Horizontal Projectile Motion Questions in English

(C) Let the initial velocity be $v$ and the angle of projection be $\theta = 60^{\circ}$.
Given,the vertical component of initial velocity $v_y = v \sin \theta = 40 \ m \ s^{-1}$.
The time of flight $T = \frac{2 v \sin \theta}{g} = \frac{2 \times 40}{10} = 8 \ s$.
The time at which we need to find the velocity is $t = \frac{T}{4} = \frac{8}{4} = 2 \ s$.
The horizontal component of velocity remains constant throughout the motion: $v_x = v \cos \theta$.
Since $v \sin \theta = 40$,we have $v = \frac{40}{\sin 60^{\circ}} = \frac{40}{\sqrt{3}/2} = \frac{80}{\sqrt{3}} \approx 46.19 \ m \ s^{-1}$.
Thus,$v_x = \frac{80}{\sqrt{3}} \times \cos 60^{\circ} = \frac{80}{\sqrt{3}} \times \frac{1}{2} = \frac{40}{\sqrt{3}} \approx 23.09 \ m \ s^{-1}$.
The vertical component of velocity at time $t$ is $v_y(t) = v \sin \theta - g t = 40 - (10 \times 2) = 20 \ m \ s^{-1}$.
The magnitude of velocity at time $t$ is $v_t = \sqrt{v_x^2 + v_y(t)^2} = \sqrt{(23.09)^2 + (20)^2} = \sqrt{533.15 + 400} = \sqrt{933.15} \approx 30.54 \ m \ s^{-1}$.

(C) The assertion $(A)$ is true. The horizontal range $R$ of a projectile is given by the formula $R = \frac{u^2 \sin(2\theta)}{g}$. For a given initial velocity $u$,the range is maximum when $\sin(2\theta)$ is maximum,i.e.,$\sin(2\theta) = 1$,which implies $2\theta = 90^{\circ}$ or $\theta = 45^{\circ}$.
The reason $(R)$ is false. The range $R$ of a projectile depends not only on the angle of projection $\theta$ but also on the initial velocity $u$ and the acceleration due to gravity $g$ $(R = \frac{u^2 \sin(2\theta)}{g})$. Therefore,it does not depend 'only' on the angle of projection.
Thus,$(A)$ is true but $(R)$ is false.

(D) Let the initial velocity be $u$ and the angle of projection be $\theta = 45^{\circ}$.
The equation of the trajectory of a projectile is given by:
$y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta}$
Given $x = 30 \,m$, $y = 20 \,m$, $\theta = 45^{\circ}$, and $g = 10 \,m/s^2$:
$20 = 30 \tan 45^{\circ} - \frac{10 \times (30)^2}{2 u^2 \cos^2 45^{\circ}}$
$20 = 30(1) - \frac{10 \times 900}{2 u^2 \times (1/2)}$
$20 = 30 - \frac{9000}{u^2}$
$\frac{9000}{u^2} = 10$
$u^2 = 900 \implies u = 30 \,m/s$
The horizontal range $R$ is given by:
$R = \frac{u^2 \sin 2 \theta}{g} = \frac{900 \times \sin 90^{\circ}}{10} = 90 \,m$
The distance $s$ from the foot of the pole is the total range minus the initial distance from the pole:
$s = R - 30 = 90 - 30 = 60 \,m$
Thus, the correct option is $D$.

(D) The maximum range of a projectile is given by $R_{\max} = \frac{u^2}{g}$.
Given that the range $R$ is half of the maximum range,we have $R = \frac{R_{\max}}{2} = \frac{u^2}{2g}$.
The formula for the range of a projectile is $R = \frac{u^2 \sin 2\theta}{g}$.
Equating the two expressions for $R$,we get $\frac{u^2 \sin 2\theta}{g} = \frac{u^2}{2g}$.
This simplifies to $\sin 2\theta = \frac{1}{2}$.
Since $\sin 30^{\circ} = \frac{1}{2}$,we have $2\theta = 30^{\circ}$,which gives $\theta = 15^{\circ}$.
Therefore,the correct option is $D$.

(D) Let the velocity of the projectile be $v$ at an angle $\frac{\theta}{2}$ with the horizontal.
Since the horizontal component of velocity remains constant throughout the motion:
$v \cos \left(\frac{\theta}{2}\right) = u \cos \theta$
$v = \frac{u \cos \theta}{\cos \left(\frac{\theta}{2}\right)}$
The centripetal force required for the curved path is provided by the component of gravity perpendicular to the velocity vector,which is $mg \cos \left(\frac{\theta}{2}\right)$.
Using the formula for centripetal force $\frac{mv^2}{r} = F_c$:
$\frac{mv^2}{r} = mg \cos \left(\frac{\theta}{2}\right)$
$r = \frac{v^2}{g \cos \left(\frac{\theta}{2}\right)}$
Substituting the value of $v$:
$r = \frac{\left(\frac{u \cos \theta}{\cos \left(\frac{\theta}{2}\right)}\right)^2}{g \cos \left(\frac{\theta}{2}\right)}$
$r = \frac{u^2 \cos^2 \theta}{g \cos^2 \left(\frac{\theta}{2}\right) \cdot \cos \left(\frac{\theta}{2}\right)}$
$r = \frac{u^2 \cos^2 \theta \sec^3 \left(\frac{\theta}{2}\right)}{g}$

(C) For a projectile projected at an angle $\theta$ with the horizontal:
Maximum height,$H_{\max} = \frac{u^2 \sin^2 \theta}{2g}$
Range,$R = \frac{u^2 \sin 2\theta}{g} = \frac{2u^2 \sin \theta \cos \theta}{g}$
The ratio of maximum height to range is given by:
$\frac{H_{\max}}{R} = \frac{u^2 \sin^2 \theta / 2g}{2u^2 \sin \theta \cos \theta / g} = \frac{\sin^2 \theta}{4 \sin \theta \cos \theta} = \frac{\tan \theta}{4}$
Given $\theta = \tan^{-1}(\frac{8}{7})$,so $\tan \theta = \frac{8}{7}$.
Substituting the value of $\tan \theta$:
$\frac{H_{\max}}{R} = \frac{8/7}{4} = \frac{8}{28} = \frac{2}{7}$.
Thus,the ratio is $2:7$.

