Horizontal Projectile Motion Questions in English

(B) The time of flight $T$ for a projectile is given by the formula: $T = \frac{2u \sin \theta}{g}$.
Given:
Initial velocity $u = 9.8 \, m/s$
Angle of projection $\theta = 30^o$
Acceleration due to gravity $g = 9.8 \, m/s^2$
Substituting the values into the formula:
$T = \frac{2 \times 9.8 \times \sin 30^o}{9.8}$
Since $\sin 30^o = 0.5$,we get:
$T = \frac{2 \times 9.8 \times 0.5}{9.8} = 1 \, s$.
Therefore,the body will hit the ground after $1 \, s$.

(A) The horizontal position is given by $x = 36t$. The horizontal velocity is $v_x = \frac{dx}{dt} = 36 \, m/s$.
The vertical position is given by $2y = 96t - 9.8t^2$,which simplifies to $y = 48t - 4.9t^2$. The vertical velocity is $v_y = \frac{dy}{dt} = 48 - 9.8t$.
At the time of projection $(t = 0)$,the initial horizontal velocity is $u_x = 36 \, m/s$ and the initial vertical velocity is $u_y = 48 \, m/s$.
The angle of projection $\theta$ is given by $\tan \theta = \frac{u_y}{u_x} = \frac{48}{36} = \frac{4}{3}$.
Using the trigonometric identity for 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,$\theta = \sin^{-1}\left(\frac{4}{5}\right)$.

(B) For a given velocity $u$,the range $R$ is the same for two angles of projection,$\theta$ and $(90^\circ - \theta)$.
The time of flight for angle $\theta$ is $t_1 = \frac{2u \sin \theta}{g}$.
The time of flight for angle $(90^\circ - \theta)$ is $t_2 = \frac{2u \sin(90^\circ - \theta)}{g} = \frac{2u \cos \theta}{g}$.
Multiplying the two times of flight:
$t_1 t_2 = \left( \frac{2u \sin \theta}{g} \right) \left( \frac{2u \cos \theta}{g} \right) = \frac{4u^2 \sin \theta \cos \theta}{g^2}$.
Using the identity $\sin 2\theta = 2 \sin \theta \cos \theta$,we get:
$t_1 t_2 = \frac{2u^2 (2 \sin \theta \cos \theta)}{g^2} = \frac{2(u^2 \sin 2\theta)}{g^2}$.
Since the range $R = \frac{u^2 \sin 2\theta}{g}$,we can substitute this into the equation:
$t_1 t_2 = \frac{2R}{g}$.
Since $g$ is constant,we conclude that $t_1 t_2 \propto R$.

(C) The initial velocity $v$ is resolved into two components:
Horizontal component: $v_x = v\cos\theta$
Vertical component: $v_y = v\sin\theta$
After $t$ seconds,the horizontal component remains constant: $v_x = v\cos\theta$
The vertical component changes due to gravity: $v_y = v\sin\theta - gt$
The magnitude of the resultant velocity $v_t$ is given by:
$v_t = \sqrt{v_x^2 + v_y^2}$
$v_t = \sqrt{(v\cos\theta)^2 + (v\sin\theta - gt)^2}$
$v_t = \sqrt{v^2\cos^2\theta + v^2\sin^2\theta + g^2t^2 - 2vgt\sin\theta}$
Since $\cos^2\theta + \sin^2\theta = 1$,we get:
$v_t = \sqrt{v^2 + g^2t^2 - 2vgt\sin\theta}$

(D) The maximum horizontal range $R_{max}$ for a projectile is given by $R_{max} = \frac{u^2}{g} = 100 \, m$.
When the ball is thrown vertically upwards,the maximum height $H$ attained is given by $H = \frac{u^2}{2g}$.
Substituting the value of $u^2$ from the range formula $(u^2 = 100g)$:
$H = \frac{100g}{2g} = \frac{100}{2} = 50 \, m$.

