Horizontal Projectile Motion Questions in English

(D) Given,
Initial speed of the ball $(u) = 20 \,m/s$
Angle of projection $(\theta) = 30^{\circ}$
Acceleration due to gravity $(g) = 10 \,m/s^2$
The formula for the maximum height $(H)$ reached by a projectile is given by:
$H = \frac{u^2 \sin^2 \theta}{2g}$
Substituting the given values into the formula:
$H = \frac{(20)^2 \times (\sin 30^{\circ})^2}{2 \times 10}$
$H = \frac{400 \times (1/2)^2}{20}$
$H = \frac{400 \times (1/4)}{20}$
$H = \frac{100}{20}$
$H = 5 \,m$
Therefore,the maximum height reached by the ball is $5 \,m$.

(C) The horizontal component of the velocity is $v_x = v \cos \theta = 2 \sqrt{gh} \cos 60^{\circ}$.
Since $\cos 60^{\circ} = \frac{1}{2}$,we have $v_x = 2 \sqrt{gh} \times \frac{1}{2} = \sqrt{gh}$.
The particle travels a horizontal distance $d = 2h$ between the two walls.
Since the horizontal component of velocity remains constant in projectile motion,the time $t$ taken to travel between the walls is given by $t = \frac{d}{v_x}$.
Substituting the values,$t = \frac{2h}{\sqrt{gh}} = 2 \sqrt{\frac{h}{g}}$.

(B) For the first body thrown vertically upwards with velocity $v_1$,the time taken to reach maximum height is $t_1 = \frac{v_1}{g}$ and the maximum height is $h_1 = \frac{v_1^2}{2g}$.
For the second body projected with velocity $v_2$ at an angle $\theta$,the time taken to reach maximum height is $t_2 = \frac{v_2 \sin \theta}{g}$ and the maximum height is $h_2 = \frac{v_2^2 \sin^2 \theta}{2g}$.
Given $t_1 = 2t_2$,we have $\frac{v_1}{g} = 2 \left( \frac{v_2 \sin \theta}{g} \right)$,which simplifies to $v_1 = 2 v_2 \sin \theta$.
Now,the ratio of heights is $\frac{h_1}{h_2} = \frac{v_1^2 / 2g}{v_2^2 \sin^2 \theta / 2g} = \frac{v_1^2}{v_2^2 \sin^2 \theta}$.
Substituting $v_1 = 2 v_2 \sin \theta$,we get $\frac{h_1}{h_2} = \frac{(2 v_2 \sin \theta)^2}{v_2^2 \sin^2 \theta} = \frac{4 v_2^2 \sin^2 \theta}{v_2^2 \sin^2 \theta} = 4$.
Thus,the ratio is $4:1$.

(D) The given equation of the trajectory is $y = ax - bx^2$.
The standard equation of a projectile trajectory is $y = x \tan \theta - \frac{gx^2}{2u^2 \cos^2 \theta}$.
Comparing the coefficients of $x$ and $x^2$ in both equations:
$\tan \theta = a \implies \theta = \tan^{-1}(a)$.
Also,$b = \frac{g}{2u^2 \cos^2 \theta}$.
The maximum height $H$ is given by $H = \frac{u^2 \sin^2 \theta}{2g}$.
From $\tan \theta = a$,we have $\sin \theta = \frac{a}{\sqrt{1+a^2}}$ and $\cos \theta = \frac{1}{\sqrt{1+a^2}}$.
Substituting $u^2 = \frac{g}{2b \cos^2 \theta}$ into the height formula:
$H = \frac{g}{2b \cos^2 \theta} \cdot \frac{\sin^2 \theta}{2g} = \frac{\tan^2 \theta}{4b} = \frac{a^2}{4b}$.
Thus,the maximum height is $\frac{a^2}{4b}$ and the angle of projection is $\tan^{-1}(a)$.

