Horizontal Projectile Motion Questions in English

(A) The horizontal range $R$ of a projectile is given by $R = \frac{u^2 \sin(2\theta)}{g} = \frac{2u^2 \sin \theta \cos \theta}{g}$.
The maximum height $H$ attained is given by $H = \frac{u^2 \sin^2 \theta}{2g}$.
According to the problem,the range is double the maximum height,so $R = 2H$.
Substituting the formulas: $\frac{2u^2 \sin \theta \cos \theta}{g} = 2 \left( \frac{u^2 \sin^2 \theta}{2g} \right)$.
Simplifying the equation: $2 \sin \theta \cos \theta = \sin^2 \theta$.
Dividing both sides by $\sin \theta$ (assuming $\sin \theta \neq 0$): $2 \cos \theta = \sin \theta$.
Therefore,$\tan \theta = \frac{\sin \theta}{\cos \theta} = 2$.
Thus,the angle of projection is $\theta = \tan^{-1}(2)$.

(D) The time of flight $T$ is given by the formula $T = \frac{2u_y}{g}$,where $u_y$ is the vertical component of the initial velocity.
Given $T = 3\,s$ and $g = 10\,m/s^2$,we have $3 = \frac{2u_y}{10}$.
Solving for $u_y$,we get $u_y = \frac{3 \times 10}{2} = 15\,m/s$.
The maximum height $H$ reached by the projectile is given by $H = \frac{u_y^2}{2g}$.
Substituting the values,$H = \frac{(15)^2}{2 \times 10} = \frac{225}{20} = 11.25\,m$.
Therefore,the correct option is $D$.

(B) The horizontal range $R$ of a projectile launched with initial velocity $u$ at an angle $\alpha$ is given by $R = \frac{u^2 \sin(2\alpha)}{g}$.
Given that the range is the same for angles $\theta$ and $2\theta$,we have:
$\sin(2\theta) = \sin(2(2\theta)) = \sin(4\theta)$.
For $\sin(A) = \sin(B)$,the general solution is $A = 180^{\circ} - B$ (since $\theta$ and $2\theta$ are acute angles in this context).
Therefore,$2\theta = 180^{\circ} - 4\theta$.
$6\theta = 180^{\circ}$.
$\theta = 30^{\circ}$.

(A) The horizontal component of the velocity of the ball is $u_x = u \cos \theta$.
Since there is no acceleration in the horizontal direction,the horizontal velocity of the ball remains constant throughout its flight.
To catch the ball,the boy must cover the same horizontal distance as the ball in the same amount of time.
Therefore,the boy must run with a constant horizontal velocity equal to the horizontal component of the ball's velocity.
Thus,the required velocity of the boy is $v = u \cos \theta$.

(B) The horizontal range of a projectile is given by the formula $R = \frac{u^2 \sin(2\theta)}{g}$.
Since the initial speed $u$ is the same for all three balls,the range depends on $\sin(2\theta)$.
For $\theta_1 = 30^{\circ}$,$R_1 \propto \sin(60^{\circ}) = \frac{\sqrt{3}}{2} \approx 0.866$.
For $\theta_2 = 45^{\circ}$,$R_2 \propto \sin(90^{\circ}) = 1$.
For $\theta_3 = 60^{\circ}$,$R_3 \propto \sin(120^{\circ}) = \sin(60^{\circ}) = \frac{\sqrt{3}}{2} \approx 0.866$.
Since $30^{\circ} + 60^{\circ} = 90^{\circ}$,the ranges for complementary angles are equal,so $R_1 = R_3$.
Since $\sin(90^{\circ})$ is the maximum value of the sine function,$R_2$ is the maximum range.
Therefore,$R_1 = R_3 < R_2$.

(B) Given:
Range $R = 200\,m$
Time of flight $T = 5\,s$
Acceleration due to gravity $g = 10\,m/s^2$
$1$. Horizontal component of velocity $(u_x)$:
The horizontal range is given by $R = u_x \times T$.
Therefore,$u_x = \frac{R}{T} = \frac{200}{5} = 40\,m/s$.
$2$. Maximum height $(H)$:
The time of flight is given by $T = \frac{2u_y}{g}$,where $u_y$ is the vertical component of velocity.
$5 = \frac{2u_y}{10} \implies 2u_y = 50 \implies u_y = 25\,m/s$.
The maximum height is given by $H = \frac{u_y^2}{2g}$.
$H = \frac{(25)^2}{2 \times 10} = \frac{625}{20} = 31.25\,m$.
Thus,the horizontal component of velocity is $40\,m/s$ and the maximum height is $31.25\,m$.

(A) The formula for the horizontal range is $R = \frac{u^2 \sin 2\theta}{g}$.
For maximum range,$\theta = 45^{\circ}$,so $R_{\max} = \frac{u^2}{g} = 160 \ m$.
Therefore,$u^2 = 160g$.
The formula for maximum height is $H = \frac{u^2 \sin^2 \theta}{2g}$.
Given $\theta = 30^{\circ}$,we have $\sin 30^{\circ} = 0.5$.
Substituting the values: $H = \frac{160g \times (\sin 30^{\circ})^2}{2g} = \frac{160 \times (0.5)^2}{2} = \frac{160 \times 0.25}{2} = \frac{40}{2} = 20 \ m$.

