu^2)$.
Since $K = \frac{1}{2}mu^2$,we have $K' = K/4$.

(C) Let the hero and the villain be at the same horizontal level at $t = 0$. At this instant,the villain fires the bullet horizontally.
Both the hero and the villain have a downward acceleration of $a_p = 2\ m/s^2$.
Since the bullet is fired horizontally,its initial vertical velocity is $0$. However,once the bullet is in the air,it is only subject to gravity (ignoring air resistance),so its vertical acceleration is $a_b = g = 10\ m/s^2$ (downwards).
Let $t$ be the time taken for the bullet to reach the hero's horizontal position.
The vertical displacement of the hero in time $t$ is $y_h = u_y t + \frac{1}{2} a_p t^2 = u_y t + \frac{1}{2} (2) t^2 = u_y t + t^2$.
The vertical displacement of the bullet in time $t$ is $y_b = u_y t + \frac{1}{2} a_b t^2 = u_y t + \frac{1}{2} (10) t^2 = u_y t + 5t^2$.
Since $5t^2 > t^2$,the bullet will have a greater downward displacement than the hero in the same amount of time.
Therefore,the bullet will pass below the hero.

(C) In projectile motion,there is no external horizontal force acting on the object (ignoring air resistance).
According to Newton's second law,if the net external force in a direction is zero,the momentum in that direction remains constant.
Since the horizontal component of velocity $v_x = V_0 \cos(30^o)$ remains constant throughout the flight,the horizontal component of momentum $p_x = m v_x$ is conserved.
Therefore,the correct statement is that the horizontal component of momentum will be conserved.

(C) The horizontal component of velocity remains constant throughout the motion of the projectile.
Let $u = 147 \ m/s$ and $\theta_1 = 60^{\circ}$. The horizontal component is $u_x = u \cos 60^{\circ} = 147 \times 0.5 = 73.5 \ m/s$.
At time $t$,the velocity is $v$ and the angle is $\theta_2 = 45^{\circ}$. The horizontal component is $v_x = v \cos 45^{\circ} = v / \sqrt{2}$.
Since $u_x = v_x$,we have $v \cos 45^{\circ} = u \cos 60^{\circ} \Rightarrow v = u \frac{\cos 60^{\circ}}{\cos 45^{\circ}} = 147 \times \frac{0.5}{1/\sqrt{2}} = 73.5 \sqrt{2} \ m/s$.
The vertical component of velocity at time $t$ is $v_y = v \sin 45^{\circ} = (73.5 \sqrt{2}) \times (1/\sqrt{2}) = 73.5 \ m/s$.
The initial vertical component is $u_y = u \sin 60^{\circ} = 147 \times \frac{\sqrt{3}}{2} \approx 127.3 \ m/s$.
Using the equation $v_y = u_y - gt$,we get $73.5 = 127.3 - 9.8t$.
$9.8t = 127.3 - 73.5 = 53.8$.
$t = 53.8 / 9.8 \approx 5.49 \ s$.

(B) The swimmer performs horizontal projectile motion.
Let $h = 10 \ m$ be the vertical height and $x = 2 \ m$ be the horizontal distance to clear.
The time taken to fall $10 \ m$ is given by $h = \frac{1}{2} g t^2$.
Using $g = 9.8 \ m/s^2$,we get $t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 10}{9.8}} = \sqrt{2.04} \approx 1.428 \ s$.
To clear the ledge,the horizontal distance covered must be at least $2 \ m$.
Using $x = v \times t$,we have $v = \frac{x}{t} = \frac{2}{1.428} \approx 1.4 \ m/s$.
Thus,the minimum speed required is $1.4 \ m/s$.

(A) The particle is projected with speed $u = 10 \ m/s$ at an angle $\theta = 60^\circ$ with the horizontal.
Due to the symmetry of projectile motion,when the particle returns to the same horizontal level,its speed remains $u = 10 \ m/s$,and the angle it makes with the horizontal is also $\theta = 60^\circ$,but directed downwards.
The radius of curvature $R$ is given by the formula $R = \frac{v^2}{a_n}$,where $v$ is the speed and $a_n$ is the normal acceleration.
The normal acceleration $a_n$ is the component of acceleration perpendicular to the velocity vector.
At the point of landing,the acceleration due to gravity $g$ acts vertically downwards.
The angle between the velocity vector and the horizontal is $60^\circ$,so the angle between the velocity vector and the vertical is $30^\circ$.
The component of $g$ perpendicular to the velocity is $a_n = g \cos(60^\circ)$.
Substituting the values: $R = \frac{u^2}{g \cos(60^\circ)} = \frac{10^2}{10 \times 0.5} = \frac{100}{5} = 20 \ m$.