(C) The range of a projectile is given by $R = \frac{u^2 \sin 2\theta}{g}$.
For maximum range,the angle of projection must be $\theta = 45^{\circ}$.
Thus,$R_{\max} = \frac{u^2 \sin(90^{\circ})}{g} = \frac{u^2}{g} \quad \dots (i)$.
The time of flight $T$ is given by $T = \frac{2u \sin \theta}{g}$.
Substituting $\theta = 45^{\circ}$,we get $T = \frac{2u \sin 45^{\circ}}{g} = \frac{2u}{g \sqrt{2}} = \frac{u \sqrt{2}}{g}$.
From this,we find $u = \frac{Tg}{\sqrt{2}}$.
Substituting the value of $u$ into equation $(i)$:
$R_{\max} = \frac{1}{g} \left( \frac{Tg}{\sqrt{2}} \right)^2 = \frac{1}{g} \cdot \frac{T^2 g^2}{2} = \frac{g T^2}{2}$.

(C) The average velocity is defined as the total displacement divided by the total time taken.
Let the particle be projected from the origin $(0,0)$. At the highest point,the coordinates are $(R/2, H)$,where $R$ is the horizontal range and $H$ is the maximum height.
Displacement vector $\vec{s} = \frac{R}{2} \hat{i} + H \hat{j}$.
The magnitude of displacement is $|\vec{s}| = \sqrt{(R/2)^2 + H^2}$.
The time taken to reach the highest point is $t = \frac{T}{2} = \frac{v \sin \theta}{g}$.
We know $R = \frac{v^2 \sin 2\theta}{g} = \frac{2v^2 \sin \theta \cos \theta}{g}$ and $H = \frac{v^2 \sin^2 \theta}{2g}$.
Thus,$R/2 = \frac{v^2 \sin \theta \cos \theta}{g}$.
$|\vec{s}| = \sqrt{\left(\frac{v^2 \sin \theta \cos \theta}{g}\right)^2 + \left(\frac{v^2 \sin^2 \theta}{2g}\right)^2} = \frac{v^2 \sin \theta}{g} \sqrt{\cos^2 \theta + \frac{\sin^2 \theta}{4}} = \frac{v^2 \sin \theta}{2g} \sqrt{4 \cos^2 \theta + \sin^2 \theta} = \frac{v^2 \sin \theta}{2g} \sqrt{3 \cos^2 \theta + 1}$.
Average velocity $v_{av} = \frac{|\vec{s}|}{t} = \frac{\frac{v^2 \sin \theta}{2g} \sqrt{3 \cos^2 \theta + 1}}{\frac{v \sin \theta}{g}} = \frac{v}{2} \sqrt{1+3 \cos^2 \theta}$.

(B) The equation of trajectory for a projectile is given by $y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta}$.
Since the particle passes through point $(r, y)$,we have $y = r \tan \theta - \frac{g r^2}{2 u^2} (1 + \tan^2 \theta)$.
Rearranging this as a quadratic in $\tan \theta$: $\frac{g r^2}{2 u^2} \tan^2 \theta - r \tan \theta + (y + \frac{g r^2}{2 u^2}) = 0$.
Let the two angles of projection be $\theta_1$ and $\theta_2$. Then $\tan \theta_1 \tan \theta_2 = \frac{y + \frac{g r^2}{2 u^2}}{\frac{g r^2}{2 u^2}} = 1 + \frac{2 u^2 y}{g r^2}$.
The time taken to reach horizontal distance $r$ is $t = \frac{r}{u \cos \theta}$.
Thus,$t_1 t_2 = \frac{r^2}{u^2 \cos \theta_1 \cos \theta_2}$.
Using the property of projectile motion,$t_1 t_2 = \frac{2 r}{g \tan \alpha}$ where $\alpha$ is the angle of elevation of point $P$. For a fixed point $(r, y)$,$t_1 t_2 = \frac{2 r}{g \tan \theta_{elevation}} = \frac{2 r}{g (y/r)} = \frac{2 r^2}{g y}$.
However,in standard problems of this type where the point is at the same level $(y=0)$,$t_1 t_2 = \frac{2 r \tan \theta}{g} \dots$ actually,for a fixed point $(r, y)$,$t_1 t_2 = \frac{2 r}{g \tan \theta_{elevation}}$. If we consider the specific case where the point is at the same horizontal level as the projection point $(y=0)$,then $t_1 t_2 = \frac{2 r}{g \tan \theta}$. Given the options provided,the product $t_1 t_2$ is proportional to $r$.