(A) For two projectiles to trace identical trajectories,their maximum heights $(H)$ and horizontal ranges $(R)$ must be equal.
The formula for maximum height is $H = \frac{u^2 \sin^2 \theta}{2g}$.
Since the trajectories are identical,the angle of projection $\theta$ is the same for both.
Equating the heights: $\frac{u_1^2 \sin^2 \theta}{2g_1} = \frac{u_2^2 \sin^2 \theta}{2g_2}$.
Substituting the given values ($u_1 = 10 \, m/s$,$u_2 = 5 \, m/s$,and $g_1 = 10 \, m/s^2$):
$\frac{10^2}{2 \times 10} = \frac{5^2}{2 \times g_2}$.
$\frac{100}{20} = \frac{25}{2g_2}$.
$5 = \frac{12.5}{g_2}$.
$g_2 = \frac{12.5}{5} = 2.5 \, m/s^2$.

(D) The bullet travels horizontally,so its initial vertical velocity $u_y = 0$.
Given: Horizontal distance $S_x = 100\,m$,vertical displacement $h = 10\,cm = 0.1\,m$,and acceleration due to gravity $g = 10\,m/s^2$.
Using the equation of motion in the vertical direction: $h = u_y t + \frac{1}{2} g t^2$.
Since $u_y = 0$,we have $h = \frac{1}{2} g t^2$.
Substituting the values: $0.1 = \frac{1}{2} \times 10 \times t^2$.
$0.1 = 5 t^2 \implies t^2 = 0.02 \implies t = \sqrt{0.02} \approx 0.1414\,s$.
In the horizontal direction,the velocity $v_x$ is constant: $v_x = \frac{S_x}{t}$.
$v_x = \frac{100}{0.1414} \approx 707\,m/s$.
Rounding to the nearest provided option,the velocity is $700\,m/s$.

(D) The initial velocity is $u = 19.6 \sqrt{2} \ m/s$ at an angle $\theta = 45^{\circ}$.
Horizontal component: $u_x = u \cos(45^{\circ}) = 19.6 \sqrt{2} \times \frac{1}{\sqrt{2}} = 19.6 \ m/s$.
Vertical component: $u_y = u \sin(45^{\circ}) = 19.6 \sqrt{2} \times \frac{1}{\sqrt{2}} = 19.6 \ m/s$.
After time $t = 2 \ s$,the horizontal velocity remains constant: $v_x = u_x = 19.6 \ m/s$.
The vertical velocity is given by $v_y = u_y - gt = 19.6 - (9.8 \times 2) = 19.6 - 19.6 = 0 \ m/s$.
Since the vertical component of velocity is $0 \ m/s$,the velocity vector is purely horizontal.
Therefore,the angle $\alpha$ with the horizontal is $0^{\circ}$.

(A) The initial momentum of the body is $p = mv$,where $m$ is the mass and $v$ is the initial velocity.
The horizontal component of the velocity remains constant throughout the motion because there is no horizontal acceleration.
The initial horizontal component of velocity is $v_x = v \cos 30^{\circ} = v \frac{\sqrt{3}}{2}$.
At the highest point of the trajectory,the vertical component of velocity becomes zero,and the velocity of the body is equal to its horizontal component,$v_h = v_x = v \frac{\sqrt{3}}{2}$.
The momentum at the highest point is $p' = m v_h = m (v \frac{\sqrt{3}}{2}) = p \frac{\sqrt{3}}{2}$.

(C) The maximum height $H$ of a projectile is given by $H = \frac{u^2 \sin^2 \theta}{2g}$.
Since the angle of projection $\theta$ and gravity $g$ are constant,$H \propto u^2$.
If $H$ increases by $10\,\%$,then $H' = 1.10 H$.
Since $H' / H = (u' / u)^2$,we have $(u' / u)^2 = 1.10$,which implies $u' / u = \sqrt{1.10} \approx 1.0488$.
The horizontal range $R$ is given by $R = \frac{u^2 \sin 2\theta}{g}$.
Since $\theta$ and $g$ are constant,$R \propto u^2$.
Therefore,the percentage increase in range is the same as the percentage increase in $H$,which is $10\,\%$.

(B) The range $R$ and maximum height $H$ of a projectile are given by:
$R = \frac{u^2 \sin(2\theta)}{g} = \frac{2u^2 \sin\theta \cos\theta}{g}$
$H = \frac{u^2 \sin^2\theta}{2g}$
Taking the ratio $R/H$:
$\frac{R}{H} = \frac{2u^2 \sin\theta \cos\theta}{g} \times \frac{2g}{u^2 \sin^2\theta} = 4 \cot\theta$
Therefore,$\frac{R}{H} = 4 \cot\theta$ or $\tan\theta = \frac{4H}{R}$.
For $A. R/H = 1$: $1 = 4 \cot\theta \implies \tan\theta = 4 \implies \theta = \tan^{-1}(4) = 76^o$ (Matches $4$).
For $B. R/H = 4$: $4 = 4 \cot\theta \implies \tan\theta = 1 \implies \theta = 45^o$ (Matches $3$).
For $C. R/H = 4\sqrt{3}$: $4\sqrt{3} = 4 \cot\theta \implies \tan\theta = 1/\sqrt{3} \implies \theta = 30^o$ (Matches $2$).
For $D. R/H = 4/\sqrt{3}$: $4/\sqrt{3} = 4 \cot\theta \implies \tan\theta = \sqrt{3} \implies \theta = 60^o$ (Matches $1$).
Thus,the correct matching is $A-4, B-3, C-2, D-1$.