(D) The horizontal component of velocity remains constant throughout the motion: $v_x = u \cos \theta = 12 \cos 60^o = 12 \times 0.5 = 6 \, m/s$.
Let the speed of the particle be $v = 10 \, m/s$ at some time $t$. The speed is given by $v = \sqrt{v_x^2 + v_y^2}$.
Substituting the values: $10 = \sqrt{6^2 + v_y^2} \implies 100 = 36 + v_y^2 \implies v_y^2 = 64 \implies v_y = \pm 8 \, m/s$.
The vertical component of velocity at time $t$ is $v_y = u \sin \theta - gt = 12 \sin 60^o - 10t = 12(\frac{\sqrt{3}}{2}) - 10t = 6\sqrt{3} - 10t$.
For $v_y = 8 \, m/s$: $6\sqrt{3} - 10t_1 = 8 \implies 10t_1 = 6\sqrt{3} - 8 \implies t_1 = \frac{6\sqrt{3}-8}{10}$.
For $v_y = -8 \, m/s$: $6\sqrt{3} - 10t_2 = -8 \implies 10t_2 = 6\sqrt{3} + 8 \implies t_2 = \frac{6\sqrt{3}+8}{10}$.
The time interval is $\Delta t = t_2 - t_1 = \frac{6\sqrt{3}+8 - (6\sqrt{3}-8)}{10} = \frac{16}{10} = 1.6 \, s$.

(C) For a projectile,the acceleration is always $g$ acting vertically downwards.
At any instant,the tangential acceleration is $a_t = g \sin \theta$ and the normal acceleration is $a_n = g \cos \theta$,where $\theta$ is the angle the velocity vector makes with the horizontal.
Given that the magnitudes of tangential and normal acceleration are equal,we have $a_t = a_n$.
Therefore,$g \sin \theta = g \cos \theta$,which implies $\tan \theta = 1$.
The angle $\theta$ is the angle of the velocity vector with the horizontal,so $\tan \theta = |v_y / v_x| = 1$.
This means $|v_y| = |v_x|$.
We know $v_x = u_x = 10\, m/s$ and $v_y = u_y - gt = 20 - 10t$.
So,$|20 - 10t| = 10$.
Case $1$: $20 - 10t = 10 \Rightarrow 10t = 10 \Rightarrow t = 1\, s$.
Case $2$: $20 - 10t = -10 \Rightarrow 10t = 30 \Rightarrow t = 3\, s$.
Thus,the magnitudes are equal at $t = 1\, s$ and $t = 3\, s$.

(D) At the maximum height $h$,the vertical component of velocity is zero,and the horizontal component of velocity is $V_x = V \cos 45^{\circ} = \frac{V}{\sqrt{2}}$.
The momentum of the particle at the maximum height is $p = m V_x = \frac{mV}{\sqrt{2}}$.
The angular momentum $L$ about the point of projection is given by $L = p \times h$,where $h$ is the perpendicular distance (maximum height) from the point of projection.
The maximum height $h$ is given by $h = \frac{V^2 \sin^2 45^{\circ}}{2g} = \frac{V^2}{4g}$.
Substituting these into the angular momentum formula:
$L = \left( \frac{mV}{\sqrt{2}} \right) \times h = \frac{mV}{\sqrt{2}} \times \frac{V^2}{4g} = \frac{mV^3}{4\sqrt{2}g}$.
Since $h = \frac{V^2}{4g}$,we have $V^2 = 4gh$,which implies $V = \sqrt{4gh} = 2\sqrt{gh}$.
Substituting $V$ into the expression for $L$:
$L = \frac{m(2\sqrt{gh})^3}{4\sqrt{2}g} = \frac{m(8g^{3/2}h^{3/2})}{4\sqrt{2}g} = \frac{2m g h \sqrt{gh}}{\sqrt{2}g} = \frac{2}{\sqrt{2}} m \sqrt{gh^3} = m\sqrt{2gh^3}$.