(C) The velocity of a projectile is always tangent to its path. At the maximum height,the vertical component of velocity becomes zero,while the horizontal component remains constant. Therefore,the velocity vector is purely horizontal at the maximum height,making an angle of $0^{\circ}$ with the horizontal.

(B) The formula for the range of a projectile is $R = \frac{u^2 \sin 2\theta}{g}$.
Given $R = 20 \ m$ and $\theta = 30^{\circ}$,we have $20 = \frac{u^2 \sin(2 \times 30^{\circ})}{g}$.
$\Rightarrow 20 = \frac{u^2 \sin 60^{\circ}}{g} = \frac{u^2}{g} \times \frac{\sqrt{3}}{2}$.
Therefore,$\frac{u^2}{g} = \frac{20 \times 2}{\sqrt{3}} = \frac{40}{\sqrt{3}}$.
The formula for maximum height is $H = \frac{u^2 \sin^2 \theta}{2g} = \frac{1}{2} \left(\frac{u^2}{g}\right) \sin^2 \theta$.
Substituting the values,$H = \frac{1}{2} \times \left(\frac{40}{\sqrt{3}}\right) \times \sin^2 30^{\circ}$.
$H = \frac{20}{\sqrt{3}} \times \left(\frac{1}{2}\right)^2 = \frac{20}{\sqrt{3}} \times \frac{1}{4} = \frac{5}{\sqrt{3}} \ m$.

(C) The standard equation of trajectory for a projectile is given by $y = x \tan \theta - \frac{g}{2u^2 \cos^2 \theta} x^2$.
Comparing this with the given equation $y = 10x - \frac{5}{9}x^2$,we get:
$\tan \theta = 10$
and
$\frac{g}{2u^2 \cos^2 \theta} = \frac{5}{9}$.
Given $g = 10 \ m/s^2$,we substitute it into the second equation:
$\frac{10}{2u^2 \cos^2 \theta} = \frac{5}{9} \implies \frac{5}{u^2 \cos^2 \theta} = \frac{5}{9} \implies u^2 \cos^2 \theta = 9$.
The range $R$ of a projectile is given by $R = \frac{2u^2 \sin \theta \cos \theta}{g}$.
We can rewrite this as $R = \frac{2(u^2 \cos^2 \theta) \tan \theta}{g}$.
Substituting the values $u^2 \cos^2 \theta = 9$,$\tan \theta = 10$,and $g = 10 \ m/s^2$:
$R = \frac{2 \times 9 \times 10}{10} = 18 \ m$.

(A) The horizontal displacement is given by $x = (u \cos \theta) t = 6t$,which implies $u \cos \theta = 6 \text{ m/s}$.
The vertical displacement is given by $y = (u \sin \theta) t - \frac{1}{2} g t^2 = 8t - 5t^2$. Comparing this with the standard equation,we get $u \sin \theta = 8 \text{ m/s}$ and $\frac{1}{2} g = 5$,so $g = 10 \text{ m/s}^2$.
The range $R$ of a projectile is given by the formula $R = \frac{u^2 \sin 2\theta}{g} = \frac{2(u \sin \theta)(u \cos \theta)}{g}$.
Substituting the values: $R = \frac{2 \times 8 \times 6}{10} = \frac{96}{10} = 9.6 \text{ m}$.

(A) The given equations are $x = 36t$ and $2y = 96t - 9.8t^2$.
Dividing the second equation by $2$,we get $y = 48t - 4.9t^2$.
Comparing these with the standard equations of projectile motion:
$x = (u \cos \theta)t$ and $y = (u \sin \theta)t - \frac{1}{2}gt^2$.
We identify the horizontal component of velocity as $u \cos \theta = 36$ and the vertical component as $u \sin \theta = 48$.
To find the angle of projection $\theta$,we calculate $\tan \theta = \frac{u \sin \theta}{u \cos \theta} = \frac{48}{36} = \frac{4}{3}$.
Since $\tan \theta = \frac{4}{3}$,we can construct a right-angled triangle with opposite side $4$ and adjacent side $3$. The hypotenuse is $\sqrt{4^2 + 3^2} = 5$.
Therefore,$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{4}{5}$.
Thus,$\theta = \sin^{-1}(\frac{4}{5})$.