(C) The maximum horizontal range of a projectile is given by the formula $R_{\text{max}} = \frac{u^2}{g}$.
Given that $R_{\text{max}} = 100\, m$ and taking the acceleration due to gravity $g = 9.8\, m/s^2$ (or $10\, m/s^2$ for approximation).
Using $g = 9.8\, m/s^2$:
$100 = \frac{u^2}{9.8}$
$u^2 = 980$
$u = \sqrt{980} \approx 31.3\, m/s$.
Rounding to the nearest integer,we get $u = 31\, m/s$.
Using $g = 10\, m/s^2$:
$100 = \frac{u^2}{10}$
$u^2 = 1000$
$u = \sqrt{1000} \approx 31.62\, m/s$.
Rounding to the nearest integer,we get $u = 32\, m/s$.
Given the options provided,the correct answer is $32\, m/s$.

(A) The time of flight $(T)$ of a projectile is the total time for which the projectile remains in the air.
It is given by the formula $T = \frac{2u \sin \theta}{g}$.
Since the vertical component of the initial velocity is $u_y = u \sin \theta$,we can rewrite the formula as $T = \frac{2u_y}{g}$.
Therefore,the time of flight is determined by the vertical component of the initial velocity,$U_{\text{vertical}}$.

(A) The person can catch the ball if their constant speed is equal to the horizontal component of the ball's velocity,as the ball will land at a distance determined by its horizontal motion.
The horizontal component of the ball's velocity is $v_x = v_0 \cos \theta$.
The person's speed is $v_p = v_0/2$.
For the person to catch the ball at the point of impact,their speeds must match: $v_0/2 = v_0 \cos \theta$.
Dividing both sides by $v_0$,we get: $\cos \theta = 1/2$.
Therefore,$\theta = \cos^{-1}(1/2) = 60^\circ$.

(B) 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 $T = \frac{2u \sin \theta}{g}$.
From the expression for $H$,we can write $u^2 \sin^2 \theta = 2gH$,which implies $u \sin \theta = \sqrt{2gH}$.
Substituting this value of $u \sin \theta$ into the expression for $T$:
$T = \frac{2}{g} \times \sqrt{2gH} = \frac{2 \sqrt{2} \sqrt{g} \sqrt{H}}{g} = 2 \sqrt{\frac{2H}{g}}$.
Thus,the correct option is $B$.

(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 $H = \frac{u^2 \sin^2 \theta}{2g}$.
Dividing $R$ by $H$,we get $\frac{R}{H} = \frac{2u^2 \sin \theta \cos \theta}{g} \times \frac{2g}{u^2 \sin^2 \theta} = 4 \cot \theta$.
Given that $R = 4\sqrt{3} H$,we have $\frac{R}{H} = 4\sqrt{3}$.
Equating the two expressions: $4 \cot \theta = 4\sqrt{3}$.
Therefore,$\cot \theta = \sqrt{3}$.
Since $\cot 30^\circ = \sqrt{3}$,the angle of projection $\theta = 30^\circ$.

(C) At the topmost point of the projectile motion,the vertical component of velocity becomes zero,and only the horizontal component of velocity remains. The acceleration due to gravity $(g)$ always acts vertically downward. Since the horizontal velocity is perpendicular to the vertical acceleration,the velocity and acceleration are perpendicular to each other at the topmost point.

(D) The horizontal range $R$ of a projectile is given by $R = \frac{u^2 \sin(2\theta)}{g}$.
The maximum height $H$ reached by the projectile is given by $H = \frac{u^2 \sin^2 \theta}{2g}$.
Taking the ratio of $R$ to $H$:
$\frac{R}{H} = \frac{u^2 \sin(2\theta) / g}{u^2 \sin^2 \theta / (2g)} = \frac{2 \sin \theta \cos \theta}{\sin^2 \theta / 2} = \frac{4 \cos \theta}{\sin \theta} = 4 \cot \theta$.
Given the angle of projection $\theta = 45^\circ$,we have:
$\frac{R}{H} = 4 \cot(45^\circ) = 4(1) = 4$.
Therefore,the ratio $R:H$ is $4:1$.