(D) The trajectory of a projectile is given by the equation:
$y = f(x) = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta}$
The slope of the trajectory at any point $(x, y)$ is given by the derivative $\frac{dy}{dx}$:
$\frac{dy}{dx} = \tan \theta - \frac{gx}{u^2 \cos^2 \theta} = \frac{v_y}{v_x}$
This slope represents the ratio of the vertical component of velocity to the horizontal component of velocity,not the magnitude of the velocity itself. Therefore,Assertion $(A)$ is incorrect.
Reason $(R)$ states that the velocity vector at any point is always along the tangent to the trajectory at that point. This is a fundamental property of motion in a plane,as the instantaneous velocity is defined as the rate of change of position,which is directed along the tangent to the path. Thus,Reason $(R)$ is correct.

(D) The equation of the trajectory of a projectile is given by:
$y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta}$
Given $\theta = 45^{\circ}$,$\tan 45^{\circ} = 1$ and $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$,so $\cos^2 45^{\circ} = \frac{1}{2}$.
The equation becomes:
$y = x - \frac{g x^2}{2 u^2 (1/2)} = x - \frac{g x^2}{u^2} \quad ... (i)$
Since the projectile passes through the point $(4, 3) \ m$,we substitute $x = 4$ and $y = 3$ into equation $(i)$:
$3 = 4 - \frac{g(4^2)}{u^2}$
$3 = 4 - \frac{16g}{u^2}$
$\frac{16g}{u^2} = 1 \Rightarrow u^2 = 16g$
Substituting $u^2 = 16g$ back into equation $(i)$:
$y = x - \frac{g x^2}{16g} = x - \frac{x^2}{16}$
The horizontal range $R$ is the value of $x$ when $y = 0$ (at the landing point):
$0 = x - \frac{x^2}{16}$
$x(1 - \frac{x}{16}) = 0$
Since $x \neq 0$ at the landing point,$1 - \frac{x}{16} = 0$,which gives $x = 16 \ m$.
Thus,the horizontal range is $16 \ m$.

(A) The initial velocity of the projectile is $\vec{u} = (1\hat{i} + 2\hat{j}) \text{ ms}^{-1}$.
Thus,the initial horizontal component is $u_x = 1 \text{ ms}^{-1}$ and the initial vertical component is $u_y = 2 \text{ ms}^{-1}$.
The acceleration components are $a_x = 0$ and $a_y = -g = -10 \text{ ms}^{-2}$.
At any time $t$,the horizontal distance covered is:
$x = u_x t = 1 \cdot t \implies t = x \quad \dots (i)$
The vertical distance covered is:
$y = u_y t + \frac{1}{2} a_y t^2$
$y = 2t + \frac{1}{2}(-10)t^2$
$y = 2t - 5t^2 \quad \dots (ii)$
Substituting the value of $t$ from equation $(i)$ into equation $(ii)$,we get:
$y = 2(x) - 5(x)^2$
$y = 2x - 5x^2$
Therefore,the correct option is $A$.

(B) Given,the height of both towers is the same,$h_1 = h_2 = h = 20 \ m$.
The time of flight $t$ for a body thrown horizontally is given by $t = \sqrt{\frac{2h}{g}}$.
$t = \sqrt{\frac{2 \times 20}{10}} = \sqrt{4} = 2 \ s$.
The horizontal displacement of the body from tower $A$ to point $P$ is $x_A = u_A t = 20 \ ms^{-1} \times 2 \ s = 40 \ m$.
The horizontal displacement of the body from tower $B$ to point $Q$ is $x_B = u_B t = 30 \ ms^{-1} \times 2 \ s = 60 \ m$.
The distance between points $P$ and $Q$ is $d = 200 \ m - (x_A + x_B) = 200 \ m - (40 \ m + 60 \ m) = 100 \ m$.
For the car starting from rest $(u = 0)$ at $P$ and reaching $Q$ in $t' = 10 \ s$,we use the equation of motion $s = ut + \frac{1}{2}at^2$:
$100 = 0 \times 10 + \frac{1}{2} \times a \times (10)^2$.
$100 = 50a$.
$a = 2 \ ms^{-2}$.

(C) Given:
Initial horizontal velocity, $u_x = 20 \,ms^{-1}$.
Initial vertical velocity, $u_y = 0$.
Acceleration due to gravity, $g = a_y = 10 \,ms^{-2}$.
Horizontal acceleration, $a_x = 0$.
For $t = 1 \,s$:
$A$. Resultant velocity $v = \sqrt{v_x^2 + v_y^2}$.
$v_x = u_x + a_x t = 20 + 0 = 20 \,ms^{-1}$.
$v_y = u_y + a_y t = 0 + 10 \times 1 = 10 \,ms^{-1}$.
$v = \sqrt{20^2 + 10^2} = \sqrt{400 + 100} = \sqrt{500} \approx 22.4 \,ms^{-1}$. (Matches $IV$)
$B$. Horizontal displacement $s_x = u_x t + \frac{1}{2} a_x t^2 = 20 \times 1 + 0 = 20 \,m$. (Matches $II$)
$C$. Vertical displacement $s_y = u_y t + \frac{1}{2} a_y t^2 = 0 + \frac{1}{2} \times 10 \times 1^2 = 5 \,m$. (Matches $I$)
$D$. Vertical velocity $v_y = 10 \,ms^{-1}$. (Matches $III$)
Thus, the correct matching is $A-IV, B-II, C-I, D-III$.