(D) The horizontal displacement is given by $x = 36t$. Comparing this with the standard equation $x = u_x t$,we get the horizontal component of initial velocity $u_x = 36 \ m/s$.
The vertical displacement is given by $y = 48t - 4.9t^2$. Comparing this with the standard equation $y = u_y t - \frac{1}{2}gt^2$ (where $g \approx 9.8 \ m/s^2$),we get the vertical component of initial velocity $u_y = 48 \ m/s$.
The initial velocity $u$ is the magnitude of the resultant vector of its components:
$u = \sqrt{u_x^2 + u_y^2}$
$u = \sqrt{36^2 + 48^2}$
$u = \sqrt{1296 + 2304}$
$u = \sqrt{3600}$
$u = 60 \ m/s$.

(D) Given: Initial velocity $u = 20 \ m/s$,angle of projection $\theta = 45^{\circ}$,and $g = 10 \ m/s^2$.
The standard equation of the trajectory of a projectile is given by $h = x \tan \theta - \frac{gx^2}{2u^2 \cos^2 \theta}$.
Comparing this with the given equation $h = Ax - Bx^2$,we get:
$A = \tan \theta = \tan 45^{\circ} = 1$.
$B = \frac{g}{2u^2 \cos^2 \theta} = \frac{10}{2 \times (20)^2 \times (\cos 45^{\circ})^2} = \frac{10}{2 \times 400 \times (1/\sqrt{2})^2} = \frac{10}{800 \times 1/2} = \frac{10}{400} = \frac{1}{40}$.
Now,the ratio $A:B$ is calculated as:
$\frac{A}{B} = \frac{1}{1/40} = 40$.
Therefore,the ratio $A:B$ is $40:1$.

(C) Let the initial velocity of projection be $u$. The initial kinetic energy is given by $KE_1 = \frac{1}{2} m u^2 = K$.
At the highest point of the trajectory,the vertical component of the velocity becomes zero,while the horizontal component remains $u \cos \theta$.
Therefore,the kinetic energy at the highest point $KE_2$ is:
$KE_2 = \frac{1}{2} m (u \cos \theta)^2$
$KE_2 = (\frac{1}{2} m u^2) \cos^2 \theta$
Substituting $KE_1 = K$ and $\theta = 60^{\circ}$:
$KE_2 = K \cos^2 60^{\circ}$
Since $\cos 60^{\circ} = \frac{1}{2}$,we have:
$KE_2 = K (\frac{1}{2})^2 = \frac{K}{4}$

(B) In projectile motion,when air resistance is neglected,there is no horizontal force acting on the projectile.
According to Newton's first law of motion,the horizontal acceleration $(a_{x})$ is zero.
Since $a_{x} = 0$,the horizontal component of velocity $(u_{x})$ remains constant throughout the motion.
Therefore,the graph of $u_{x}$ versus $t$ must be a horizontal straight line parallel to the time axis.
This corresponds to the figure shown in option $B$.

(A) Given:
Horizontal velocity of the plane,$u = 360 \text{ km/h} = 360 \times \frac{5}{18} \text{ m/s} = 100 \text{ m/s}$.
Altitude (vertical height),$h = 500 \text{ m}$.
Acceleration due to gravity,$g = 10 \text{ m/s}^2$.
When the bomb is dropped,it follows a projectile path. The time taken to reach the ground is given by $t = \sqrt{\frac{2h}{g}}$.
$t = \sqrt{\frac{2 \times 500}{10}} = \sqrt{100} = 10 \text{ s}$.
The horizontal distance covered by the bomb during this time is $R = u \times t$.
$R = 100 \text{ m/s} \times 10 \text{ s} = 1000 \text{ m}$.
Therefore,the bomb should be dropped at a distance of $1000 \text{ m}$ ahead of the target.