(B) The maximum horizontal range is given by $R_{\max} = \frac{u^2}{g} = 400\, m$ (at $\theta = 45^\circ$).
The maximum height attained by a projectile is given by $H_{\max} = \frac{u^2 \sin^2 \theta}{2g}$.
For the maximum height,the projectile must be projected at $\theta = 90^\circ$,so $H_{\max} = \frac{u^2}{2g}$.
Substituting $\frac{u^2}{g} = 400\, m$ into the equation:
$H_{\max} = \frac{1}{2} \times \left( \frac{u^2}{g} \right) = \frac{1}{2} \times 400 = 200\, m$.

(D) The time of flight $T$ for a projectile is given by $T = \frac{2u_y}{g}$,where $u_y$ is the initial vertical velocity component. Since all four paths reach the same maximum height $H = \frac{u_y^2}{2g}$,they all have the same initial vertical velocity component $u_y$ and thus the same time of flight $T$.
The horizontal range $R$ is given by $R = u_x T$,where $u_x$ is the initial horizontal velocity component.
Since $T$ is constant for all paths,the range $R$ is directly proportional to the initial horizontal velocity component $u_x$ $(R \propto u_x)$.
From the figure,the ranges are $R_4 > R_3 > R_2 > R_1$. Therefore,the initial horizontal velocity components are ranked as $4, 3, 2, 1$.

(B) When air resistance is considered,it acts as a resistive force opposite to the direction of motion of the projectile.
This resistive force continuously does negative work on the projectile,leading to a loss of mechanical energy.
As a result,both the horizontal velocity and the vertical velocity components decrease over time compared to the ideal case.
Consequently,the projectile will reach a lower maximum height and cover a shorter horizontal range than it would in a vacuum.
Looking at the provided figure,the dotted line represents the ideal path. Path $A$ shows a trajectory with a smaller maximum height and a shorter range compared to the dotted line. Therefore,path $A$ is the appropriate path when air resistance is taken into account.

(D) The formula for the horizontal range of a projectile is given by $R = \frac{u^2 \sin 2\theta}{g}$.
Given that the velocity of projection $u$ and the angle of projection $\theta$ are constant,the range $R$ is inversely proportional to the acceleration due to gravity $g$,i.e.,$R \propto \frac{1}{g}$.
Let $R_e$ and $g_e$ be the range and acceleration due to gravity on Earth,and $R_m$ and $g_m$ be the range and acceleration due to gravity on the Moon.
We are given $g_m = 0.2 \, g_e$.
Using the proportionality $R_m \, g_m = R_e \, g_e$,we get:
$R_m = R_e \left( \frac{g_e}{g_m} \right)$
$R_m = R_e \left( \frac{g_e}{0.2 \, g_e} \right)$
$R_m = \frac{R_e}{0.2} = 5 \, R_e$.
Therefore,the maximum range on the surface of the moon is $5 \, R_e$.

(B) The initial kinetic energy is given by $K = \frac{1}{2}mu^2$, where $u$ is the initial velocity.
At the highest point of the trajectory, the vertical component of velocity becomes zero, and the particle only possesses the horizontal component of velocity, which is $v_x = u \cos \theta$.
The kinetic energy at the highest point $K'$ is given by $K' = \frac{1}{2}m(u \cos \theta)^2 = \frac{1}{2}mu^2 \cos^2 \theta$.
Substituting $\theta = 45^{\circ}$, we get $K' = K \cos^2(45^{\circ}) = K \times (1/\sqrt{2})^2 = K/2$.

(C) Let the initial kinetic energy of the projectile be $E_0 = \frac{1}{2}mv_0^2$,where $v_0$ is the initial velocity.
At maximum height,the potential energy is $U = \frac{3}{4}E_0$.
By the law of conservation of energy,the total energy remains constant. The kinetic energy at maximum height is $K = E_0 - U = E_0 - \frac{3}{4}E_0 = \frac{1}{4}E_0$.
At maximum height,the vertical component of velocity is zero,so the kinetic energy is due to the horizontal component $v_x = v_0 \cos \theta$.
Thus,$K = \frac{1}{2}m(v_0 \cos \theta)^2 = \frac{1}{2}mv_0^2 \cos^2 \theta = E_0 \cos^2 \theta$.
Equating the two expressions for $K$: $E_0 \cos^2 \theta = \frac{1}{4}E_0$.
$\cos^2 \theta = \frac{1}{4} \implies \cos \theta = \frac{1}{2}$.
Therefore,$\theta = 60^o$.