(D) Initial velocity $u = 10 \sqrt{3} \ m/s$ at $\theta = 60^{\circ}$.
Horizontal component: $u_x = u \cos 60^{\circ} = 10 \sqrt{3} \times (1/2) = 5 \sqrt{3} \ m/s$.
Vertical component: $u_y = u \sin 60^{\circ} = 10 \sqrt{3} \times (\sqrt{3}/2) = 15 \ m/s$.
Initial velocity vector: $\vec{v}_i = 5 \sqrt{3} \hat{i} + 15 \hat{j}$.
After $t = 2 \ s$,horizontal velocity $v_x = u_x = 5 \sqrt{3} \ m/s$.
Vertical velocity $v_y = u_y - gt = 15 - (10 \times 2) = 15 - 20 = -5 \ m/s$.
Final velocity vector: $\vec{v}_f = 5 \sqrt{3} \hat{i} - 5 \hat{j}$.
Calculate the dot product: $\vec{v}_i \cdot \vec{v}_f = (5 \sqrt{3})(5 \sqrt{3}) + (15)(-5) = 75 - 75 = 0$.
Since the dot product is $0$,the angle between the vectors is $90^{\circ}$.

(D) For horizontal projectile motion, the horizontal velocity $v_x = u = 30 \,ms^{-1}$ remains constant.
The vertical velocity at time $t$ is $v_y = gt = 10t$.
At time $t_1$, $v_x = v_y$, so $30 = 10t_1$, which gives $t_1 = 3 \,s$.
The horizontal displacement at time $t$ is $x = ut = 30t$.
The vertical displacement at time $t$ is $y = \frac{1}{2}gt^2 = \frac{1}{2} \times 10 \times t^2 = 5t^2$.
At time $t_2$, $x = y$, so $30t_2 = 5t_2^2$. Since $t_2 \neq 0$, we have $t_2 = \frac{30}{5} = 6 \,s$.
Therefore, $t_2 - t_1 = 6 \,s - 3 \,s = 3 \,s$.

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

(B) Initial velocity $u = 60 \sqrt{2} \ m/s$ at angle $\theta = 45^{\circ}$.
Horizontal component of velocity: $u_x = u \cos(45^{\circ}) = 60 \sqrt{2} \times \frac{1}{\sqrt{2}} = 60 \ m/s$.
Vertical component of velocity: $u_y = u \sin(45^{\circ}) = 60 \sqrt{2} \times \frac{1}{\sqrt{2}} = 60 \ m/s$.
After time $t = 6 \ s$,the horizontal velocity remains constant: $v_x = u_x = 60 \ m/s$.
The vertical velocity changes due to gravity $(g = 10 \ m/s^2)$: $v_y = u_y - gt = 60 - (10 \times 6) = 0 \ m/s$.
The angle $\alpha$ made by the velocity with the horizontal is given by $\tan(\alpha) = \frac{v_y}{v_x} = \frac{0}{60} = 0$.
Therefore,$\alpha = 0^{\circ}$.

(C) Let the initial velocity of the projectile be $u$ and the angle of projection be $\theta$. The initial kinetic energy is $K_i = \frac{1}{2}mu^2$.
At the maximum height, the vertical component of velocity becomes zero, and the velocity of the projectile is only the horizontal component, $v_x = u \cos \theta$.
The kinetic energy at the maximum height is $K_f = \frac{1}{2}m(u \cos \theta)^2 = \frac{1}{2}mu^2 \cos^2 \theta$.
According to the problem, $K_f = \frac{1}{2}K_i$.
Substituting the expressions, we get $\frac{1}{2}mu^2 \cos^2 \theta = \frac{1}{2} (\frac{1}{2}mu^2)$.
This simplifies to $\cos^2 \theta = \frac{1}{2}$, which means $\cos \theta = \frac{1}{\sqrt{2}}$.
Therefore, $\theta = 45^{\circ}$.

(C) The standard equation of a projectile is $y = x \tan \theta - \frac{gx^2}{2u^2 \cos^2 \theta}$.
Comparing this with $y = ax - bx^2$,we get $\tan \theta = a$ and $b = \frac{g}{2u^2 \cos^2 \theta}$.
$i)$ Initial velocity $u$: Since $\tan \theta = a$,$\sec^2 \theta = 1 + a^2$,so $\cos^2 \theta = \frac{1}{1+a^2}$. Substituting into $b$,$b = \frac{g(1+a^2)}{2u^2} \implies u = \sqrt{\frac{g(1+a^2)}{2b}}$. Thus,$i-d$.
$ii)$ Range $R$: $y=0 \implies x(a-bx)=0 \implies R = \frac{a}{b}$. Thus,$ii-a$.
$iii)$ Max height $H$: $H = \frac{u^2 \sin^2 \theta}{2g} = \frac{u^2 \tan^2 \theta \cos^2 \theta}{2g} = \frac{a^2}{4b}$. Thus,$iii-c$.
$iv)$ Time of flight $T$: $T = \frac{2u \sin \theta}{g} = \frac{2u \tan \theta \cos \theta}{g} = a \sqrt{\frac{2}{bg}}$. Thus,$iv-b$.
Therefore,the correct match is $i-d, ii-a, iii-c, iv-b$.

(D) The initial velocity is $u = 10\sqrt{2} \ m/s$ at $\theta = 45^{\circ}$.
The horizontal component is $u_x = u \cos 45^{\circ} = 10\sqrt{2} \times \frac{1}{\sqrt{2}} = 10 \ m/s$.
The vertical component is $u_y = u \sin 45^{\circ} = 10\sqrt{2} \times \frac{1}{\sqrt{2}} = 10 \ m/s$.
At any time $t$,the velocity components are $v_x = 10 \ m/s$ and $v_y = 10 - 10t$.
The speed $v$ is given by $v^2 = v_x^2 + v_y^2$.
Given $v = \sqrt{125} \ m/s$,so $v^2 = 125$.
$125 = 10^2 + (10 - 10t)^2$.
$125 = 100 + (10 - 10t)^2$.
$(10 - 10t)^2 = 25$.
Taking the square root,$10 - 10t = \pm 5$.
Case $1$: $10 - 10t = 5 \implies 10t = 5 \implies t_1 = 0.5 \ s$.
Case $2$: $10 - 10t = -5 \implies 10t = 15 \implies t_2 = 1.5 \ s$.
The time interval is $\Delta t = t_2 - t_1 = 1.5 - 0.5 = 1.0 \ s$.