(A) Given,initial velocity $u = 10 \ m/s$.
Range $R = 5 \ m$.
Acceleration due to gravity $g = 10 \ m/s^2$.
The formula for the range of a projectile is $R = \frac{u^2 \sin(2\alpha)}{g}$.
Substituting the given values:
$5 = \frac{(10)^2 \sin(2\alpha)}{10}$
$5 = \frac{100 \sin(2\alpha)}{10}$
$5 = 10 \sin(2\alpha)$
$\sin(2\alpha) = \frac{5}{10} = 0.5$.
Since $\sin(30^{\circ}) = 0.5$,we have $2\alpha = 30^{\circ}$ or $2\alpha = 150^{\circ}$.
Therefore,$\alpha = 15^{\circ}$ or $\alpha = 75^{\circ}$.
Comparing with the given options,the correct value is $15^{\circ}$.

(A) The vertical displacement $y$ of a projectile at time $t$ is given by $y = (u \sin \theta)t - \frac{1}{2}gt^2$.
For heights $h_1$ and $h_2$ at times $t_1$ and $t_2$:
$h_1 = (u \sin \theta)t_1 - \frac{1}{2}gt_1^2 \implies \frac{h_1}{t_1} = u \sin \theta - \frac{1}{2}gt_1 \quad (1)$
$h_2 = (u \sin \theta)t_2 - \frac{1}{2}gt_2^2 \implies \frac{h_2}{t_2} = u \sin \theta - \frac{1}{2}gt_2 \quad (2)$
Subtracting $(2)$ from $(1)$:
$\frac{h_1}{t_1} - \frac{h_2}{t_2} = \frac{1}{2}g(t_2 - t_1) \implies \frac{h_1 t_2 - h_2 t_1}{t_1 t_2} = \frac{1}{2}g(t_2 - t_1)$
$\frac{g}{2} = \frac{h_1 t_2 - h_2 t_1}{t_1 t_2 (t_2 - t_1)} \quad (3)$
The time of flight $T$ is given by $T = \frac{2u \sin \theta}{g}$. From $(1)$,$u \sin \theta = \frac{h_1}{t_1} + \frac{1}{2}gt_1$.
$T = \frac{2}{g} \left( \frac{h_1}{t_1} + \frac{1}{2}gt_1 \right) = \frac{2h_1}{gt_1} + t_1$.
Substituting $\frac{g}{2}$ from $(3)$ into the expression for $T$:
$T = \frac{h_1}{t_1} \left( \frac{t_1 t_2 (t_2 - t_1)}{h_1 t_2 - h_2 t_1} \right) + t_1 = \frac{h_1 t_2 (t_2 - t_1) + t_1 (h_1 t_2 - h_2 t_1)}{h_1 t_2 - h_2 t_1}$
$T = \frac{h_1 t_2^2 - h_1 t_1 t_2 + h_1 t_1 t_2 - h_2 t_1^2}{h_1 t_2 - h_2 t_1} = \frac{h_1 t_2^2 - h_2 t_1^2}{h_1 t_2 - h_2 t_1}$.

(A) Given,initial velocity vector $v = (3 \hat{i} + 10 \hat{j}) \text{ m/s}$.
Comparing with $v = v_x \hat{i} + v_y \hat{j}$,we get $v_x = 3 \text{ m/s}$ and $v_y = 10 \text{ m/s}$.
The maximum height $H$ attained by a projectile is given by the formula $H = \frac{v_y^2}{2g}$.
Substituting the values: $H = \frac{10^2}{2 \times 10} = \frac{100}{20} = 5 \text{ m}$.
The horizontal range $R$ is given by the formula $R = v_x \times T$,where $T = \frac{2v_y}{g}$ is the time of flight.
Calculating time of flight: $T = \frac{2 \times 10}{10} = 2 \text{ s}$.
Calculating range: $R = 3 \times 2 = 6 \text{ m}$.
Thus,the maximum height is $5 \text{ m}$ and the range is $6 \text{ m}$.

