Horizontal Projectile Motion Questions in English

(C) The vertical motion of both balls is governed by the equation of motion $h = ut + \frac{1}{2}gt^2$.
For both balls,the initial vertical velocity $u_y = 0$.
Thus,the time taken to reach the ground is $t = \sqrt{\frac{2h}{g}}$.
Since both $h$ and $g$ are the same for both balls,the time taken to reach the ground is identical for both balls,regardless of their horizontal velocity.

(B) The time taken by the bullet to reach the target is $t = \frac{\text{distance}}{\text{speed}} = \frac{100 \, m}{1000 \, m/s} = 0.1 \, s$.
During this time,the bullet undergoes a vertical downward displacement due to gravity,given by $h = \frac{1}{2}gt^2$.
Substituting the values: $h = \frac{1}{2} \times 10 \, m/s^2 \times (0.1 \, s)^2 = 5 \times 0.01 \, m = 0.05 \, m = 5 \, cm$.
To compensate for this downward drop,the gun must be aimed $5 \, cm$ above the target.

(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 muzzle velocity and $g$ is the acceleration due to gravity.
Given: $R_{\max} = 16 \, km = 16,000 \, m$ and $g = 10 \, m/s^2$.
Substituting the values into the formula:
$16,000 = \frac{u^2}{10}$
$u^2 = 16,000 \times 10 = 160,000$
$u = \sqrt{160,000} = 400 \, m/s$.
Therefore,the muzzle velocity of the shell is $400 \, m/s$.

(C) When the stone is released from the moving train,it possesses an initial horizontal velocity equal to the velocity of the train.
After release,there is no horizontal force acting on the stone (ignoring air resistance),so its horizontal velocity remains constant.
Simultaneously,the stone is subjected to a constant vertical downward acceleration due to gravity $(g)$.
The combination of constant horizontal velocity and constant vertical acceleration results in a trajectory that is a parabola.
Therefore,the stone will hit the ground following a parabolic path.

(B) The motion of both bullets can be analyzed by separating the vertical and horizontal components. For both bullets,the initial vertical velocity $u_y = 0$. The vertical displacement $h$ is given by the equation of motion $h = u_y t + \frac{1}{2} g t^2$. Substituting $u_y = 0$,we get $h = \frac{1}{2} g t^2$,which simplifies to $t = \sqrt{\frac{2h}{g}}$. Since the time $t$ depends only on the height $h$ and the acceleration due to gravity $g$,both bullets will hit the ground simultaneously.

(C) When the pilot releases the ball,it possesses the same horizontal velocity as the aeroplane $(600\, km/hr)$.
For the pilot inside the aeroplane,the relative horizontal velocity of the ball is zero. Therefore,the pilot observes the ball falling vertically in a straight line.
However,for an observer standing on the ground,the ball has both a constant horizontal velocity and a vertical velocity component due to gravity. This combination of motions results in a parabolic trajectory.
Thus,the ball appears to follow a parabolic path to an observer on the ground.

(C) The horizontal velocity of the bomb is $u = 600 \, km/h = 600 \times \frac{5}{18} \, m/s = \frac{500}{3} \, m/s$.
The height from which the bomb is released is $h = 1960 \, m$.
The time taken by the bomb to reach the ground is given by $t = \sqrt{\frac{2h}{g}}$.
Taking $g = 9.8 \, m/s^2$,we have $t = \sqrt{\frac{2 \times 1960}{9.8}} = \sqrt{\frac{3920}{9.8}} = \sqrt{400} = 20 \, s$.
The horizontal distance $AB$ is the horizontal displacement,which is given by $AB = u \times t$.
$AB = \frac{500}{3} \times 20 = \frac{10000}{3} \, m = 3333.33 \, m = 3.33 \, km$.

(C) For a body projected horizontally from a height $h$,the time taken to reach the ground is given by $t = \sqrt{\frac{2h}{g}}$.
Given $h = 5 \, m$ and $g = 10 \, m/s^2$,we have $t = \sqrt{\frac{2 \times 5}{10}} = \sqrt{1} = 1 \, s$.
The horizontal distance (range) $R$ is given by $R = u \times t$,where $u$ is the initial horizontal velocity.
Given $R = 10 \, m$,we have $10 = u \times 1$.
Therefore,$u = 10 \, m/s$.