(B) The initial velocity is given by $\vec{u} = u_x \hat{i} + u_y \hat{j} = (1\hat{i} + 2\hat{j}) \text{ ms}^{-1}$.
Thus,the horizontal component is $u_x = 1 \text{ ms}^{-1}$ and the vertical component is $u_y = 2 \text{ ms}^{-1}$.
At any time $t$,the horizontal position is $x = u_x t = 1 \cdot t$,which implies $t = x$.
The vertical position is $y = u_y t - \frac{1}{2}gt^2$.
Substituting $g = 10 \text{ ms}^{-2}$,$u_y = 2 \text{ ms}^{-1}$,and $t = x$ into the equation:
$y = 2x - \frac{1}{2}(10)x^2$.
$y = 2x - 5x^2$.
Therefore,the equation of the trajectory is $y = 2x - 5x^2$.

(D) The equation of the trajectory of a projectile is given by:
$y = x \tan \theta \left(1 - \frac{x}{R}\right) = x \tan \theta - \frac{x^2 \tan \theta}{R}$
Comparing this with the given equation $y = Px - Qx^2$,we get:
$P = \tan \theta$
$Q = \frac{\tan \theta}{R} \implies R = \frac{\tan \theta}{Q} = \frac{P}{Q}$
We know that the maximum height $H$ is given by:
$H = \frac{R \tan \theta}{4}$
Substituting the values of $R$ and $\tan \theta$:
$H = \frac{(P/Q) \cdot P}{4} = \frac{P^2}{4Q}$
Therefore,the ratio of maximum height to range is:
$\frac{H}{R} = \frac{P^2 / 4Q}{P / Q} = \frac{P}{4}$

(A) The initial velocity is given by $\vec{u} = 10 \hat{i} + p \hat{j}$. Thus,$u_x = 10 \ m/s$ and $u_y = p \ m/s$.
The maximum height $H$ is given by $H = \frac{u_y^2}{2g} = \frac{p^2}{2g}$.
The range $R$ is given by $R = \frac{2 u_x u_y}{g} = \frac{2(10)(p)}{g} = \frac{20p}{g}$.
According to the problem,$H = 0.5 R$.
Substituting the expressions: $\frac{p^2}{2g} = 0.5 \times \frac{20p}{g}$.
$\frac{p^2}{2g} = \frac{10p}{g}$.
Since $p \neq 0$,we can divide both sides by $p/g$: $\frac{p}{2} = 10$.
Therefore,$p = 20$.

(D) The horizontal range $R$ of a projectile is given by $R = \frac{u^2 \sin 2\theta}{g}$.
Let $R$ be the distance to the target.
For $\theta_1 = 15^{\circ}$,the range is $R_1 = R - 10$. So,$R - 10 = \frac{u^2 \sin 30^{\circ}}{g} = \frac{u^2}{2g}$.
For $\theta_2 = 45^{\circ}$,the range is $R_2 = R + 15$. So,$R + 15 = \frac{u^2 \sin 90^{\circ}}{g} = \frac{u^2}{g}$.
Dividing the two equations: $\frac{R - 10}{R + 15} = \frac{u^2/2g}{u^2/g} = \frac{1}{2}$.
$2R - 20 = R + 15 \Rightarrow R = 35 \ m$.
Now,$R + 15 = \frac{u^2}{g} \Rightarrow 35 + 15 = \frac{u^2}{g} \Rightarrow \frac{u^2}{g} = 50 \ m$.
To hit the target,the range must be $R = 35 \ m$. Thus,$35 = 50 \sin 2\theta$.
$\sin 2\theta = \frac{35}{50} = \frac{7}{10}$.
$\theta = \frac{1}{2} \sin^{-1}\left(\frac{7}{10}\right)$.

(B) The equation of the trajectory of 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 equation 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 roots be $\tan \theta_1$ and $\tan \theta_2$. Then $\tan \theta_1 \tan \theta_2 = \frac{y + g r^2 / 2 u^2}{g r^2 / 2 u^2} = 1 + \frac{2 u^2 y}{g r^2}$.
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 relation for projectile motion,it is a standard result that $t_1 t_2 = \frac{2 r}{g \tan \alpha}$ where $\alpha$ is the angle of elevation of the point. For a fixed point $(r, y)$,$t_1 t_2 = \frac{2 r}{g \tan \theta_{elevation}}$.
Since $g$ is constant,$t_1 t_2 \propto r$.

(A) The maximum height $H$ and time of flight $T$ of a projectile are given by:
$H = \frac{u^2 \sin^2 \theta}{2g}$ and $T = \frac{2u \sin \theta}{g}$.
From these equations,we can see that $H \propto u^2$ and $T \propto u$ (since $\theta$ is constant).
Therefore,$H \propto T^2$.
If the height increases by $10 \%$,the new height $H_2 = 1.10 H_1$.
Since $H \propto T^2$,we have $\frac{H_2}{H_1} = \left(\frac{T_2}{T_1}\right)^2$.
$1.10 = \left(\frac{T_2}{T_1}\right)^2 \implies \frac{T_2}{T_1} = \sqrt{1.10} \approx 1.0488$.
This corresponds to an increase of approximately $4.88 \%$,which is closest to $5 \%$.

(D) Given: Mass $m = 50 \ kg$,Range $R = 8 \ m$,Angle $\theta = 45^{\circ}$,Acceleration due to gravity $g = 9.8 \ m/s^2$.
The formula for the horizontal range of a projectile is $R = \frac{u^2 \sin(2\theta)}{g}$.
Substituting the given values:
$8 = \frac{u^2 \sin(2 \times 45^{\circ})}{9.8} = \frac{u^2 \sin(90^{\circ})}{9.8} = \frac{u^2}{9.8}$.
Therefore,$u^2 = 8 \times 9.8 = 78.4 \ m^2/s^2$.
The initial kinetic energy $(K.E.)$ is given by $\frac{1}{2} m u^2$.
$(K.E.) = \frac{1}{2} \times 50 \times 78.4 = 25 \times 78.4 = 1960 \ J$.