(B) The angular momentum is given by $\overrightarrow{L} = \overrightarrow{r} \times \overrightarrow{p} = \overrightarrow{r} \times m\overrightarrow{v}$.
At the maximum height $h$,the vertical component of velocity is zero,and the velocity is purely horizontal,given by $v_x = v \cos(30^{\circ}) = \frac{\sqrt{3}v}{2}$.
The position vector $\overrightarrow{r}$ at maximum height has a horizontal component equal to the horizontal range at half-time,but the cross product $\overrightarrow{r} \times \overrightarrow{v}$ simplifies to $L = m \cdot v_x \cdot h$.
The maximum height $h$ is given by $h = \frac{v^2 \sin^2(30^{\circ})}{2g} = \frac{v^2 (1/4)}{2g} = \frac{v^2}{8g}$.
Substituting these values into the angular momentum formula:
$L = m \cdot \left( \frac{\sqrt{3}v}{2} \right) \cdot \left( \frac{v^2}{8g} \right) = \frac{\sqrt{3}mv^3}{16g}$.

(C) The horizontal position is given by $x = u_x t$. Given $x = 40\,m$ at $t = 2\,s$,we have $40 = u_x \times 2$,so $u_x = 20\,m/s$.
The vertical position is given by $y = u_y t - \frac{1}{2} g t^2$. Given $y = 50\,m$ at $t = 2\,s$ and $g = 10\,m/s^2$,we have $50 = u_y(2) - \frac{1}{2}(10)(2)^2$.
$50 = 2u_y - 20$,which implies $2u_y = 70$,so $u_y = 35\,m/s$.
The angle of projection $\theta$ is given by $\tan \theta = \frac{u_y}{u_x}$.
$\tan \theta = \frac{35}{20} = \frac{7}{4}$.
Therefore,$\theta = \tan^{-1}(\frac{7}{4})$.

(C) Let $t$ be the time taken by the bullet to hit the target.
Horizontal distance $d = 700 \; m$ and horizontal velocity $v_x = 630 \; m/s$.
Since there is no horizontal acceleration,$t = \frac{d}{v_x} = \frac{700}{630} = \frac{10}{9} \; s$.
For vertical motion,the bullet is subject to gravity. The vertical displacement $h$ (drop) is given by $h = u_y t + \frac{1}{2} g t^2$.
Since the initial vertical velocity $u_y = 0$,we have $h = \frac{1}{2} g t^2$.
Substituting the values: $h = \frac{1}{2} \times 10 \times \left( \frac{10}{9} \right)^2$.
$h = 5 \times \frac{100}{81} = \frac{500}{81} \approx 6.17 \; m$.
Therefore,the rifle must be aimed $6.17 \; m$ above the center of the target to compensate for the gravitational drop.

(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.
Given: $\theta = 45^o$,$x = 4 \, m$ (distance of the wall from the projection point),and the total range $R = 4 \, m + 6 \, m = 10 \, m$.
Substituting these values into the trajectory equation:
$y = 4 \tan(45^o) \left( 1 - \frac{4}{10} \right)$
$y = 4 \times 1 \times (1 - 0.4)$
$y = 4 \times 0.6 = 2.4 \, m$.
Thus,the height of the wall is $2.4 \, m$.

(A) The maximum horizontal range $R$ of a projectile fired with speed $v$ is given by $R = \frac{v^2}{g}$.
Since the bullets are fired in all possible directions,the area $A$ covered on the ground is a circle with radius $R$.
Thus,the area $A = \pi R^2 = \pi \left( \frac{v^2}{g} \right)^2 = \frac{\pi v^4}{g^2}$.
This implies that the area $A \propto v^4$.
Given speeds are $v_A = 1 \ km/s$ and $v_B = 2 \ km/s$.
The ratio of the areas is $\frac{A_A}{A_B} = \left( \frac{v_A}{v_B} \right)^4 = \left( \frac{1}{2} \right)^4 = \frac{1}{16}$.
Therefore,the ratio is $1:16$.

(B) Initial velocity components: $u_x = 10 \cos 60^\circ = 5\,ms^{-1}$ and $u_y = 10 \sin 60^\circ = 5\sqrt{3}\,ms^{-1}$.
At $t = 1\,s$,the velocity components are:
$v_x = u_x = 5\,ms^{-1}$
$v_y = u_y - gt = 5\sqrt{3} - 10(1) = 5\sqrt{3} - 10\,ms^{-1}$.
The speed $v$ at $t = 1\,s$ is $v = \sqrt{v_x^2 + v_y^2} = \sqrt{5^2 + (5\sqrt{3} - 10)^2} = \sqrt{25 + (75 + 100 - 100\sqrt{3})} = \sqrt{200 - 100\sqrt{3}} \approx \sqrt{200 - 173.2} = \sqrt{26.8} \approx 5.17\,ms^{-1}$.
The radius of curvature is given by $R = \frac{v^2}{a_{\perp}}$,where $a_{\perp}$ is the component of acceleration perpendicular to the velocity.
The angle $\theta$ that the velocity vector makes with the horizontal is $\tan \theta = \frac{v_y}{v_x} = \frac{5\sqrt{3} - 10}{5} = \sqrt{3} - 2 \approx -0.268$. Thus,$\theta \approx -15^\circ$.
The acceleration $g$ acts vertically downwards. The component of $g$ perpendicular to the velocity is $a_{\perp} = g \cos \theta$.
$R = \frac{v^2}{g \cos \theta} = \frac{200 - 100\sqrt{3}}{10 \cos(-15^\circ)} = \frac{26.8}{10 \times 0.966} \approx 2.77\,m \approx 2.8\,m$.