(B) The general equation of a projectile is $y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta}$.
Comparing this with the given equation $y = 12x - \frac{5}{9}x^2$:
$1) \tan \theta = 12$
$2) \frac{g}{2 u^2 \cos^2 \theta} = \frac{5}{9}$
Given the horizontal component of velocity $v_x = u \cos \theta = 3 \ ms^{-1}$.
Substitute $u \cos \theta = 3$ into the second equation:
$\frac{10}{2 (3)^2} = \frac{5}{9} \implies \frac{10}{18} = \frac{5}{9}$,which is consistent.
To find the range $R$,we set $y = 0$ in the equation of motion:
$0 = 12x - \frac{5}{9}x^2$
$x(12 - \frac{5}{9}x) = 0$
Since $x \neq 0$,$12 = \frac{5}{9}x$
$x = \frac{12 \times 9}{5} = \frac{108}{5} = 21.6 \ m$.
Thus,the range of the projectile is $21.6 \ m$.

(B) The equations of motion for a projectile are:
Horizontal displacement: $d = (u \cos \theta) t$
Vertical displacement: $h = (u \sin \theta) t - \frac{1}{2} g t^2$
From the horizontal equation,we find the time of flight to reach distance $d$:
$t = \frac{d}{u \cos \theta}$
Substitute this expression for $t$ into the vertical displacement equation:
$h = (u \sin \theta) \left( \frac{d}{u \cos \theta} \right) - \frac{1}{2} g \left( \frac{d}{u \cos \theta} \right)^2$
$h = d \tan \theta - \frac{g d^2}{2 u^2 \cos^2 \theta}$
Rearranging to solve for $u^2$:
$\frac{g d^2}{2 u^2 \cos^2 \theta} = d \tan \theta - h$
$u^2 = \frac{g d^2}{2 \cos^2 \theta (d \tan \theta - h)}$
Taking the square root of both sides:
$u = \frac{d}{\cos \theta} \sqrt{\frac{g}{2(d \tan \theta - h)}}$

(C) Let the initial speed be $u$ and the angle of projection be $\theta$.
The speed at maximum height is $u \cos \theta$.
Given that $u \cos \theta = \frac{\sqrt{3}}{2} u$,so $\cos \theta = \frac{\sqrt{3}}{2}$,which implies $\theta = 30^{\circ}$.
The maximum height $H$ is given by $H = \frac{u^2 \sin^2 \theta}{2g}$ and the range $R$ is given by $R = \frac{u^2 \sin 2\theta}{g}$.
We are given $R = P \times H$,so $P = \frac{R}{H}$.
Substituting the formulas: $P = \frac{u^2 (2 \sin \theta \cos \theta) / g}{u^2 \sin^2 \theta / (2g)} = \frac{4 \cos \theta}{\sin \theta} = 4 \cot \theta$.
For $\theta = 30^{\circ}$,$P = 4 \cot 30^{\circ} = 4 \times \sqrt{3} = 4\sqrt{3}$.

(C) For a projectile,the range $R$ is given by $R = \frac{u^2 \sin(2\theta)}{g}$.
To achieve the maximum range,the angle of projection $\theta$ must be $45^\circ$,which makes $\sin(2\theta) = \sin(90^\circ) = 1$.
Thus,the formula for maximum range is $R_{\max} = \frac{u^2}{g}$.
Given initial velocity $u = 20\; m/s$ and acceleration due to gravity $g = 10\; m/s^2$.
Substituting these values: $R_{\max} = \frac{(20)^2}{10} = \frac{400}{10} = 40\; m$.