(C) The standard equation of a projectile trajectory is given by $y = x \tan \theta - \frac{gx^2}{2u^2 \cos^2 \theta}$.
Comparing this with the given equation $y = \sqrt{3} x - \frac{gx^2}{2}$,we can identify the term corresponding to $\tan \theta$.
Here,$\tan \theta = \sqrt{3}$.
Since $\tan 60^\circ = \sqrt{3}$,the angle of projection $\theta$ is $60^\circ$.

(B) The standard equation of a projectile is given by $y = x \tan \theta - \frac{gx^2}{2u^2 \cos^2 \theta}$.
Comparing this with the given equation $y = \sqrt{3}x - \frac{x^2}{2}$,we get:
$\tan \theta = \sqrt{3} \implies \theta = 60^\circ$.
Also,$\frac{g}{2u^2 \cos^2 \theta} = \frac{1}{2}$.
Taking $g = 10 \, m/s^2$ and $\theta = 60^\circ$,we have $\cos 60^\circ = 1/2$,so $\cos^2 60^\circ = 1/4$.
Substituting these values: $\frac{10}{2u^2(1/4)} = \frac{1}{2}$.
$\frac{10}{u^2/2} = \frac{1}{2} \implies \frac{20}{u^2} = \frac{1}{2}$.
$u^2 = 40 \implies u = \sqrt{40} = 2\sqrt{10} \, m/s$.

(D) The standard equation of a projectile trajectory is given by $y = x \tan \theta \left( 1 - \frac{x}{R} \right)$,where $R$ is the horizontal range.
Given equation: $y = 16x - \frac{5x^2}{4}$.
Factor out $16x$ from the expression: $y = 16x \left( 1 - \frac{5x^2}{4 \cdot 16x} \right) = 16x \left( 1 - \frac{5x}{64} \right)$.
Rewrite the term inside the bracket to match the standard form: $y = 16x \left( 1 - \frac{x}{64/5} \right)$.
Comparing this with the standard equation $y = x \tan \theta \left( 1 - \frac{x}{R} \right)$,we get $R = \frac{64}{5}$.
Calculating the value: $R = 12.8 \ m$.

(A) The horizontal component of velocity remains constant throughout the motion because there is no acceleration in the $x$-direction.
$v_{x} = u \cos \theta = 30 \cos 30^{\circ} = 30 \times \frac{\sqrt{3}}{2} = 15\sqrt{3}\; m/s$.
The vertical component of velocity after time $t$ is given by $v_{y} = u \sin \theta - gt$.
Substituting the values: $v_{y} = 30 \sin 30^{\circ} - 10 \times 1 = 30 \times \frac{1}{2} - 10 = 15 - 10 = 5\; m/s$.
The magnitude of the resultant velocity $v$ is given by $v = \sqrt{v_{x}^{2} + v_{y}^{2}}$.
$v = \sqrt{(15\sqrt{3})^{2} + (5)^{2}} = \sqrt{225 \times 3 + 25} = \sqrt{675 + 25} = \sqrt{700}$.
$v = 10\sqrt{7}\; m/s$.

(B) Given the vertical component of velocity $u_y = u \sin \theta = 80 \; ms^{-1}$ and $\theta = 30^{\circ}$.
Since $\sin 30^{\circ} = 0.5$,we have $u = \frac{80}{0.5} = 160 \; ms^{-1}$.
The horizontal component of velocity is $u_x = u \cos \theta = 160 \cos 30^{\circ} = 160 \times \frac{\sqrt{3}}{2} = 80\sqrt{3} \; ms^{-1}$.
At time $t = \frac{T}{2}$,the projectile is at its maximum height,where the vertical component of velocity $v_y = 0$.
Therefore,the resultant velocity at $t = \frac{T}{2}$ is equal to the horizontal component $u_x$,which remains constant throughout the motion.
Thus,$v = u_x = 80\sqrt{3} \; ms^{-1}$.