(A) The standard equation of the trajectory of a projectile is given by $y = x \tan \theta_0 - \frac{g x^2}{2 v_0^2 \cos^2 \theta_0}$.
Comparing this with the given equation $y = 2x - 9x^2$:
$1$. $\tan \theta_0 = 2$. Since $\tan \theta_0 = \frac{\text{opposite}}{\text{adjacent}} = \frac{2}{1}$,the hypotenuse is $\sqrt{2^2 + 1^2} = \sqrt{5}$. Thus,$\cos \theta_0 = \frac{1}{\sqrt{5}}$ and $\sin \theta_0 = \frac{2}{\sqrt{5}}$.
$2$. $\frac{g}{2 v_0^2 \cos^2 \theta_0} = 9$.
Substituting $g = 10$ and $\cos^2 \theta_0 = \left( \frac{1}{\sqrt{5}} \right)^2 = \frac{1}{5}$:
$\frac{10}{2 v_0^2 (1/5)} = 9 \implies \frac{10}{2 v_0^2 / 5} = 9 \implies \frac{25}{v_0^2} = 9 \implies v_0^2 = \frac{25}{9} \implies v_0 = \frac{5}{3} \, ms^{-1}$.
Therefore,$\theta_0 = \cos^{-1} \left( \frac{1}{\sqrt{5}} \right) = \sin^{-1} \left( \frac{2}{\sqrt{5}} \right)$ and $v_0 = \frac{5}{3} \, ms^{-1}$. Comparing with options,option $A$ is correct.

(C) For the same range $R$,the angles of projection must be complementary,i.e.,$\theta$ and $(90^\circ - \theta)$.
The range $R$ is given by: $R = \frac{u^2 \sin(2\theta)}{g} = \frac{2u^2 \sin\theta \cos\theta}{g}$.
The maximum heights for the two particles are:
$h_1 = \frac{u^2 \sin^2\theta}{2g}$
$h_2 = \frac{u^2 \sin^2(90^\circ - \theta)}{2g} = \frac{u^2 \cos^2\theta}{2g}$
Multiplying $h_1$ and $h_2$:
$h_1 h_2 = \left( \frac{u^2 \sin^2\theta}{2g} \right) \left( \frac{u^2 \cos^2\theta}{2g} \right) = \frac{u^4 \sin^2\theta \cos^2\theta}{4g^2}$
$h_1 h_2 = \frac{1}{16} \left( \frac{4u^4 \sin^2\theta \cos^2\theta}{g^2} \right) = \frac{1}{16} R^2$
Therefore,$R^2 = 16 h_1 h_2$.

(B) The equation of the trajectory of a projectile is given by $y = 12x - \frac{3}{4}x^2$.
The range $R$ of a projectile is the horizontal distance covered when the projectile returns to the ground,i.e.,when the vertical displacement $y = 0$.
Setting $y = 0$ in the given equation:
$0 = 12x - \frac{3}{4}x^2$
Factoring out $x$:
$0 = x(12 - \frac{3}{4}x)$
Since $x = 0$ represents the starting point,the range corresponds to the non-zero solution:
$12 - \frac{3}{4}x = 0$
$\frac{3}{4}x = 12$
$x = 12 \times \frac{4}{3}$
$x = 16\,m$
Therefore,the range of the projectile is $16\,m$.

(C) In projectile motion,the horizontal component of velocity remains constant throughout the flight because there is no acceleration in the horizontal direction.
Let the initial velocity be $u$ at an angle $\theta$ with the horizontal. The horizontal component is $u_x = u \cos \theta$.
When the particle is at a point where its velocity is $v$ at an angle $\phi$ with the horizontal,the horizontal component of this velocity is $v_x = v \cos \phi$.
Since the horizontal component remains constant,we have $v_x = u_x$.
Therefore,$v \cos \phi = u \cos \theta$.
Solving for $v$,we get $v = \frac{u \cos \theta}{\cos \phi} = u \cos \theta \sec \phi$.

(D) For maximum range,the angle of projection is $\theta = 45^{\circ}$.
The range $R$ is given by $R = \frac{u^2 \sin(2\theta)}{g}$. Since $\theta = 45^{\circ}$,$R = \frac{u^2}{g}$.
Given $R = 100 \, km = 10^5 \, m$ and $g = 10 \, m/s^2$,we have $u^2 = gR = 10 \times 10^5 = 10^6 \, m^2/s^2$,so $u = 1000 \, m/s$.
The total time of flight is $T = \frac{2u \sin \theta}{g} = \frac{2 \times 1000 \times \sin(45^{\circ})}{10} = 200 \times \frac{1}{\sqrt{2}} = 100\sqrt{2} \, s$.
The missile is detected at its half-way point,which corresponds to the peak of its trajectory (since the range is symmetric for $45^{\circ}$ projection).
The time taken to reach the peak is $t_{peak} = \frac{T}{2} = \frac{100\sqrt{2}}{2} = 50\sqrt{2} \, s$.
The remaining time until impact is $T - t_{peak} = 100\sqrt{2} - 50\sqrt{2} = 50\sqrt{2} \, s$.
Wait,re-evaluating the detection point: The half-way point in terms of horizontal distance is the peak. The time remaining is the time from the peak to the ground,which is $t = \frac{u \sin \theta}{g} = \frac{1000 \times (1/\sqrt{2})}{10} = \frac{100}{\sqrt{2}} \, s$.