(A) For both particles,the vertical motion is governed by the equation of motion $h = ut + \frac{1}{2}gt^2$.
Since both particles are released from the same height $h$ and their initial vertical velocity component $u_y$ is $0$ in both cases,the time taken to reach the ground is given by $t = \sqrt{\frac{2h}{g}}$.
Since $h$ and $g$ are the same for both,the time taken $t$ is identical for both particles.
Therefore,both particles will reach the ground simultaneously.

(C) particle moves with constant acceleration in a direction different from the initial velocity. Let the initial velocity be $v$ and the constant acceleration be $a$,with an angle $\theta$ between them $(0 < \theta < 180^{\circ})$.
Resolve the velocity into two components: one parallel to the acceleration $(v \cos \theta)$ and one perpendicular to the acceleration $(v \sin \theta)$.
Let the direction perpendicular to the acceleration be the $x$-axis and the direction of acceleration be the $y$-axis.
In the $x$-direction,there is no acceleration,so the displacement is $x = (v \sin \theta) t$,which gives $t = \frac{x}{v \sin \theta}$.
In the $y$-direction,the initial velocity is $v \cos \theta$ and the acceleration is $a$. The displacement is $y = (v \cos \theta) t + \frac{1}{2} a t^2$.
Substituting the expression for $t$ into the equation for $y$:
$y = (v \cos \theta) \left( \frac{x}{v \sin \theta} \right) + \frac{1}{2} a \left( \frac{x}{v \sin \theta} \right)^2$
$y = x \cot \theta + \frac{a}{2 v^2 \sin^2 \theta} x^2$.
This equation is of the form $y = Ax^2 + Bx$,which represents a parabola.

(B) The bullets fired in all directions with speed $v$ will cover a circular area on the ground.
The radius of this circle is equal to the maximum horizontal range $R_{\max}$ of the projectile.
The formula for the horizontal range is $R = \frac{v^2 \sin(2\theta)}{g}$.
The maximum range occurs at $\theta = 45^\circ$,where $R_{\max} = \frac{v^2}{g}$.
The area $A$ covered by the bullets is the area of a circle with radius $R_{\max}$,given by $A = \pi R_{\max}^2$.
Substituting the value of $R_{\max}$,we get $A = \pi \left( \frac{v^2}{g} \right)^2 = \frac{\pi v^4}{g^2}$.

(D) 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 the range is directly proportional to the square of the initial velocity,i.e.,$R \propto u^2$.
If the initial velocity $u$ is doubled (i.e.,$u' = 2u$),the new range $R'$ becomes:
$R' = \frac{(2u)^2 \sin(2\theta)}{g} = 4 \times \frac{u^2 \sin(2\theta)}{g} = 4R$.
Therefore,the range becomes $4R$.

(C) The formula for the maximum height $(H)$ reached by a projectile is given by:
$H = \frac{u^2 \sin^2 \theta}{2g}$
From this expression,we can see that the maximum height is directly proportional to the square of the initial velocity $(u)$:
$H \propto u^2$
If the initial velocity is doubled $(u' = 2u)$,the new maximum height $(H')$ becomes:
$H' \propto (2u)^2 = 4u^2$
Therefore,$H' = 4H$.
Thus,the maximum height reached by the projectile will be quadrupled.

(A) In the motion of a projectile under gravity,the only force acting on the body is the gravitational force,which is a conservative force.
Since the gravitational force is conservative,the total mechanical energy (kinetic energy + potential energy) of the projectile remains constant throughout its flight.
However,because there is an external gravitational force acting on the projectile,the net external force is not zero.
According to Newton's second law,the rate of change of momentum is equal to the external force. Since the external force is non-zero,the linear momentum of the projectile is not conserved.
Therefore,only the total energy is conserved.

(A) 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.
To find the minimum range,we look for the angle $\theta$ that makes $\sin(2\theta)$ minimum.
The range $R$ is minimum when $\sin(2\theta) = 0$,which occurs at $2\theta = 0^\circ$ or $2\theta = 180^\circ$.
For $2\theta = 180^\circ$,we get $\theta = 90^\circ$.
At $\theta = 90^\circ$,the projectile is thrown vertically upwards,and it returns to the point of projection,resulting in a horizontal range of $R = 0$.