(B) For the particles to collide,they must reach the same point at the same time.
Particle $P$ reaches its maximum height at the horizontal distance equal to half of its range.
At the maximum height,the vertical component of the velocity of particle $P$ becomes zero.
However,for the particles to collide at the peak,the vertical velocity of particle $Q$ must match the vertical velocity of particle $P$ at the point of collision.
Since particle $Q$ is projected from the point directly below the maximum height of $P$,it will reach the maximum height of $P$ at the same time $P$ reaches it if their vertical velocities are equal at that point.
The vertical component of the initial velocity of $P$ is $u_{1y} = u_1 \sin 30^\circ$.
For collision at the maximum height,the vertical velocity of $Q$ must be equal to the vertical component of $P$'s velocity at the start,or more precisely,the time taken to reach the peak must be the same.
Since $v_y = u_y - gt$,at the peak $v_y = 0$,so $t = \frac{u_y}{g}$.
For $P$,$t_P = \frac{u_1 \sin 30^\circ}{g}$.
For $Q$,$t_Q = \frac{u_2}{g}$.
Equating $t_P = t_Q$,we get $u_1 \sin 30^\circ = u_2$.
$u_1 \times \frac{1}{2} = u_2$.
Therefore,$u_1 = 2u_2$.

(C) The initial velocity of the particle is $\vec{u} = u \cos \theta \hat{i} + u \sin \theta \hat{j}$.
At the maximum height,the vertical component of the velocity becomes zero,so the velocity is $\vec{v} = u \cos \theta \hat{i}$.
The change in velocity is $\Delta \vec{v} = \vec{v} - \vec{u}$.
$\Delta \vec{v} = (u \cos \theta \hat{i}) - (u \cos \theta \hat{i} + u \sin \theta \hat{j}) = -u \sin \theta \hat{j}$.
The magnitude of the change in velocity is $|\Delta \vec{v}| = u \sin \theta$.

(B) Let the initial velocity be $\vec{u} = u \cos \alpha \hat{i} + u \sin \alpha \hat{j}$.
At point $P$,the velocity $\vec{v}$ is perpendicular to $\vec{u}$,so $\vec{v} \cdot \vec{u} = 0$.
The velocity at any time $t$ is $\vec{v} = u \cos \alpha \hat{i} + (u \sin \alpha - gt) \hat{j}$.
Taking the dot product: $(u \cos \alpha \hat{i} + (u \sin \alpha - gt) \hat{j}) \cdot (u \cos \alpha \hat{i} + u \sin \alpha \hat{j}) = 0$.
$u^2 \cos^2 \alpha + u \sin \alpha (u \sin \alpha - gt) = 0$.
$u^2 \cos^2 \alpha + u^2 \sin^2 \alpha - u \sin \alpha gt = 0$.
$u^2 (\cos^2 \alpha + \sin^2 \alpha) = u \sin \alpha gt$.
$u^2 = u \sin \alpha gt$.
$t = \frac{u}{g \sin \alpha} = \frac{u \csc \alpha}{g}$.

(A) The time of flight $T$ for a projectile is given by $T = \frac{2u \sin \theta}{g}$,where $\theta$ is the angle of projection with the horizontal.
For the first particle,the angle with the horizontal is $\theta_1 = \alpha$. Thus,$T_1 = \frac{2u \sin \alpha}{g}$.
For the second particle,the angle with the vertical is $\alpha$,so the angle with the horizontal is $\theta_2 = 90^\circ - \alpha$. Thus,$T_2 = \frac{2u \sin(90^\circ - \alpha)}{g} = \frac{2u \cos \alpha}{g}$.
The ratio of their times of flight is $\frac{T_1}{T_2} = \frac{\frac{2u \sin \alpha}{g}}{\frac{2u \cos \alpha}{g}} = \frac{\sin \alpha}{\cos \alpha} = \tan \alpha$.
Therefore,the ratio is $\tan \alpha : 1$.