(C) For maximum range,the angle of projection is $\theta = 45^{\circ}$. The range $R = \frac{u^2 \sin(2\theta)}{g} = \frac{u^2}{g} = 10^5 \, m$. Thus,$u^2 = gR = 10 \times 10^5 = 10^6$,so $u = 1000 \, m/s$.
The horizontal component of velocity is $v_x = u \cos(45^{\circ}) = 1000 \times \frac{1}{\sqrt{2}} = 500\sqrt{2} \, m/s$.
At the maximum height,the vertical component is $0$. At the halfway point of the range,the missile is at its maximum height. Therefore,the speed of the missile when detected is equal to its horizontal component,which is $500\sqrt{2} \, m/s$.

(A) For maximum range,the angle of projection is $\theta = 45^\circ$. The range $R$ is given as $100\, km = 10^5\, m$.
The formula for range is $R = \frac{u^2 \sin(2\theta)}{g}$. Since $\sin(90^\circ) = 1$,we have $R = \frac{u^2}{g}$,so $u^2 = Rg$.
The formula for maximum height is $H = \frac{u^2 \sin^2 \theta}{2g}$.
Substituting $u^2 = Rg$ and $\theta = 45^\circ$ $(\sin^2 45^\circ = 1/2)$:
$H = \frac{(Rg) \times (1/2)}{2g} = \frac{R}{4}$.
Given $R = 10^5\, m$,we get $H = \frac{10^5}{4} = 2.5 \times 10^4\, m$.

(C) The projectile reaches its maximum height $H$ at a horizontal distance $x = \sqrt{3} H$ from the point of projection.
For a projectile,the horizontal distance to the highest point is half the range,i.e.,$x = R/2$.
We know the formulas for range $R$ and maximum height $H$:
$R = \frac{v_0^2 \sin(2\theta)}{g} = \frac{2 v_0^2 \sin \theta \cos \theta}{g}$
$H = \frac{v_0^2 \sin^2 \theta}{2g}$
Given $x = R/2 = \sqrt{3} H$,we have:
$\frac{R}{2H} = \sqrt{3}$
Substituting the expressions for $R$ and $H$:
$\frac{\frac{2 v_0^2 \sin \theta \cos \theta}{g}}{2 \left(\frac{v_0^2 \sin^2 \theta}{2g}\right)} = \sqrt{3}$
$\frac{2 v_0^2 \sin \theta \cos \theta}{g} \cdot \frac{g}{v_0^2 \sin^2 \theta} = \sqrt{3}$
$2 \cot \theta = \sqrt{3}$
$\cot \theta = \frac{\sqrt{3}}{2}$
$\tan \theta = \frac{2}{\sqrt{3}}$
Therefore,$\theta = \tan ^{-1}\left(\frac{2}{\sqrt{3}}\right)$.

(A) Given: Mass $m = 2\,kg$,initial velocity $\vec{u} = (4 \hat{i} + 4 \hat{j})\,m/s$,force $\vec{F} = -20 \hat{j}\,N$.
Acceleration $\vec{a} = \frac{\vec{F}}{m} = \frac{-20 \hat{j}}{2} = -10 \hat{j}\,m/s^2$.
For $y$-motion: $u_y = 4\,m/s$,$a_y = -10\,m/s^2$,initial $y_0 = 0$. The $y$-coordinate is zero when displacement $y = 0$.
Using $y = u_y t + \frac{1}{2} a_y t^2$,we get $0 = 4t - 5t^2$.
Solving for $t$,$t(4 - 5t) = 0$. Since $t \neq 0$,$t = \frac{4}{5} = 0.8\,s$.
For $x$-motion: $u_x = 4\,m/s$,$a_x = 0$.
The $x$-coordinate is $x = u_x t = 4 \times 0.8 = 3.2\,m$.

(B) The horizontal component of velocity is $u_x = u \cos \theta = 20 \cos 45^{\circ} = 20 \times \frac{1}{\sqrt{2}} = 10\sqrt{2}\,m/s$.
The vertical component of velocity is $u_y = u \sin \theta = 20 \sin 45^{\circ} = 20 \times \frac{1}{\sqrt{2}} = 10\sqrt{2}\,m/s$.
The time $t$ taken to reach the wall at a distance $x = 10\,m$ is given by $t = \frac{x}{u_x} = \frac{10}{10\sqrt{2}} = \frac{1}{\sqrt{2}}\,s$.
The height $y$ at which the ball hits the wall is given by the equation of trajectory: $y = u_y t - \frac{1}{2} g t^2$.
Substituting the values (taking $g = 10\,m/s^2$): $y = (10\sqrt{2}) \times \frac{1}{\sqrt{2}} - \frac{1}{2} \times 10 \times (\frac{1}{\sqrt{2}})^2$.
$y = 10 - 5 \times \frac{1}{2} = 10 - 2.5 = 7.5\,m$.

(A) Let the horizontal direction be the $x$-axis and the vertical direction be the $y$-axis.
At time $t$,the velocity vector of the first particle is $\vec{v}_1 = u_1 \hat{i} - gt \hat{j}$.
The velocity vector of the second particle is $\vec{v}_2 = -u_2 \hat{i} - gt \hat{j}$.
Since the velocities are mutually perpendicular,their dot product must be zero: $\vec{v}_1 \cdot \vec{v}_2 = 0$.
$(u_1 \hat{i} - gt \hat{j}) \cdot (-u_2 \hat{i} - gt \hat{j}) = 0$.
$-u_1 u_2 + g^2 t^2 = 0$.
$g^2 t^2 = u_1 u_2$.
$t = \frac{\sqrt{u_1 u_2}}{g}$.