(C) At the ground,the initial kinetic energy is given by $E = \frac{1}{2} m u^{2}$,where $u$ is the initial velocity.
At the highest point of the trajectory,the vertical component of the velocity becomes zero,and the velocity of the particle is only the horizontal component,which is $v_x = u \cos \theta$.
Given the angle of projection $\theta = 60^{\circ}$,the horizontal velocity at the highest point is $v_x = u \cos 60^{\circ} = u \times \frac{1}{2} = \frac{u}{2}$.
The kinetic energy at the highest point $E^{\prime}$ is given by $E^{\prime} = \frac{1}{2} m v_x^{2}$.
Substituting the value of $v_x$,we get $E^{\prime} = \frac{1}{2} m \left( \frac{u}{2} \right)^{2} = \frac{1}{2} m \left( \frac{u^{2}}{4} \right) = \frac{1}{4} \left( \frac{1}{2} m u^{2} \right)$.
Since $E = \frac{1}{2} m u^{2}$,we have $E^{\prime} = \frac{E}{4}$.

(C) Given: Height $h = 80 \ m$,initial horizontal velocity $v = 8 \ ms^{-1}$,and acceleration due to gravity $g = 10 \ ms^{-2}$.
For vertical motion,the time taken to reach the ground is given by the equation $h = \frac{1}{2}gt^2$.
Substituting the values: $80 = \frac{1}{2} \times 10 \times t^2$.
$80 = 5t^2 \implies t^2 = 16 \implies t = 4 \ s$.
For horizontal motion,the distance $d$ covered is given by $d = v \times t$.
Substituting the values: $d = 8 \times 4 = 32 \ m$.
Thus,the time $t$ is $4 \ s$ and the distance $d$ is $32 \ m$.

(A) The initial kinetic energy of the body is $K = \frac{1}{2} m u^2$.
At the highest point of the trajectory,the vertical component of velocity is zero,and the horizontal component remains $u \cos \beta$.
Thus,the kinetic energy at the highest point is $K' = \frac{1}{2} m (u \cos \beta)^2 = K \cos^2 \beta$.
Given that $K' = \frac{3}{4} K$,we have $K \cos^2 \beta = \frac{3}{4} K$.
This simplifies to $\cos^2 \beta = \frac{3}{4}$,which means $\cos \beta = \frac{\sqrt{3}}{2}$.
Therefore,$\beta = 30^{\circ}$.

(B) The motion of the ball is a horizontal projectile motion.
For vertical motion, the initial vertical velocity $u_y = 0 \text{ m/s}$.
The vertical displacement is $H = 19.6 \text{ m}$.
The acceleration due to gravity is $g = 9.8 \text{ m/s}^2$.
Using the equation of motion $H = u_y t + \frac{1}{2} g t^2$, we get:
$19.6 = 0 \times t + \frac{1}{2} \times 9.8 \times t^2$
$19.6 = 4.9 \times t^2$
$t^2 = \frac{19.6}{4.9} = 4$
$t = \sqrt{4} = 2 \text{ s}$.
Thus, the ball will take $2 \text{ s}$ to hit the ground.

(D) In projectile motion,the horizontal component of velocity remains constant throughout the flight because there is no acceleration acting in the horizontal direction.
Given the initial velocity is $u$ and the angle of projection is $\theta = 60^{\circ}$.
The horizontal component of the initial velocity is given by $u_x = u \cos(\theta)$.
Substituting the values,we get $u_x = u \cos(60^{\circ})$.
Since $\cos(60^{\circ}) = 1/2$,the horizontal component is $u_x = u \times (1/2) = u/2$.
At the maximum height,the vertical component of velocity becomes zero,but the horizontal component remains unchanged.
Therefore,the horizontal component of the velocity at the maximum height is $u/2$.

(C) The initial kinetic energy of the particle is given by $K = \frac{1}{2} m v^2$,where $v$ is the initial velocity.
At the highest point of the trajectory,the vertical component of the velocity becomes zero,and the velocity of the particle is equal to the horizontal component,which is $v_x = v \cos \theta$.
Given $\theta = 60^{\circ}$,the velocity at the highest point is $v_x = v \cos 60^{\circ} = \frac{v}{2}$.
The kinetic energy at the highest point $K'$ is given by $K' = \frac{1}{2} m (v_x)^2$.
Substituting the value of $v_x$,we get $K' = \frac{1}{2} m (\frac{v}{2})^2 = \frac{1}{2} m (\frac{v^2}{4}) = \frac{1}{4} (\frac{1}{2} m v^2)$.
Since $K = \frac{1}{2} m v^2$,we have $K' = \frac{K}{4}$.