(B) Let the initial velocity be $u$ at an angle $\theta$ with the horizontal. The horizontal component is $v_x = u \cos \theta$ and the vertical component is $v_y(t) = u \sin \theta - gt$.
At $t = 3 \, s$,the projectile is at its highest point,so $v_y(3) = 0$.
$u \sin \theta - g(3) = 0 \implies u \sin \theta = 3g = 30 \, m/s$ (taking $g = 10 \, m/s^2$).
At $t = 2 \, s$,the angle of inclination is $30^o$,so $\tan 30^o = \frac{v_y(2)}{v_x}$.
$\frac{1}{\sqrt{3}} = \frac{u \sin \theta - 2g}{u \cos \theta} = \frac{30 - 20}{u \cos \theta} = \frac{10}{u \cos \theta}$.
$u \cos \theta = 10\sqrt{3} \, m/s$.
The magnitude of initial velocity $u = \sqrt{(u \cos \theta)^2 + (u \sin \theta)^2} = \sqrt{(10\sqrt{3})^2 + 30^2} = \sqrt{300 + 900} = \sqrt{1200} = 20\sqrt{3} \, m/s$.
The direction $\theta = \tan^{-1}(\frac{u \sin \theta}{u \cos \theta}) = \tan^{-1}(\frac{30}{10\sqrt{3}}) = \tan^{-1}(\sqrt{3}) = 60^o$.

(A) Given: Horizontal velocity $u = 4\, m/s$,Time $t = 0.4\, s$,Acceleration due to gravity $g = 10\, m/s^2$.
Horizontal distance (Range) $R = u \times t = 4 \times 0.4 = 1.6\, m$.
Vertical height $h = \frac{1}{2}gt^2 = 0.5 \times 10 \times (0.4)^2 = 5 \times 0.16 = 0.8\, m$.
Final vertical velocity $v_y = gt = 10 \times 0.4 = 4\, m/s$.
Final speed $v = \sqrt{u^2 + v_y^2} = \sqrt{4^2 + 4^2} = 4\sqrt{2} \approx 5.66\, m/s$.
Angle with horizontal $\tan \theta = \frac{v_y}{u} = \frac{4}{4} = 1 \implies \theta = 45^o$.
Comparing with options,option $A$ is correct.

(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 of the projectile.
Given the total horizontal distance covered is $R = d_1 + d_2$ and the angle of projection is $\theta = 45^{\circ}$.
At horizontal distance $x = d_1$,the height of the ball is $y = h$.
Substituting these values into the trajectory equation:
$h = d_1 \tan 45^{\circ} \left(1 - \frac{d_1}{d_1 + d_2}\right)$
Since $\tan 45^{\circ} = 1$:
$h = d_1 \left(\frac{d_1 + d_2 - d_1}{d_1 + d_2}\right)$
$h = d_1 \left(\frac{d_2}{d_1 + d_2}\right)$
$h = \frac{d_1 d_2}{d_1 + d_2}$

(C) For maximum horizontal range,the angle of projection is $\theta = 45^{\circ}$.
The horizontal range is given by $R = \frac{u^2 \sin(2\theta)}{g} = \frac{u^2}{g}$.
The speed of a projectile is minimum at the highest point of its trajectory,where the vertical component of velocity becomes zero and only the horizontal component $u_x = u \cos \theta$ remains.
The coordinates of the highest point are $(x, y) = \left( \frac{R}{2}, H \right)$.
The maximum height $H$ is given by $H = \frac{u^2 \sin^2 \theta}{2g}$.
Substituting $\theta = 45^{\circ}$,we get $H = \frac{u^2 \sin^2 45^{\circ}}{2g} = \frac{u^2 (1/2)}{2g} = \frac{u^2}{4g}$.
Since $R = \frac{u^2}{g}$,we can write $H = \frac{R}{4}$.
Therefore,the coordinates of the point where the speed is minimum are $\left( \frac{R}{2}, \frac{R}{4} \right)$.

(D) The maximum height $H$ attained by a projectile is given by the formula $H = \frac{u^2 \sin^2 \theta}{2g}$,where $u$ is the initial speed,$\theta$ is the angle of projection,and $g$ is the acceleration due to gravity.
For the first stone,the angle of projection is $\alpha$,so its maximum height is $H_1 = \frac{u^2 \sin^2 \alpha}{2g}$.
For the second stone,the angle of projection is $(90^{\circ}-\alpha)$,so its maximum height is $H_2 = \frac{u^2 \sin^2(90^{\circ}-\alpha)}{2g}$.
Since $\sin(90^{\circ}-\alpha) = \cos \alpha$,we have $H_2 = \frac{u^2 \cos^2 \alpha}{2g}$.
The ratio of their maximum heights is $\frac{H_1}{H_2} = \frac{\frac{u^2 \sin^2 \alpha}{2g}}{\frac{u^2 \cos^2 \alpha}{2g}} = \frac{\sin^2 \alpha}{\cos^2 \alpha} = \tan^2 \alpha$.
Thus,the ratio is $\tan^2 \alpha : 1$.