(A) The ball is projected from a height $h = 10 \ m$ and returns to the same vertical level ($10 \ m$ height). This is equivalent to the horizontal range of a projectile on a level plane.
The formula for the horizontal range $R$ is given by:
$R = \frac{u^2 \sin(2\theta)}{g}$
Given:
Initial velocity $u = 10 \ m/s$
Angle of projection $\theta = 30^\circ$
Acceleration due to gravity $g = 10 \ m/s^2$
Substituting the values:
$R = \frac{(10)^2 \sin(2 \times 30^\circ)}{10}$
$R = \frac{100 \times \sin(60^\circ)}{10}$
$R = 10 \times \frac{\sqrt{3}}{2}$
$R = 5\sqrt{3} \ m$
Using $\sqrt{3} \approx 1.732$:
$R = 5 \times 1.732 = 8.66 \ m$

(A) The horizontal range $R$ of a projectile is given by the formula: $R = \frac{u^2 \sin(2\theta)}{g}$.
Since the initial velocity $u$ and acceleration due to gravity $g$ are constant for all objects,the range $R$ depends only on $\sin(2\theta)$.
For the given angles:
For $P$ $(\theta = 15^o)$: $2\theta = 30^o$,$\sin(30^o) = 0.5$
For $Q$ $(\theta = 30^o)$: $2\theta = 60^o$,$\sin(60^o) \approx 0.866$
For $R$ $(\theta = 45^o)$: $2\theta = 90^o$,$\sin(90^o) = 1.0$
For $S$ $(\theta = 60^o)$: $2\theta = 120^o$,$\sin(120^o) = \sin(60^o) \approx 0.866$
Comparing the values,$\sin(2\theta)$ is minimum for $\theta = 15^o$ (which is $P$) and $\theta = 75^o$ (not given). Among the choices,$P$ has the smallest value of $\sin(2\theta)$,therefore $P$ has the minimum range.

(C) For maximum range,the angle of projection is $\theta = 45^\circ$.
At the highest point of the trajectory,the vertical component of velocity becomes zero,and the velocity is minimum,equal to $u \cos \theta$.
The coordinates of the highest point are $(x, y) = (R/2, H)$,where $R$ is the range and $H$ is the maximum height.
The formula for maximum height is $H = \frac{u^2 \sin^2 \theta}{2g}$ and for range is $R = \frac{u^2 \sin 2\theta}{g}$.
For $\theta = 45^\circ$,$H = \frac{u^2 (1/\sqrt{2})^2}{2g} = \frac{u^2}{4g}$ and $R = \frac{u^2 \sin 90^\circ}{g} = \frac{u^2}{g}$.
Thus,$H = R/4$.
The coordinates of the highest point are $(R/2, R/4)$.

(A) Let the initial velocity be $u$ and the angle of projection be $\theta$.
At maximum height,the vertical component of velocity is $0$,so the velocity is only the horizontal component,$v_x = u \cos \theta$.
Given that $v_x = \frac{u}{\sqrt{2}}$,we have $u \cos \theta = \frac{u}{\sqrt{2}}$.
This implies $\cos \theta = \frac{1}{\sqrt{2}}$,so $\theta = 45^o$.
The formula for the range $R$ of a projectile is $R = \frac{u^2 \sin(2\theta)}{g}$.
Substituting $\theta = 45^o$,we get $R = \frac{u^2 \sin(2 \times 45^o)}{g} = \frac{u^2 \sin(90^o)}{g}$.
Since $\sin(90^o) = 1$,the range is $R = \frac{u^2}{g}$.

(B) Given velocity of the projectile is $\vec{u} = (6\hat{i} + 8\hat{j}) \ m/s$.
Here,the horizontal component of velocity is $u_x = 6 \ m/s$ and the vertical component is $u_y = 8 \ m/s$.
The time of flight $T$ is given by $T = \frac{2u_y}{g} = \frac{2 \times 8}{10} = 1.6 \ s$.
The horizontal range $R$ is given by $R = u_x \times T$.
Substituting the values,$R = 6 \times 1.6 = 9.6 \ m$.
Therefore,the horizontal range is $9.6 \ m$.