(C) 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 $H = \frac{u^2 \sin^2\theta}{2g}$.
Given that the horizontal range is equal to the maximum height,$R = H$.
Substituting the formulas: $\frac{2u^2 \sin\theta \cos\theta}{g} = \frac{u^2 \sin^2\theta}{2g}$.
Simplifying the equation: $2 \cos\theta = \frac{\sin\theta}{2}$.
Rearranging for $\tan\theta$: $\frac{\sin\theta}{\cos\theta} = 4$,which means $\tan\theta = 4$.
Therefore,$\theta = \tan^{-1}(4)$,which is approximately $76^\circ$.

(C) In projectile motion,the only force acting on the object is gravity,which acts vertically downwards.
There is no force acting in the horizontal direction,meaning the horizontal acceleration is $a_x = 0$.
Since $a_x = 0$,the horizontal component of velocity,$v_x = u \cos \theta$,remains constant throughout the flight.
The vertical component of velocity changes due to the acceleration due to gravity $(g)$,and consequently,the speed and kinetic energy also change.
Therefore,the correct option is $C$.

(A) The direction of velocity is always tangent to the path of the projectile. At the highest point of the trajectory,the vertical component of velocity becomes zero,so the velocity vector is purely horizontal.
Acceleration due to gravity $(g)$ always acts in the vertically downward direction throughout the motion.
Since the velocity is horizontal and the acceleration is vertical,the angle between the velocity vector $\vec{v}$ and the acceleration vector $\vec{g}$ is $90^o$.
Therefore,they are perpendicular to each other.

(D) The horizontal range $R$ of a projectile is given by $R = \frac{u^2 \sin(2\theta)}{g}$.
The maximum vertical height $H$ is given by $H = \frac{u^2 \sin^2 \theta}{2g}$.
Dividing $R$ by $H$,we get $\frac{R}{H} = \frac{u^2 \sin(2\theta) / g}{u^2 \sin^2 \theta / (2g)} = \frac{2 \sin \theta \cos \theta}{\sin^2 \theta / 2} = 4 \cot \theta$.
Therefore,$R = 4H \cot \theta$.
Given the angle of projection $\theta = 45^\circ$,we substitute this into the equation:
$R = 4H \cot(45^\circ)$.
Since $\cot(45^\circ) = 1$,we have $R = 4H$.
Thus,the horizontal range is four times the vertical height.

(C) The velocity components are obtained by differentiating the position equations with respect to time $t$.
$v_x = \frac{dx}{dt} = \frac{d}{dt}(6t) = 6 \ m/s$.
$v_y = \frac{dy}{dt} = \frac{d}{dt}(8t - 5t^2) = 8 - 10t \ m/s$.
At the time of projection,$t = 0$.
Substituting $t = 0$ into the velocity components:
$v_x = 6 \ m/s$.
$v_y = 8 - 10(0) = 8 \ m/s$.
The magnitude of the initial velocity $v$ is given by $v = \sqrt{v_x^2 + v_y^2}$.
$v = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \ m/s$.

(B) The velocity components are obtained by differentiating the position equations with respect to time $t$.
$v_y = \frac{dy}{dt} = \frac{d}{dt}(8t - 5t^2) = 8 - 10t \ m/s$
$v_x = \frac{dx}{dt} = \frac{d}{dt}(6t) = 6 \ m/s$
At the time of projection,$t = 0$.
Substituting $t = 0$ into the velocity components:
$v_{y0} = 8 - 10(0) = 8 \ m/s$
$v_{x0} = 6 \ m/s$
The angle of projection $\theta$ is given by $\tan \theta = \frac{v_{y0}}{v_{x0}}$.
$\tan \theta = \frac{8}{6} = \frac{4}{3}$
Therefore,$\theta = \tan^{-1}(4/3)$.

(A) The position coordinates of the projectile are given by $x = 6t$ and $y = 8t - 5t^2$.
To find the acceleration,we first find the velocity components by differentiating the position with respect to time $t$.
Velocity in the $x$-direction: $v_x = \frac{dx}{dt} = \frac{d}{dt}(6t) = 6 \text{ m/s}$.
Velocity in the $y$-direction: $v_y = \frac{dy}{dt} = \frac{d}{dt}(8t - 5t^2) = 8 - 10t \text{ m/s}$.
Now,we find the acceleration components by differentiating the velocity with respect to time $t$.
Acceleration in the $x$-direction: $a_x = \frac{dv_x}{dt} = \frac{d}{dt}(6) = 0 \text{ m/s}^2$.
Acceleration in the $y$-direction: $a_y = \frac{dv_y}{dt} = \frac{d}{dt}(8 - 10t) = -10 \text{ m/s}^2$.
The negative sign indicates that the acceleration is acting downwards. The magnitude of the acceleration due to gravity is $g = |a_y| = 10 \text{ m/s}^2$.