(D) 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 speed and $\theta$ is the angle of projection.
Since $u$ is constant for all three balls,$R \propto \sin(2\theta)$.
For $\theta_1 = 15^{\circ}$,$R_1 \propto \sin(30^{\circ}) = 0.5$.
For $\theta_2 = 45^{\circ}$,$R_2 \propto \sin(90^{\circ}) = 1$.
For $\theta_3 = 75^{\circ}$,$R_3 \propto \sin(150^{\circ}) = \sin(180^{\circ} - 30^{\circ}) = \sin(30^{\circ}) = 0.5$.
Thus,$R_1 = R_3 = 0.5 \times \frac{u^2}{g}$ and $R_2 = 1 \times \frac{u^2}{g}$.
Comparing these,we get $R_1 = R_3 < R_2$.

(C) The horizontal range 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 with the horizontal,and $g$ is the acceleration due to gravity.
For the first case,the angle with the horizontal is $\theta$,so the range is $R_1 = \frac{u^2 \sin(2\theta)}{g}$.
For the second case,the angle with the vertical is $\theta$,which means the angle with the horizontal is $(90^{\circ} - \theta)$. Therefore,the range is $R_2 = \frac{u^2 \sin(2(90^{\circ} - \theta))}{g} = \frac{u^2 \sin(180^{\circ} - 2\theta)}{g}$.
Since $\sin(180^{\circ} - 2\theta) = \sin(2\theta)$,we have $R_2 = \frac{u^2 \sin(2\theta)}{g}$.
Comparing the two expressions,we find that $R_1 = R_2$.

(D) Let the speed of projection be $u$ and the angle of projection be $\theta$.
At the point of projection,the speed is $v_1 = u$.
At the highest point of the trajectory,the vertical component of velocity becomes zero,so the speed is $v_2 = u \cos \theta$.
The ratio of the speed at the point of projection to the speed at the top is given as $x$.
Therefore,$\frac{v_1}{v_2} = x$.
Substituting the values,we get $\frac{u}{u \cos \theta} = x$.
This simplifies to $\frac{1}{\cos \theta} = x$,which means $\cos \theta = \frac{1}{x}$.
Thus,the angle of projection is $\theta = \cos^{-1}(\frac{1}{x})$.

(A) Let the initial velocity be $\vec{u} = u \cos \theta \hat{i} + u \sin \theta \hat{j}$.
After time $t$,the velocity vector is $\vec{v} = u \cos \theta \hat{i} + (u \sin \theta - gt) \hat{j}$.
Given that the velocity vector $\vec{v}$ is perpendicular to the initial velocity $\vec{u}$,their dot product must be zero: $\vec{v} \cdot \vec{u} = 0$.
$(u \cos \theta \hat{i} + (u \sin \theta - gt) \hat{j}) \cdot (u \cos \theta \hat{i} + u \sin \theta \hat{j}) = 0$.
$u^2 \cos^2 \theta + (u \sin \theta - gt)(u \sin \theta) = 0$.
$u^2 \cos^2 \theta + u^2 \sin^2 \theta - ugt \sin \theta = 0$.
$u^2 (\cos^2 \theta + \sin^2 \theta) = ugt \sin \theta$.
$u^2 = ugt \sin \theta$.
$t = \frac{u}{g \sin \theta}$.

(A) In projectile motion,the only force acting on the projectile is gravity,which acts vertically downwards.
According to Newton's second law of motion,the acceleration of the projectile is given by $\vec{a} = \frac{\vec{F}}{m} = \frac{m\vec{g}}{m} = \vec{g}$.
The rate of change of velocity is defined as acceleration,i.e.,$\frac{d\vec{v}}{dt} = \vec{a} = \vec{g}$.
The modulus of the rate of change of velocity is $|\frac{d\vec{v}}{dt}| = |\vec{g}| = g$.
Since the acceleration due to gravity $g$ is constant throughout the motion,the modulus of the rate of change of velocity is also constant.