(C) The formula for the horizontal range $R$ of a projectile is given by $R = \frac{u^2 \sin(2\theta)}{g}$.
From this,we can see that $R \propto u^2 \sin(2\theta)$.
Let the initial range be $R_1 = R$,initial velocity be $u_1 = u$,and initial angle be $\theta_1 = 15^o$.
Let the final range be $R_2$,final velocity be $u_2 = 2u$,and final angle be $\theta_2 = 45^o$.
Taking the ratio:
$\frac{R_2}{R_1} = \left( \frac{u_2}{u_1} \right)^2 \left( \frac{\sin(2\theta_2)}{\sin(2\theta_1)} \right)$
Substituting the values:
$\frac{R_2}{R} = \left( \frac{2u}{u} \right)^2 \left( \frac{\sin(2 \times 45^o)}{\sin(2 \times 15^o)} \right)$
$\frac{R_2}{R} = (2)^2 \left( \frac{\sin(90^o)}{\sin(30^o)} \right)$
$\frac{R_2}{R} = 4 \left( \frac{1}{0.5} \right) = 4 \times 2 = 8$
Therefore,$R_2 = 8R$.

(A) At the maximum height,the vertical component of velocity is zero,so the velocity of the projectile is equal to its horizontal component,$v_x = u \cos \theta$.
Given that this velocity is half of the initial velocity $u$,we have $u \cos \theta = \frac{u}{2}$.
This implies $\cos \theta = \frac{1}{2}$,so the angle of projection is $\theta = 60^\circ$.
The horizontal range $R$ of a projectile is given by the formula $R = \frac{u^2 \sin(2\theta)}{g}$.
Substituting $\theta = 60^\circ$,we get $R = \frac{u^2 \sin(2 \times 60^\circ)}{g} = \frac{u^2 \sin(120^\circ)}{g}$.
Since $\sin(120^\circ) = \frac{\sqrt{3}}{2}$,the range is $R = \frac{\sqrt{3} u^2}{2g}$.

(D) The horizontal velocity is given as $u_x = 9.8 \; m/s$ and the vertical velocity is $u_y = 19.6 \; m/s$.
The time of flight $T$ is given by $T = \frac{2u_y}{g} = \frac{2 \times 19.6}{9.8} = 4 \; s$.
The horizontal range $R$ is given by $R = u_x \times T$.
Substituting the values,$R = 9.8 \times 4 = 39.2 \; m$.
Alternatively,using the formula $R = \frac{2u_x u_y}{g} = \frac{2 \times 9.8 \times 19.6}{9.8} = 39.2 \; m$.

(A) The formula for the range $R$ is $R = \frac{u^2 \sin(2\theta)}{g} = \frac{2u^2 \sin \theta \cos \theta}{g}$.
The formula for maximum height $H$ is $H = \frac{u^2 \sin^2 \theta}{2g}$.
Given that the range is twice the maximum height,$R = 2H$.
Substituting the formulas: $\frac{2u^2 \sin \theta \cos \theta}{g} = 2 \left( \frac{u^2 \sin^2 \theta}{2g} \right)$.
Simplifying,we get $2 \sin \theta \cos \theta = \sin^2 \theta$,which implies $\tan \theta = 2$.
If $\tan \theta = 2$,then $\sin \theta = \frac{2}{\sqrt{5}}$ and $\cos \theta = \frac{1}{\sqrt{5}}$.
Substituting these into the range formula: $R = \frac{2u^2}{g} \left( \frac{2}{\sqrt{5}} \right) \left( \frac{1}{\sqrt{5}} \right) = \frac{4u^2}{5g}$.

(C) For maximum horizontal range,the angle of projection is $\theta = 45^\circ$.
The maximum range is given by $R_{\max} = \frac{u^2}{g} = 1.6 \; m$.
Taking $g = 10 \; m/s^2$,we have $u^2 = 1.6 \times 10 = 16$,so $u = 4 \; m/s$.
The horizontal component of velocity is $v_x = u \cos \theta = 4 \cos 45^\circ = 4 \times \frac{1}{\sqrt{2}} = 2\sqrt{2} \; m/s$.
Since the time spent on the ground is negligible,the grasshopper is effectively in the air for the entire duration of $10 \; s$.
The total horizontal distance covered in time $t = 10 \; s$ is $S = v_x \times t$.
$S = 2\sqrt{2} \times 10 = 20\sqrt{2} \; m$.