(B) The formula for the horizontal range of a projectile is $R = \frac{u^2 \sin(2\theta)}{g}$.
For $\theta_1 = 15^\circ$,the range is $R_1 = \frac{u^2 \sin(30^\circ)}{g} = \frac{u^2}{2g} = 1.5 \, km$.
From this,we find $\frac{u^2}{g} = 1.5 \times 2 = 3.0 \, km$.
For $\theta_2 = 45^\circ$,the range is $R_2 = \frac{u^2 \sin(90^\circ)}{g} = \frac{u^2}{g}$.
Substituting the value,$R_2 = 3.0 \, km$.

(A) The horizontal component of the initial velocity is $v_x = v \cos \theta = 25 \cos 60^\circ = 25 \times 0.5 = 12.5\,m/s$.
The vertical component of the initial velocity is $v_y = v \sin \theta = 25 \sin 60^\circ = 25 \times \frac{\sqrt{3}}{2} = 12.5\sqrt{3}\,m/s$.
The time $t$ taken to cover a horizontal distance of $x = 50\,m$ is $t = \frac{x}{v_x} = \frac{50}{12.5} = 4\,s$.
The vertical height $y$ at time $t$ is given by the equation of motion $y = v_y t - \frac{1}{2} g t^2$.
Substituting the values: $y = (12.5\sqrt{3}) \times 4 - \frac{1}{2} \times 9.8 \times (4)^2$.
$y = 50\sqrt{3} - 4.9 \times 16 = 50 \times 1.732 - 78.4 = 86.6 - 78.4 = 8.2\,m$.

(A) The vertical component of the initial velocity is $u_y = u \sin \theta = 25 \sin \theta$.
Using the equation of motion for the vertical direction: $h = u_y t - \frac{1}{2} g t^2$.
Given $h = 5\,m$,$t = 2\,s$,and $g = 10\,m/s^2$,we substitute these values:
$5 = (25 \sin \theta) \times 2 - \frac{1}{2} \times 10 \times (2)^2$.
$5 = 50 \sin \theta - 20$.
$25 = 50 \sin \theta$.
$\sin \theta = \frac{25}{50} = 0.5$.
Therefore,$\theta = \arcsin(0.5) = 30^\circ$.

(C) 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,we see that $R_1 = R_2$.
Therefore,the ratio of the horizontal ranges is $R_1 : R_2 = 1 : 1$.

(B) The horizontal range $R$ of a projectile is given by the formula $R = \frac{v^2 \sin(2\theta)}{g}$.
Here,$v$ is the initial speed,$\theta$ is the angle of projection,and $g$ is the acceleration due to gravity.
On the surface of the Earth,the range is $R = \frac{v^2 \sin(2\theta)}{g_e}$,where $g_e = g$.
On the surface of the Moon,the acceleration due to gravity is $g_m = \frac{g_e}{6} = \frac{g}{6}$.
Since the speed $v$ and the angle $\theta$ remain the same,the new range $R_m$ on the Moon is given by $R_m = \frac{v^2 \sin(2\theta)}{g_m} = \frac{v^2 \sin(2\theta)}{g/6} = 6 \times \left( \frac{v^2 \sin(2\theta)}{g} \right)$.
Substituting the value of $R$,we get $R_m = 6R$.

(B) The initial kinetic energy of the ball is $E = \frac{1}{2}mv^2$,where $v$ is the initial velocity.
At the highest point of the trajectory,the vertical component of the velocity becomes zero,while the horizontal component remains constant.
The horizontal component of velocity is $v_x = v \cos \theta$.
At the highest point,the velocity of the ball is $v_h = v \cos \theta$.
The kinetic energy at the highest point is $E' = \frac{1}{2}m(v_h)^2 = \frac{1}{2}m(v \cos \theta)^2$.
$E' = (\frac{1}{2}mv^2) \cos^2 \theta = E \cos^2 \theta$.
Given $\theta = 45^\circ$,we have $\cos 45^\circ = \frac{1}{\sqrt{2}}$.
Therefore,$E' = E (\frac{1}{\sqrt{2}})^2 = E (\frac{1}{2}) = \frac{E}{2}$.