(B) Given: Velocity of helicopter $u = 288 \ km/h = 288 \times \frac{5}{18} \ m/s = 80 \ m/s$. Acceleration due to gravity $g = 10 \ m/s^2$. Let $h$ be the height of the helicopter and $R$ be the horizontal range of the bomb. The time taken to hit the ground is $t = \sqrt{\frac{2h}{g}}$. The horizontal range is $R = u \times t = u \sqrt{\frac{2h}{g}}$. The angle $\theta$ made by the line joining the dropping point and the impact point with the horizontal is given by $\tan \theta = \frac{h}{R}$. Given $\theta = 45^{\circ}$,so $\tan 45^{\circ} = 1$,which implies $h = R$. Substituting $R = u \sqrt{\frac{2h}{g}}$,we get $h = u \sqrt{\frac{2h}{g}}$. Squaring both sides: $h^2 = u^2 \frac{2h}{g} \implies h = \frac{2u^2}{g}$. Substituting the values: $h = \frac{2 \times (80)^2}{10} = \frac{2 \times 6400}{10} = 1280 \ m$.

(B) Given: Initial velocity $u = 19.6 \ m/s$,Maximum height $H = 9.8 \ m$.
We know the formula for maximum height is $H = \frac{u^2 \sin^2 \theta}{2g}$.
Assuming $g = 9.8 \ m/s^2$,we have $9.8 = \frac{(19.6)^2 \sin^2 \theta}{2 \times 9.8}$.
$9.8 = \frac{384.16 \sin^2 \theta}{19.6} \implies 9.8 = 19.6 \sin^2 \theta$.
$\sin^2 \theta = \frac{9.8}{19.6} = 0.5$,so $\sin \theta = \frac{1}{\sqrt{2}}$,which means $\theta = 45^\circ$.
The range $R$ is given by $R = \frac{u^2 \sin(2\theta)}{g}$.
Since $\theta = 45^\circ$,$\sin(2\theta) = \sin(90^\circ) = 1$.
$R = \frac{(19.6)^2}{9.8} = \frac{384.16}{9.8} = 39.2 \ m$.

(A) Let the ball be projected with an initial velocity $u$ at an angle $\theta = 45^{\circ}$.
The vertical displacement $y$ at time $t$ is given by $y = (u \sin \theta)t - \frac{1}{2}gt^2$.
Since the ball is at the same height $y$ at $t_1 = 2 \ s$ and $t_2 = 8 \ s$,we have:
$(u \sin \theta)t_1 - \frac{1}{2}gt_1^2 = (u \sin \theta)t_2 - \frac{1}{2}gt_2^2$
$u \sin \theta (t_2 - t_1) = \frac{1}{2}g(t_2^2 - t_1^2)$
$u \sin \theta = \frac{g(t_1 + t_2)}{2} = \frac{10(2 + 8)}{2} = 50 \ m/s$.
Since $\theta = 45^{\circ}$,$u \sin 45^{\circ} = u \cos 45^{\circ} = 50 \ m/s$,so $u_x = 50 \ m/s$.
The horizontal distance $d$ between the two points is the horizontal displacement between $t_1$ and $t_2$:
$d = u_x(t_2 - t_1) = 50 \times (8 - 2) = 50 \times 6 = 300 \ m$.

(A) For a projectile motion,if a body passes through a point at time $t_1$ and reaches the ground after a further time $t_2$,the total time of flight is $T = t_1 + t_2 = 2.3 \ s + 5.7 \ s = 8.0 \ s$.
The time taken to reach the maximum height is $t_{max} = \frac{T}{2} = \frac{8.0 \ s}{2} = 4.0 \ s$.
The maximum height $H$ is given by the formula $H = \frac{1}{2} g t_{max}^2$.
Substituting the values: $H = \frac{1}{2} \times 10 \ ms^{-2} \times (4.0 \ s)^2$.
$H = 5 \times 16 = 80 \ m$.
Thus,the maximum height reached by the body is $80 \ m$.

(B) The equation of the trajectory of a projectile is given by $y = x \tan \theta - \frac{gx^2}{2u^2 \cos^2 \theta}$.
Comparing this with the given equation $y = \sqrt{3}x - 0.2x^2$:
$1$. $\tan \theta = \sqrt{3} \implies \theta = 60^\circ$.
$2$. $\frac{g}{2u^2 \cos^2 \theta} = 0.2$.
Given $g = 10 \ ms^{-2}$ and $\cos 60^\circ = 0.5$,we have $\frac{10}{2u^2 (0.5)^2} = 0.2$.
$\frac{10}{2u^2 (0.25)} = 0.2 \implies \frac{10}{0.5u^2} = 0.2 \implies \frac{20}{u^2} = 0.2 \implies u^2 = 100 \implies u = 10 \ ms^{-1}$.
The time of flight $T$ is given by $T = \frac{2u \sin \theta}{g}$.
$T = \frac{2 \times 10 \times \sin 60^\circ}{10} = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3} \ s$.