(C) The initial velocity is given by $\vec{u} = a \hat{i} + b \hat{j}$,where $u_x = a$ and $u_y = b$.
The range $R$ of a projectile is given by $R = \frac{2 u_x u_y}{g} = \frac{2ab}{g}$.
The maximum height $H$ reached by the projectile is given by $H = \frac{u_y^2}{2g} = \frac{b^2}{2g}$.
According to the problem,the range is twice the maximum height: $R = 2H$.
Substituting the expressions for $R$ and $H$:
$\frac{2ab}{g} = 2 \left( \frac{b^2}{2g} \right)$.
Simplifying the equation:
$\frac{2ab}{g} = \frac{b^2}{g}$.
Since $b \neq 0$,we can divide both sides by $b/g$:
$2a = b$ or $b = 2a$.

(D) The first ball is projected vertically upwards,so its maximum height is given by $h = \frac{u^2}{2g}$.
The second ball is projected at an angle $\theta = 30^{\circ}$ with the horizontal,so its maximum height is given by $h' = \frac{u^2 \sin^2 \theta}{2g} = \frac{u^2 \sin^2 30^{\circ}}{2g} = \frac{u^2}{2g} \times (\frac{1}{2})^2 = \frac{1}{4} \times \frac{u^2}{2g} = \frac{h}{4}$.
The potential energy at the highest point is $PE = mgh$.
The ratio of their potential energies is $\frac{PE_1}{PE_2} = \frac{mgh}{mgh'} = \frac{h}{h/4} = \frac{4}{1}$.

(C) Let the projectile $A$ be thrown with velocity $v_1$ at an angle $\theta = 30^{\circ}$ with the horizontal.
The time taken by projectile $A$ to reach its highest point is $t = \frac{v_1 \sin \theta}{g}$.
At the highest point,the vertical velocity of projectile $A$ becomes zero,and it only has a horizontal velocity component $v_x = v_1 \cos \theta$.
For the two projectiles to collide,they must reach the same point at the same time. Since $B$ is thrown from a point $Q$ vertically below the highest point of $A$,the horizontal distance between them is covered by $A$ in time $t$.
For a collision to occur at the highest point of $A$,the vertical displacement of $B$ in time $t$ must equal the maximum height $H$ of projectile $A$.
Maximum height of $A$ is $H = \frac{v_1^2 \sin^2 \theta}{2g}$.
For projectile $B$,the displacement in time $t$ is $y = v_2 t - \frac{1}{2} g t^2$.
Substituting $t = \frac{v_1 \sin \theta}{g}$ into the equation for $y$:
$y = v_2 \left( \frac{v_1 \sin \theta}{g} \right) - \frac{1}{2} g \left( \frac{v_1 \sin \theta}{g} \right)^2 = \frac{v_2 v_1 \sin \theta}{g} - \frac{v_1^2 \sin^2 \theta}{2g}$.
Setting $y = H$:
$\frac{v_2 v_1 \sin \theta}{g} - \frac{v_1^2 \sin^2 \theta}{2g} = \frac{v_1^2 \sin^2 \theta}{2g}$.
$\frac{v_2 v_1 \sin \theta}{g} = \frac{v_1^2 \sin^2 \theta}{g}$.
$v_2 = v_1 \sin \theta$.
Given $\theta = 30^{\circ}$,$v_2 = v_1 \sin 30^{\circ} = v_1 \left( \frac{1}{2} \right)$.
Therefore,$\frac{v_2}{v_1} = \frac{1}{2}$.

(A) The equation of the projectile path is given by $y = Ax - Bx^2$.
At the horizontal range $R$,the vertical displacement $y$ is $0$.
Substituting $y = 0$ and $x = R$ into the equation:
$0 = AR - BR^2$
$0 = R(A - BR)$
Since $R \neq 0$ for the landing point,we have:
$A - BR = 0$
$BR = A$
$R = A/B$
Thus,the horizontal range of the projectile is $A/B$.

(A) 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 $v_y = b - ct$.
The particle is projected from a horizontal plane,so the vertical displacement is zero at the time of landing.
The time of flight $T$ is the time when the particle returns to the horizontal plane,which occurs when the vertical displacement $y = 0$.
Integrating $v_y$ with respect to time: $y = \int (b - ct) dt = bt - \frac{1}{2}ct^2$.
Setting $y = 0$,we get $t(b - \frac{1}{2}ct) = 0$. Thus,$T = \frac{2b}{c}$.
The range $R$ is the horizontal distance covered in time $T$: $R = v_x \times T = a \times \frac{2b}{c} = \frac{2ab}{c}$.

(A) The formula for the horizontal range is $R = \frac{u^2 \sin(2\theta)}{g}$.
Given the maximum horizontal range $R_{\max} = \frac{u^2}{g} = 16 \ km$.
The formula for the maximum height is $H = \frac{u^2 \sin^2(\theta)}{2g}$.
Given the angle of projection $\theta = 30^o$,we substitute the values:
$H = \frac{u^2 \sin^2(30^o)}{2g} = \frac{u^2}{2g} \times (\frac{1}{2})^2$.
$H = \frac{u^2}{2g} \times \frac{1}{4} = \frac{u^2}{8g}$.
Since $\frac{u^2}{g} = 16 \ km$,we have $H = \frac{16}{8} \ km = 2 \ km$.