(B) At the highest point,the vertical component of velocity is zero,and the horizontal component is $v_x = v \cos 45^{\circ} = v / \sqrt{2}$.
The momentum of the particle at this point is $p = m v_x = mv / \sqrt{2}$.
The angular momentum $L$ about the point of projection is given by $L = p \times h$,where $h$ is the maximum height.
The maximum height $h$ is given by $h = (v^2 \sin^2 45^{\circ}) / (2g) = (v^2 \cdot (1/2)) / (2g) = v^2 / (4g)$.
Substituting these values,we get $L = (mv / \sqrt{2}) \times (v^2 / 4g) = mv^3 / (4\sqrt{2}g)$.

(C) In projectile motion,the horizontal displacement $x$ is given by $x = u_x t$,which is a linear function of time. The vertical displacement $y$ is given by $y = u_y t - \frac{1}{2} g t^2$,which is a downward-opening parabola. At the highest point,the vertical velocity $v_y = \frac{dy}{dt} = 0$. On a displacement-time graph for vertical motion,the slope is $\frac{dy}{dt}$. At the highest point,the slope is $0$. Since the path is a downward-opening parabola ($y = at^2 + bt + c$ with $a < 0$),the second derivative $\frac{d^2y}{dt^2} = -g$ is negative. However,the question asks for the displacement-time graph of the particle's motion. The vertical displacement $y(t)$ reaches a maximum at the peak,where the slope is $0$ and the curvature (concavity) is negative. Thus,the correct description is zero slope and negative curvature.

(D) During the entire flight of a projectile,the only force acting on it is the gravitational force,which acts vertically downwards.
According to Newton's second law,$F = ma$,the acceleration $a = F/m = mg/m = g$.
Therefore,the acceleration remains constant and equal to $g$ throughout the projectile motion,including at the highest point of the trajectory.

(C) The horizontal component of the initial velocity is $u_x = u \cos \theta$.
Since there is no acceleration in the horizontal direction,the horizontal velocity remains constant throughout the motion.
The time of flight $T$ for a projectile is given by $T = \frac{2u \sin \theta}{g}$.
The horizontal range $R$ is the product of the horizontal velocity and the time of flight:
$R = u_x \times T$
$R = (u \cos \theta) \times \left( \frac{2u \sin \theta}{g} \right)$
Using the trigonometric identity $\sin 2\theta = 2 \sin \theta \cos \theta$,we get:
$R = \frac{u^2 (2 \sin \theta \cos \theta)}{g} = \frac{u^2 \sin 2\theta}{g}$.

(C) The time of flight $T$ for a projectile is given by the formula:
$T = \frac{2u \sin \theta}{g}$
Given:
Initial velocity $u = 50 \ m/s$
Angle of projection $\theta = 30^{\circ}$
Acceleration due to gravity $g = 10 \ m/s^2$
Substituting the values:
$T = \frac{2 \times 50 \times \sin 30^{\circ}}{10}$
Since $\sin 30^{\circ} = 0.5$:
$T = \frac{100 \times 0.5}{10} = \frac{50}{10} = 5 \ s$
Therefore,the ball remains in the air for $5 \ s$.

(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 $H$ is given by $H = \frac{u^2 \sin^2 \theta}{2g}$.
According to the problem,the horizontal range is three times the maximum height,so $R = 3H$.
Substituting the formulas:
$\frac{2u^2 \sin \theta \cos \theta}{g} = 3 \left( \frac{u^2 \sin^2 \theta}{2g} \right)$.
Simplifying the equation:
$2 \cos \theta = \frac{3}{2} \sin \theta$.
Rearranging to find $\tan \theta$:
$\tan \theta = \frac{4}{3}$.
Therefore,$\theta = \tan^{-1}(1.333) \approx 53^\circ 8'$.

(C) Let the initial velocity of the bullet be $u$ and the angle of projection be $\theta$. The target is at a distance $d$ from the gun. The time taken for the bullet to reach the horizontal distance $d$ is $t = d / (u \cos \theta)$.
In this time $t$,the vertical displacement of the bullet due to gravity is $y_1 = (1/2) g t^2$.
The vertical position of the bullet relative to the line of sight (which is at angle $\theta$) is $y_2 = (u \sin \theta) t - (1/2) g t^2$.
Since the target is in the line of the barrel,the initial height of the target is $h = d \tan \theta = (u \sin \theta) t$.
The actual position of the target at time $t$ is $h' = h - (1/2) g t^2 = (u \sin \theta) t - (1/2) g t^2$.
Since the vertical position of the bullet $y_2$ is equal to the vertical position of the target $h'$,the bullet will hit the target.