(D) For projectile motion, the angle of projection is $\theta = \tan^{-1}(\sqrt{7})$, so $\tan \theta = \sqrt{7}$.
At half of the maximum height $(y = H/2)$, the vertical component of velocity is given by $v_y^2 = u_y^2 - 2g(H/2) = u^2 \sin^2 \theta - gH$.
The horizontal component of velocity remains constant: $v_x = u_x = u \cos \theta$.
The speed $v$ at this height is $v = \sqrt{v_x^2 + v_y^2} = \sqrt{u^2 \cos^2 \theta + u^2 \sin^2 \theta - gH} = \sqrt{u^2 - gH}$.
Since $H = \frac{u^2 \sin^2 \theta}{2g}$, we have $gH = \frac{u^2 \sin^2 \theta}{2}$.
Substituting this into the expression for $v^2$:
$v^2 = u^2 - \frac{u^2 \sin^2 \theta}{2} = u^2 (1 - \frac{\sin^2 \theta}{2})$.
Given $\tan \theta = \sqrt{7}$, we have $\sin^2 \theta = \frac{\tan^2 \theta}{1 + \tan^2 \theta} = \frac{7}{1 + 7} = \frac{7}{8}$.
Thus, $v^2 = u^2 (1 - \frac{7/8}{2}) = u^2 (1 - \frac{7}{16}) = u^2 (\frac{9}{16})$.
Since $v = nu$, we have $n^2 = \frac{9}{16}$, which gives $n = \frac{3}{4}$.

(B) Given: Velocity of the aeroplane,$u = 60 \ km/h = 60 \times \frac{5}{18} \ m/s = \frac{50}{3} \ m/s$.
Height of the aeroplane,$h = 490 \ m$.
Acceleration due to gravity,$g = 9.8 \ m/s^2$.
The time taken by the bomb to reach the ground is given by $h = \frac{1}{2}gt^2$,so $t = \sqrt{\frac{2h}{g}}$.
$t = \sqrt{\frac{2 \times 490}{9.8}} = \sqrt{\frac{980}{9.8}} = \sqrt{100} = 10 \ s$.
The horizontal distance covered by the bomb in this time is $R = u \times t$.
$R = \left(\frac{50}{3}\right) \times 10 = \frac{500}{3} \ m$.

(C) Initial velocity $u = 50 \ m/s$,angle $\theta = 30^{\circ}$,$g = 10 \ m/s^2$.
Horizontal component of velocity $u_x = u \cos \theta = 50 \cos 30^{\circ} = 50 \times \frac{\sqrt{3}}{2} = 25\sqrt{3} \ m/s$.
Vertical component of velocity $u_y = u \sin \theta = 50 \sin 30^{\circ} = 50 \times 0.5 = 25 \ m/s$.
Total range $R = \frac{u^2 \sin 2\theta}{g} = \frac{50^2 \sin 60^{\circ}}{10} = \frac{2500 \times \sqrt{3}}{20} = 125\sqrt{3} \ m \approx 216.5 \ m$.
Horizontal distance covered in $t = 3 \ s$ is $x = u_x \times t = 25\sqrt{3} \times 3 = 75\sqrt{3} \ m \approx 129.9 \ m$.
The distance beyond the wall where the stone strikes the ground is $R - x = 125\sqrt{3} - 75\sqrt{3} = 50\sqrt{3} \ m$.
$50 \times 1.732 = 86.6 \ m$.

(D) The maximum horizontal range $R_{\max}$ for a projectile is given by $R_{\max} = \frac{u^2}{g}$.
Given $R_{\max} = 80 \,m$, we have $80 = \frac{u^2}{g}$, which implies $u^2 = 80g$.
When the ball is thrown vertically upward with the same velocity $u$, the maximum height $H$ reached is given by the kinematic equation $v^2 = u^2 - 2gH$.
At maximum height, the final velocity $v = 0$.
Substituting the values: $0 = u^2 - 2gH$.
$2gH = u^2$.
Since $u^2 = 80g$, we get $2gH = 80g$.
$H = \frac{80g}{2g} = 40 \,m$.

(B) The Cartesian equation of the path of a projectile is given by $y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta}$.
Given the initial velocity vector $\vec{u} = \hat{i} + 2 \hat{j} \,ms^{-1}$.
Here, the horizontal component $u_x = 1 \,ms^{-1}$ and the vertical component $u_y = 2 \,ms^{-1}$.
The horizontal displacement is $x = u_x t = 1 \cdot t$, so $t = x$.
The vertical displacement is $y = u_y t - \frac{1}{2} g t^2$.
Substituting $t = x$ into the equation for $y$:
$y = 2(x) - \frac{1}{2} (10) (x)^2$.
$y = 2x - 5x^2$.

(B) The initial velocity components are $u_x = 3 \text{ m s}^{-1}$ and $u_y = 4 \text{ m s}^{-1}$.
Using the equations of motion: $x = u_x t = 3t \Rightarrow t = \frac{x}{3}$.
$y = u_y t - \frac{1}{2} g t^2 = 4t - \frac{1}{2} (10) t^2 = 4t - 5t^2$.
Substituting $t = \frac{x}{3}$ into the equation for $y$:
$y = 4 \left( \frac{x}{3} \right) - 5 \left( \frac{x}{3} \right)^2 = \frac{4x}{3} - \frac{5x^2}{9}$.
To match the form $\frac{1}{9} [\beta x + \gamma x^2]$,we multiply and divide by $9$:
$y = \frac{1}{9} [12x - 5x^2]$.
Comparing this with the given form $\frac{1}{9} [\beta x + \gamma x^2]$,we get $\beta = 12$ and $\gamma = -5$.

(B) The horizontal component of velocity is $u_x = u \cos \theta = 20 \cos 30^{\circ} = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \ m/s$.
The vertical component of velocity is $u_y = u \sin \theta = 20 \sin 30^{\circ} = 20 \times \frac{1}{2} = 10 \ m/s$.
After time $t = 0.1 \ s$,the horizontal displacement is $x = u_x t = 10\sqrt{3} \times 0.1 = \sqrt{3} \ m$.
The vertical displacement is $y = u_y t - \frac{1}{2}gt^2 = 10 \times 0.1 - \frac{1}{2} \times 10 \times (0.1)^2 = 1 - 0.05 = 0.95 \ m = \frac{19}{20} \ m$.
The angle $\theta$ between the displacement vector and the horizontal is given by $\tan \theta = \frac{y}{x}$.
Substituting the values,$\tan \theta = \frac{19/20}{\sqrt{3}} = \frac{19}{20\sqrt{3}}$.
Thus,$\theta = \tan^{-1}\left(\frac{19}{20\sqrt{3}}\right)$. The question asks for the value of $\tan \theta$ (implied by options). Therefore,the correct value is $\frac{19}{20\sqrt{3}}$.