(D) Given: Initial velocity $u = 30\,m/s$,angle $\theta_0 = \tan^{-1}(3/4)$.
From $\tan\theta_0 = 3/4$,we have $\sin\theta_0 = 3/5$ and $\cos\theta_0 = 4/5$.
Initial horizontal component: $u_x = u \cos\theta_0 = 30 \times (4/5) = 24\,m/s$.
Initial vertical component: $u_y = u \sin\theta_0 = 30 \times (3/5) = 18\,m/s$.
After $t = 1\,s$,the horizontal velocity remains constant: $v_x = u_x = 24\,m/s$.
The vertical velocity changes due to gravity: $v_y = u_y - gt = 18 - (10 \times 1) = 8\,m/s$.
The angle $\theta$ with the horizontal is given by $\tan\theta = v_y / v_x$.
Substituting the values: $\tan\theta = 8 / 24 = 1/3$.

(C) In horizontal projectile motion,the horizontal component of velocity $(v_x)$ remains constant throughout the flight.
Given,initial horizontal velocity $v_x = 18 \, m s^{-1}$.
Let $v_y$ be the vertical component of velocity when the body strikes the ground.
The angle of impact $\theta$ with the horizontal is given by $\tan \theta = \frac{v_y}{v_x}$.
Given $\theta = 45^{\circ}$,so $\tan 45^{\circ} = \frac{v_y}{18}$.
Since $\tan 45^{\circ} = 1$,we have $1 = \frac{v_y}{18}$.
Therefore,$v_y = 18 \, m s^{-1}$.

(C) Let $H$ be the maximum height and $R$ be the horizontal range of the projectile.
From the geometry of the trajectory,the highest point is at a horizontal distance of $R/2$ from the launch point.
The elevation angle $\phi$ is defined as the angle whose tangent is the ratio of the maximum height to the horizontal distance of the highest point from the launch point.
Therefore,$\tan \phi = \frac{H}{R/2}$.
Substituting the standard formulas $H = \frac{u^2 \sin^2 \theta}{2g}$ and $R = \frac{u^2 \sin 2\theta}{g}$,we get:
$\tan \phi = \frac{u^2 \sin^2 \theta / 2g}{(u^2 \sin 2\theta / g) / 2} = \frac{u^2 \sin^2 \theta / 2g}{u^2 \sin 2\theta / 2g} = \frac{\sin^2 \theta}{\sin 2\theta}$.
Using the identity $\sin 2\theta = 2 \sin \theta \cos \theta$,we have:
$\tan \phi = \frac{\sin^2 \theta}{2 \sin \theta \cos \theta} = \frac{1}{2} \frac{\sin \theta}{\cos \theta} = \frac{1}{2} \tan \theta$.

(B) The horizontal range $R$ is given by $R = \frac{v^2 \sin(2\theta)}{g}$.
For maximum horizontal range,the angle of projection $\theta$ must be $45^{\circ}$.
The initial kinetic energy is $K = \frac{1}{2}mv^2$.
At the highest point of the trajectory,the vertical component of velocity is zero,and the horizontal component is $v_x = v \cos(\theta)$.
Substituting $\theta = 45^{\circ}$,we get $v_x = v \cos(45^{\circ}) = \frac{v}{\sqrt{2}}$.
The kinetic energy at the highest point is $K' = \frac{1}{2}mv_x^2 = \frac{1}{2}m\left(\frac{v}{\sqrt{2}}\right)^2$.
$K' = \frac{1}{2}m\left(\frac{v^2}{2}\right) = \frac{1}{2} \left(\frac{1}{2}mv^2\right) = \frac{K}{2} = 0.5K$.

(C) The horizontal velocity $u_x = u \cos \theta$ is equal to the velocity acquired from rest with acceleration $a = 3 \, ms^{-2}$ for time $t = 0.5 \, minutes = 30 \, s$.
$u_x = a \times t = 3 \times 30 = 90 \, ms^{-1}$.
The maximum height $H$ is given by $H = \frac{u_y^2}{2g}$,where $u_y = u \sin \theta$.
$80 = \frac{(u \sin \theta)^2}{2 \times 10} \implies (u \sin \theta)^2 = 1600 \implies u_y = 40 \, ms^{-1}$.
The angle of projection $\theta$ is given by $\tan \theta = \frac{u_y}{u_x}$.
$\tan \theta = \frac{40}{90} = \frac{4}{9}$.
Therefore,$\theta = \tan^{-1}(4/9)$.

(C) The kinetic energy of a projectile is maximum at the point of projection and at the ground level (where speed is $v$),and minimum at the highest point (where speed is $v \cos \theta$).
Given $\frac{K_{\max}}{K_{\min}} = 9$,we have $\frac{\frac{1}{2}mv^2}{\frac{1}{2}mv^2 \cos^2 \theta} = 9$.
This simplifies to $\frac{1}{\cos^2 \theta} = 9$,which means $\cos^2 \theta = \frac{1}{9}$,so $\cos \theta = \frac{1}{3}$.
Since $\cos \theta = \frac{1}{3}$,we find $\sin \theta = \sqrt{1 - (\frac{1}{3})^2} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}$.
Thus,$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{2\sqrt{2}/3}{1/3} = 2\sqrt{2}$.
The range $R$ is given by $R = \frac{v^2 \sin 2\theta}{g} = \frac{2v^2 \sin \theta \cos \theta}{g}$ and the maximum height $H$ is $H = \frac{v^2 \sin^2 \theta}{2g}$.
The ratio $\frac{R}{H} = \frac{2v^2 \sin \theta \cos \theta / g}{v^2 \sin^2 \theta / 2g} = 4 \cot \theta$.
Substituting $\cot \theta = \frac{1}{\tan \theta} = \frac{1}{2\sqrt{2}}$,we get $\frac{R}{H} = 4 \times \frac{1}{2\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