(D) The time taken to reach the maximum height is given by $t_m = \frac{u_y}{g} = 2 \text{ s}$.
Given $g = 10 \text{ m/s}^2$,we have $u_y = 2 \times 10 = 20 \text{ m/s}$.
The vertical displacement $y$ at any time $t$ is given by the equation $y = u_y t - \frac{1}{2} g t^2$.
At $t = 3 \text{ s}$,the ball is at height $H$,so:
$H = (20 \text{ m/s}) \times (3 \text{ s}) - \frac{1}{2} \times (10 \text{ m/s}^2) \times (3 \text{ s})^2$
$H = 60 \text{ m} - 5 \times 9 \text{ m}$
$H = 60 \text{ m} - 45 \text{ m} = 15 \text{ m}$.

(A) The time of flight for a projectile is given by $T = \frac{2v \sin \theta}{g}$.
For two complementary angles $\theta$ and $(90^\circ - \theta)$,the horizontal range $R$ is the same.
The times of flight are $T_1 = \frac{2v \sin \theta}{g} = 5 \text{ s}$ and $T_2 = \frac{2v \sin(90^\circ - \theta)}{g} = \frac{2v \cos \theta}{g} = 10 \text{ s}$.
The horizontal range is $R = \frac{v^2 \sin(2\theta)}{g} = \frac{2v^2 \sin \theta \cos \theta}{g}$.
Multiplying $T_1$ and $T_2$,we get $T_1 T_2 = \frac{4v^2 \sin \theta \cos \theta}{g^2} = \frac{2}{g} \left( \frac{2v^2 \sin \theta \cos \theta}{g} \right) = \frac{2R}{g}$.
Therefore,$R = \frac{1}{2} g T_1 T_2 = \frac{1}{2} \times 10 \times 5 \times 10 = 250 \text{ m}$.

(B) 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 and $\theta$ is the angle of projection.
Since both projectiles have the same initial velocity $u$,the range is directly proportional to $\sin(2\theta)$,i.e.,$R \propto \sin(2\theta)$.
For the first projectile,$\theta_1 = 15^{\circ}$,so $R_1 \propto \sin(2 \times 15^{\circ}) = \sin(30^{\circ}) = 1/2$.
For the second projectile,$\theta_2 = 30^{\circ}$,so $R_2 \propto \sin(2 \times 30^{\circ}) = \sin(60^{\circ}) = \sqrt{3}/2$.
The ratio of their ranges is $\frac{R_1}{R_2} = \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}}$.
Given the ratio is $1:x$,we have $1:x = 1:\sqrt{3}$.
Therefore,$x = \sqrt{3}$.

(B) The horizontal position is given by $x = 24t$. The horizontal velocity component is $v_x = \frac{dx}{dt} = 24 \text{ m/s}$.
The vertical position is given by $y = 43.6t - 4.9t^2$. The vertical velocity component is $v_y = \frac{dy}{dt} = 43.6 - 9.8t$.
At time $t = 2 \text{ s}$,the vertical velocity is $v_y = 43.6 - 9.8(2) = 43.6 - 19.6 = 24 \text{ m/s}$.
The angle $\theta$ made by the projectile with the horizontal is given by $\tan \theta = \frac{v_y}{v_x}$.
Substituting the values,$\tan \theta = \frac{24}{24} = 1$.
Therefore,$\theta = \tan^{-1}(1) = 45^{\circ}$.

(A) The farthest distance reached by the bullets is the maximum horizontal range,denoted by $R_{max}$.
The formula for the maximum horizontal range is $R_{max} = \frac{u^2}{g}$,where $u$ is the initial speed and $g$ is the acceleration due to gravity.
Given values are $R_{max} = 6.4 \text{ m}$ and $g = 10 \text{ m/s}^2$.
Substituting these values into the formula: $6.4 = \frac{u^2}{10}$.
Multiplying both sides by $10$,we get $u^2 = 64$.
Taking the square root of both sides,we find $u = 8 \text{ m/s}$.