(C) The maximum height $h$ is given by $h = \frac{u^2 \sin^2 \alpha}{2g}$.
The range $R$ is given by $R = \frac{u^2 \sin 2\alpha}{g} = \frac{2 u^2 \sin \alpha \cos \alpha}{g}$.
Dividing $h$ by $R$:
$\frac{h}{R} = \frac{u^2 \sin^2 \alpha}{2g} \cdot \frac{g}{2 u^2 \sin \alpha \cos \alpha} = \frac{\tan \alpha}{4}$.
Therefore,$\tan \alpha = \frac{4h}{R}$.
The time of flight $T$ is given by $T = \frac{2u \sin \alpha}{g}$.
Squaring both sides,$T^2 = \frac{4u^2 \sin^2 \alpha}{g^2}$.
Thus,$\frac{g T^2}{8} = \frac{g}{8} \cdot \frac{4u^2 \sin^2 \alpha}{g^2} = \frac{u^2 \sin^2 \alpha}{2g} = h$.
So,$h = \frac{g T^2}{8}$.

(A) Let the initial speed be $u$ and the angle of projection be $\theta$. The speed at the maximum height is $u \cos \theta$.
According to the problem,the initial speed is twice the speed at the maximum height:
$u = 2(u \cos \theta)$
$\Rightarrow \cos \theta = \frac{1}{2}$
$\Rightarrow \theta = 60^{\circ}$.
The range $R$ is given by $R = \frac{u^2 \sin(2\theta)}{g}$ and the maximum height $H$ is given by $H = \frac{u^2 \sin^2 \theta}{2g}$.
The ratio $\frac{R}{H}$ is:
$\frac{R}{H} = \frac{u^2 \sin(2\theta) / g}{u^2 \sin^2 \theta / (2g)} = \frac{2 \sin(2\theta)}{\sin^2 \theta} = \frac{4 \sin \theta \cos \theta}{\sin^2 \theta} = 4 \cot \theta$.
Substituting $\theta = 60^{\circ}$:
$\frac{R}{H} = 4 \cot(60^{\circ}) = 4 \times \frac{1}{\sqrt{3}} = \frac{4}{\sqrt{3}}$.

(A) The horizontal component of velocity remains constant throughout the projectile motion. Let the initial velocity be $v_0 = 140 \ m/s$ at angle $\theta_0 = 60^{\circ}$. Let the velocity at time $t$ be $v$ at angle $\theta = 30^{\circ}$.
Horizontal component: $v_x = v_0 \cos 60^{\circ} = v \cos 30^{\circ}$.
$140 \times \frac{1}{2} = v \times \frac{\sqrt{3}}{2} \implies v = \frac{140}{\sqrt{3}} \ m/s$.
Vertical component at time $t$: $v_y = v_0 \sin 60^{\circ} - gt = v \sin 30^{\circ}$.
$140 \times \frac{\sqrt{3}}{2} - 10t = \frac{140}{\sqrt{3}} \times \frac{1}{2}$.
$70\sqrt{3} - 10t = \frac{70}{\sqrt{3}}$.
$10t = 70\sqrt{3} - \frac{70}{\sqrt{3}} = 70 \left( \frac{3-1}{\sqrt{3}} \right) = \frac{140}{\sqrt{3}}$.
$t = \frac{14}{\sqrt{3}} \ s$.

(B) Let the angle of projection be $\theta = 45^{\circ}$.
At the highest point $P$,the height is $H = \frac{u^2 \sin^2 \theta}{2g}$ and the horizontal distance from the point of projection $O$ is $R/2 = \frac{u^2 \sin \theta \cos \theta}{g}$.
The elevation angle $\alpha$ is the angle subtended by the highest point $P$ at the point of projection $O$ with the horizontal.
In the right-angled triangle $\triangle POM$,we have:
$\tan \alpha = \frac{H}{R/2} = \frac{\frac{u^2 \sin^2 \theta}{2g}}{\frac{u^2 \sin \theta \cos \theta}{g}} = \frac{\sin \theta}{2 \cos \theta} = \frac{1}{2} \tan \theta$.
Substituting $\theta = 45^{\circ}$:
$\tan \alpha = \frac{1}{2} \tan 45^{\circ} = \frac{1}{2} (1) = \frac{1}{2}$.
Therefore,$\alpha = \tan^{-1}\left(\frac{1}{2}\right)$.

(B) Let $d$ be the distance of the target.
The horizontal range of a projectile is given by $R = \frac{u^2 \sin 2\theta}{g}$.
For $\theta = 15^{\circ}$,the range is $R_1 = \frac{u^2 \sin(30^{\circ})}{g} = \frac{u^2}{2g}$.
Given that it falls short by $10 \ m$,we have $R_1 = d - 10$,so $\frac{u^2}{2g} = d - 10$ (Equation $i$).
For $\theta = 45^{\circ}$,the range is $R_2 = \frac{u^2 \sin(90^{\circ})}{g} = \frac{u^2}{g}$.
Given that it is away (beyond) by $10 \ m$,we have $R_2 = d + 10$,so $\frac{u^2}{g} = d + 10$ (Equation $ii$).
Dividing Equation $i$ by Equation $ii$,we get $\frac{1}{2} = \frac{d - 10}{d + 10}$.
Solving for $d$,$d + 10 = 2d - 20$,which gives $d = 30 \ m$.
Substituting $d = 30$ into Equation $ii$,we get $\frac{u^2}{g} = 30 + 10 = 40$.
To hit the target at $d = 30 \ m$,we set $R = 30 = \frac{u^2 \sin 2\theta}{g} = 40 \sin 2\theta$.
Thus,$\sin 2\theta = \frac{30}{40} = \frac{3}{4}$,which implies $\theta = \frac{1}{2} \sin^{-1}\left(\frac{3}{4}\right)$.