(B) The horizontal range $R$ is given by $R = u \cos \theta \cdot T$,where $u$ is the initial velocity and $\theta$ is the angle of projection.
The time of flight $T$ is given by $T = \frac{2u \sin \theta}{g}$,which implies $u \sin \theta = \frac{gT}{2}$.
Dividing the vertical component by the horizontal component:
$\tan \theta = \frac{u \sin \theta}{u \cos \theta} = \frac{gT/2}{R/T} = \frac{gT^2}{2R}$.
Therefore,the angle of projection is $\theta = \tan^{-1} \left( \frac{gT^2}{2R} \right)$.

(B) Let the angle of projection of particle $B$ with the horizontal be $\alpha$.
For particle $B$ to hit particle $A$ moving on the ground,the horizontal component of the velocity of $B$ must be equal to the velocity of $A$ (since $A$ moves with constant speed $v$ on the horizontal surface).
Thus,$v_B \cos \alpha = v$.
Given $v_B = \frac{2v}{\sqrt{3}}$,we have $\frac{2v}{\sqrt{3}} \cos \alpha = v$.
$\cos \alpha = \frac{\sqrt{3}}{2}$.
Therefore,$\alpha = 30^{\circ}$ with the horizontal.
The question asks for the angle of projection with the vertical.
Angle with vertical = $90^{\circ} - \alpha = 90^{\circ} - 30^{\circ} = 60^{\circ}$.

(B) The initial velocity is given by $\vec{u} = u_0 \hat{i} + v_0 \hat{j}$.
Here,the horizontal component of velocity is $u_x = u_0$ and the vertical component is $u_y = v_0$.
The acceleration in the $x$-direction is $a_x = 0$ and in the $y$-direction is $a_y = -g$.
The time of flight $T$ is the time taken for the vertical displacement to become zero: $y = u_y T - \frac{1}{2} g T^2 = 0$.
$v_0 T - \frac{1}{2} g T^2 = 0 \implies T = \frac{2v_0}{g}$.
The maximum displacement in the $x$-direction (horizontal range $R$) is given by $R = u_x \times T$.
Substituting the values,$R = u_0 \times \left( \frac{2v_0}{g} \right) = \frac{2u_0v_0}{g}$.

(D) The vertical displacement of a projectile is given by $h(t) = u_y t - \frac{1}{2} g t^2$,where $u_y$ is the initial vertical velocity.
Given the total time of flight is $T$,we have $T = \frac{2 u_y}{g}$,which implies $u_y = \frac{g T}{2}$.
At $t_A = \frac{T}{3}$,the height $h_A$ is:
$h_A = \left(\frac{g T}{2}\right) \left(\frac{T}{3}\right) - \frac{1}{2} g \left(\frac{T}{3}\right)^2 = \frac{g T^2}{6} - \frac{g T^2}{18} = \frac{3 g T^2 - g T^2}{18} = \frac{2 g T^2}{18} = \frac{g T^2}{9}$.
At $t_B = \frac{5T}{6}$,the height $h_B$ is:
$h_B = \left(\frac{g T}{2}\right) \left(\frac{5T}{6}\right) - \frac{1}{2} g \left(\frac{5T}{6}\right)^2 = \frac{5 g T^2}{12} - \frac{25 g T^2}{72} = \frac{30 g T^2 - 25 g T^2}{72} = \frac{5 g T^2}{72}$.
The difference in heights is $H = h_A - h_B = \frac{g T^2}{9} - \frac{5 g T^2}{72} = \frac{8 g T^2 - 5 g T^2}{72} = \frac{3 g T^2}{72} = \frac{g T^2}{24}$.

(C) The time taken by the cart to cover $80\,m$ is given by $t = \frac{d}{v} = \frac{80}{30} = \frac{8}{3}\,s$.
For the projectile to return to the same point on the cart,it must be fired vertically relative to the cart. The total time of flight $T$ must be equal to the time taken by the cart to travel $80\,m$,so $T = \frac{8}{3}\,s$.
The time taken to reach the maximum height is $t_h = \frac{T}{2} = \frac{8/3}{2} = \frac{4}{3}\,s$.
Using the equation of motion $v = u + at$,where $v = 0$ at maximum height,$a = -g = -10\,m/s^2$,and $t = \frac{4}{3}\,s$:
$0 = u - 10 \times \frac{4}{3}$
$u = \frac{40}{3}\,m/s$.
Thus,the projectile must be fired with a velocity of $\frac{40}{3}\,m/s$ relative to the cart.

(B) For maximum range,the angle of projection $\theta$ must be $45^o$.
Initial kinetic energy $KE_i = \frac{1}{2}mv^2 = 1000\,J$.
At the highest point,the vertical component of velocity is zero,so the velocity is only the horizontal component $v_x = v \cos(45^o) = \frac{v}{\sqrt{2}}$.
The kinetic energy at the highest point is $KE_f = \frac{1}{2}m(v_x)^2 = \frac{1}{2}m(\frac{v}{\sqrt{2}})^2 = \frac{1}{2}m(\frac{v^2}{2}) = \frac{1}{2} KE_i$.
Therefore,$KE_f = \frac{1000}{2} = 500\,J$.

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