(B) The formula for the maximum height $H$ of a projectile is given by $H = \frac{u^2 \sin^2 \theta}{2g}$.
Since both bodies are projected with the same velocity $u$,the ratio of the maximum heights $H_1$ and $H_2$ is given by:
$\frac{H_1}{H_2} = \frac{\sin^2 \theta_1}{\sin^2 \theta_2}$.
Given $\theta_1 = 30^\circ$ and $\theta_2 = 60^\circ$,we have:
$\frac{H_1}{H_2} = \frac{\sin^2 30^\circ}{\sin^2 60^\circ} = \frac{(1/2)^2}{(\sqrt{3}/2)^2} = \frac{1/4}{3/4} = \frac{1}{3}$.
Therefore,the ratio is $1:3$.

(D) The horizontal range $R$ of a projectile fired with muzzle speed $V$ at an angle of elevation $\theta$ is given by the formula:
$R = \frac{V^2 \sin(2\theta)}{g}$
Rearranging the formula to solve for $\theta$:
$\sin(2\theta) = \frac{Rg}{V^2}$
Taking the inverse sine on both sides:
$2\theta = \sin^{-1}\left(\frac{gR}{V^2}\right)$
Dividing by $2$:
$\theta = \frac{1}{2} \sin^{-1}\left(\frac{gR}{V^2}\right)$
Thus,the correct option is $D$.

(A) The formula for the time of flight of a projectile is $T = \frac{2u \sin \theta}{g}$.
Given $T = 10 \ s$ and taking $g = 10 \ m/s^2$,we have $10 = \frac{2u \sin \theta}{10}$,which gives $u \sin \theta = 50 \ m/s$.
The formula for the maximum height attained by a projectile is $H = \frac{u^2 \sin^2 \theta}{2g}$.
Substituting the value of $u \sin \theta$,we get $H = \frac{(u \sin \theta)^2}{2g} = \frac{50^2}{2 \times 10} = \frac{2500}{20} = 125 \ m$.

(B) The horizontal range $R$ of a projectile is given by the formula: $R = \frac{v^2 \sin(2\theta)}{g}$.
For body $A$,the angle of projection is $\theta_A = 30^{\circ}$. Thus,$R_A = \frac{v^2 \sin(2 \times 30^{\circ})}{g} = \frac{v^2 \sin(60^{\circ})}{g}$.
For body $B$,the angle of projection is $\theta_B = 60^{\circ}$. Thus,$R_B = \frac{v^2 \sin(2 \times 60^{\circ})}{g} = \frac{v^2 \sin(120^{\circ})}{g}$.
Since $\sin(120^{\circ}) = \sin(180^{\circ} - 60^{\circ}) = \sin(60^{\circ})$,it follows that $R_A = R_B$.
Therefore,the ratio of the horizontal range of $A$ to $B$ is $R_A : R_B = 1:1$.

(B) The formula for the horizontal range $R$ of a projectile is given by $R = \frac{u^2 \sin(2\theta)}{g}$.
Given values are:
Initial velocity $u = 500 \, m/s$
Angle of projection $\theta = 15^\circ$
Acceleration due to gravity $g = 10 \, m/s^2$
Substituting these values into the formula:
$R = \frac{(500)^2 \times \sin(2 \times 15^\circ)}{10}$
$R = \frac{250000 \times \sin(30^\circ)}{10}$
Since $\sin(30^\circ) = 0.5$,we have:
$R = \frac{250000 \times 0.5}{10} = 25000 \times 0.5 = 12500 \, m$
This can be written as $12.5 \times 10^3 \, m$.
Therefore,the correct option is $B$.

(A) The time of flight $T$ for a projectile is given by the formula: $T = \frac{2u \sin \theta}{g}$.
Given: $T = 2 \ s$,$\theta = 30^o$,and $g = 9.8 \ m/s^2$.
Rearranging the formula to solve for the initial velocity $u$:
$u = \frac{T \times g}{2 \sin \theta}$.
Substituting the values:
$u = \frac{2 \times 9.8}{2 \times \sin 30^o} = \frac{19.6}{2 \times 0.5} = \frac{19.6}{1} = 19.6 \ m/s$.