(B) Given, initial velocity $u = 5 \text{ m/s}$.
Horizontal range $R$ is twice the maximum height $H$, i.e., $R = 2H$.
We know that $R = \frac{u^2 \sin 2\theta}{g}$ and $H = \frac{u^2 \sin^2 \theta}{2g}$.
Substituting these into the given condition:
$\frac{u^2 \sin 2\theta}{g} = 2 \left( \frac{u^2 \sin^2 \theta}{2g} \right)$
$\frac{u^2 (2 \sin \theta \cos \theta)}{g} = \frac{u^2 \sin^2 \theta}{g}$
$2 \sin \theta \cos \theta = \sin^2 \theta$
$2 \cos \theta = \sin \theta \Rightarrow \tan \theta = 2$.
For $\tan \theta = 2$, we have a right-angled triangle with perpendicular $= 2$ and base $= 1$. 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 $R$:
$R = \frac{u^2 (2 \sin \theta \cos \theta)}{g} = \frac{5^2 \times 2 \times (\frac{2}{\sqrt{5}}) \times (\frac{1}{\sqrt{5}})}{10}$
$R = \frac{25 \times 2 \times 2}{10 \times 5} = \frac{100}{50} = 2 \text{ m}$.
Therefore, the correct option is $B$.

(C) The velocity of the truck is $v = 30 \,m/s$. The time taken by the truck to travel a distance of $100 \,m$ is given by $t = \frac{d}{v} = \frac{100}{30} = \frac{10}{3} \,s$.
Since the boy catches the projectile after this time, the total time of flight of the projectile is $T = \frac{10}{3} \,s$.
For a projectile thrown vertically relative to the truck, the time of flight is given by $T = \frac{2u_y}{g}$, where $u_y$ is the vertical component of the velocity.
Substituting the values: $\frac{10}{3} = \frac{2u_y}{10}$.
Solving for $u_y$: $u_y = \frac{10 \times 10}{3 \times 2} = \frac{100}{6} = \frac{50}{3} \,m/s$.
Since the projectile is thrown relative to the truck, the horizontal component of velocity relative to the truck is $0$. Thus, the speed relative to the truck is simply $u_y = \frac{50}{3} \,m/s$.

(D) Let the object be at point $B(x, y)$ after time $t = \sqrt{2} \,s$.
The horizontal displacement is $x = u_x \times t = (v_0 \cos 45^{\circ}) \times \sqrt{2} = v_0 \times \frac{1}{\sqrt{2}} \times \sqrt{2} = v_0$.
The vertical displacement is $y = u_y t - \frac{1}{2} g t^2 = (v_0 \sin 45^{\circ}) \sqrt{2} - \frac{1}{2} (10) (\sqrt{2})^2 = v_0 - 10$.
The displacement vector $\vec{OB}$ has magnitude $OB = \sqrt{x^2 + y^2} = \sqrt{v_0^2 + (v_0 - 10)^2}$.
The average velocity magnitude is given by $v_{\text{avg}} = \frac{OB}{t} = |v_0|$.
Thus,$OB = |v_0| t = v_0 \sqrt{2}$.
Squaring both sides: $v_0^2 + (v_0 - 10)^2 = (v_0 \sqrt{2})^2$.
$v_0^2 + v_0^2 - 20v_0 + 100 = 2v_0^2$.
$2v_0^2 - 20v_0 + 100 = 2v_0^2$.
$20v_0 = 100$,which gives $v_0 = 5 \,m/s$.

(B) The maximum range of a projectile is achieved when the angle of projection is $\theta = 45^{\circ}$.
The formula for the range $R$ is given by $R = \frac{u^2 \sin(2\theta)}{g}$.
For maximum range, $\sin(2\theta) = \sin(90^{\circ}) = 1$.
Thus, $R_{max} = \frac{u^2}{g}$.
Given $u = 30 \,m/s$ and $g = 10 \,m/s^2$, we have:
$R_{max} = \frac{(30)^2}{10} = \frac{900}{10} = 90 \,m$.

(A) Given: Initial speed $u = 20 \,m/s$, angle of projection $\theta = 34^{\circ}$, and acceleration due to gravity $g = 9.8 \,m/s^2$.
The formula for the maximum height $H$ of a projectile is given by $H = \frac{u^2 \sin^2 \theta}{2g}$.
Substituting the given values:
$H = \frac{(20)^2 \times (\sin 34^{\circ})^2}{2 \times 9.8}$
$H = \frac{400 \times (0.56)^2}{19.6}$
$H = \frac{400 \times 0.3136}{19.6}$
$H = \frac{125.44}{19.6} = 6.4 \,m$.
Rounding to the nearest provided option, the correct answer is $6.3 \,m$.

(B) Initial velocity $u = 2\sqrt{2} \ m/s$ at angle $\theta = 45^{\circ}$.
Horizontal component $u_x = u \cos 45^{\circ} = 2\sqrt{2} \times \frac{1}{\sqrt{2}} = 2 \ m/s$.
Vertical component $u_y = u \sin 45^{\circ} = 2\sqrt{2} \times \frac{1}{\sqrt{2}} = 2 \ m/s$.
Horizontal displacement in $t = 2 \ s$ is $x = u_x \times t = 2 \times 2 = 4 \ m$.
Vertical displacement in $t = 2 \ s$ is $y = u_y t - \frac{1}{2}gt^2 = 2(2) - \frac{1}{2}(10)(2^2) = 4 - 20 = -16 \ m$.
Total displacement $S = \sqrt{x^2 + y^2} = \sqrt{4^2 + (-16)^2} = \sqrt{16 + 256} = \sqrt{272} \approx 16.49 \ m$.
Mean velocity $v_{avg} = \frac{S}{t} = \frac{16.49}{2} = 8.245 \ m/s \approx 8.2 \ m/s$.