(C) The horizontal range $R$ of a projectile is given by the formula $R = \frac{u^2 \sin(2\theta)}{g}$.
Assuming the angle of projection $\theta$ and the acceleration due to gravity $g$ remain constant,the range is directly proportional to the square of the initial speed: $R \propto u^2$.
Let the initial speed be $u_1$ and the initial range be $R_1 = 50\, m$.
When the speed is doubled,$u_2 = 2u_1$.
The new range $R_2$ is given by $R_2 = R_1 \times (\frac{u_2}{u_1})^2$.
Substituting the values: $R_2 = 50 \times (\frac{2u_1}{u_1})^2 = 50 \times 2^2 = 50 \times 4 = 200\, m$.

(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.
For the first case,$\theta_1 = 60^o$,so $R_1 = \frac{u^2 \sin(120^o)}{g} = 90 \,m$.
For the second case,$\theta_2 = 30^o$,so $R_2 = \frac{u^2 \sin(60^o)}{g}$.
Since $\sin(120^o) = \sin(180^o - 60^o) = \sin(60^o)$,it follows that $\sin(2 \times 60^o) = \sin(2 \times 30^o)$.
Therefore,the range is the same for complementary angles (angles that add up to $90^o$).
Since $60^o + 30^o = 90^o$,the range $R_2$ will be equal to $R_1$,which is $90 \,m$.

(A) The maximum height $H$ reached by a projectile is given by $H = \frac{u^2 \sin^2 \theta}{2g}$.
The time of flight $T$ is given by $T = \frac{2u \sin \theta}{g}$.
Squaring the time of flight,we get $T^2 = \frac{4u^2 \sin^2 \theta}{g^2}$.
Now,the ratio of maximum height to the square of flight time is $\frac{H}{T^2} = \frac{u^2 \sin^2 \theta / 2g}{4u^2 \sin^2 \theta / g^2}$.
Simplifying this expression: $\frac{H}{T^2} = \frac{u^2 \sin^2 \theta}{2g} \times \frac{g^2}{4u^2 \sin^2 \theta} = \frac{g}{8}$.
Given $g = 10 \ m/s^2$,we have $\frac{H}{T^2} = \frac{10}{8} = \frac{5}{4}$.

(B) The horizontal distance covered in a long-jump is equivalent to the range of a projectile. The formula for the horizontal range $R$ is given by $R = \frac{u^2 \sin(2\theta)}{g}$,where $u$ is the initial speed,$\theta$ is the angle of projection,and $g$ is the acceleration due to gravity.
From this formula,it is clear that the range $R$ depends on the initial speed $u$ and the angle of projection $\theta$ (the direction in which the athlete leaps).
Therefore,the correct set of factors is the direction in which he leaps and the initial speed.

(C) The time of flight $T$ for a projectile is given by $T = \frac{2u \sin \theta}{g}$.
Given $T = 2 \ s$ and $g = 10 \ m/s^2$,we have $2 = \frac{2u \sin \theta}{10}$.
This simplifies to $u \sin \theta = 10 \ m/s$.
The maximum height $H$ attained by the projectile is given by $H = \frac{u^2 \sin^2 \theta}{2g}$.
Substituting the value of $u \sin \theta$,we get $H = \frac{(10)^2}{2 \times 10} = \frac{100}{20} = 5 \ m$.

(B) In projectile motion,the velocity of the projectile can be resolved into two components: horizontal $(u_x = u \cos \theta)$ and vertical $(u_y = u \sin \theta)$.
At the maximum height,the vertical component of velocity becomes zero $(v_y = 0)$.
However,the horizontal component of velocity remains constant throughout the motion because there is no acceleration acting in the horizontal direction.
Therefore,the velocity at the maximum height is equal to the horizontal component,which is $u \cos \theta$.

(A) The horizontal range $R$ of a projectile is given by the formula $R = \frac{u^2 \sin(2\theta)}{g}$.
Given that the initial velocity $u$ is the same for both projectiles.
For the first projectile,$\theta_1 = 60^o$,so $R_1 = \frac{u^2 \sin(120^o)}{g} = \frac{u^2 \sin(60^o)}{g}$.
For the second projectile,$\theta_2 = 30^o$,so $R_2 = \frac{u^2 \sin(60^o)}{g}$.
Since $\sin(120^o) = \sin(60^o)$,the horizontal range $R$ is the same for both angles.
Angles $\theta$ and $(90^o - \theta)$ are complementary angles,and for any pair of complementary angles,the horizontal range is identical.