(C) The initial velocity components are $v_x = a$ and $v_y = b$.
The angle of projection $\theta$ is given by $\tan \theta = \frac{v_y}{v_x} = \frac{b}{a}$.
We know the formulas for range $R = \frac{2v_x v_y}{g} = \frac{2ab}{g}$ and maximum height $H = \frac{v_y^2}{2g} = \frac{b^2}{2g}$.
Given the condition $R = 2H$,we substitute the expressions:
$\frac{2ab}{g} = 2 \left( \frac{b^2}{2g} \right)$.
Simplifying the equation:
$\frac{2ab}{g} = \frac{b^2}{g}$.
Dividing both sides by $b/g$ (assuming $b \neq 0$):
$2a = b$ or $b = 2a$.

(D) The maximum horizontal range is given by $R_{\max} = \frac{u^2}{g} = 100\,m$ (at $\theta = 45^\circ$).
Therefore,$u^2 = 100 \times g = 1000\,m^2/s^2$ (taking $g = 10\,m/s^2$).
The maximum height reached when thrown vertically is $H_{\max} = \frac{u^2}{2g}$ (at $\theta = 90^\circ$).
Substituting the value of $u^2$,we get $H_{\max} = \frac{1000}{2 \times 10} = 50\,m$.

(C) The time of flight $T$ is given by $T = \frac{2u \sin \theta}{g} = 2 \, s$.
From this,we get $u \sin \theta = \frac{g \times T}{2} = \frac{10 \times 2}{2} = 10 \, m/s$.
The maximum height $H$ attained by the projectile is given by the formula $H = \frac{u^2 \sin^2 \theta}{2g}$.
Substituting the value of $u \sin \theta = 10 \, m/s$:
$H = \frac{(10)^2}{2 \times 10} = \frac{100}{20} = 5 \, m$.
Therefore,the maximum height attained by the ball is $5 \, m$.

(D) The formula for the maximum height of a projectile is given by $H = \frac{u^2 \sin^2 \theta}{2g}$.
Given the initial velocity $u$ is the same for both cases.
For the first projectile,$\theta_1 = \pi/3 = 60^\circ$,so $H_1 = Y = \frac{u^2 \sin^2(60^\circ)}{2g} = \frac{u^2}{2g} \left( \frac{\sqrt{3}}{2} \right)^2 = \frac{3u^2}{8g}$.
For the second projectile,$\theta_2 = \pi/6 = 30^\circ$,so $H_2 = \frac{u^2 \sin^2(30^\circ)}{2g} = \frac{u^2}{2g} \left( \frac{1}{2} \right)^2 = \frac{u^2}{8g}$.
Comparing the two heights: $\frac{H_2}{H_1} = \frac{u^2/8g}{3u^2/8g} = \frac{1}{3}$.
Therefore,$H_2 = \frac{H_1}{3} = \frac{Y}{3}$.

(A) The maximum horizontal range is given by $R_{\max} = \frac{u^2}{g} = 400\;m$.
The maximum height attained by a projectile is given by $H = \frac{u^2 \sin^2 \theta}{2g}$.
For maximum horizontal range,the angle of projection is $\theta = 45^\circ$.
Substituting $\theta = 45^\circ$ into the height formula:
$H = \frac{u^2 \sin^2 45^\circ}{2g} = \frac{u^2 (1/\sqrt{2})^2}{2g} = \frac{u^2}{2g \times 2} = \frac{u^2}{4g}$.
Since $R_{\max} = \frac{u^2}{g} = 400\;m$,we have:
$H = \frac{1}{4} \times R_{\max} = \frac{1}{4} \times 400\;m = 100\;m$.

(D) The maximum horizontal range $R_{\max}$ is given by the formula $R_{\max} = \frac{u^2}{g} = 80 \, m$,where $u$ is the initial velocity and $g$ is the acceleration due to gravity.
The maximum height $H_{\max}$ reached when thrown vertically upwards is given by $H_{\max} = \frac{u^2}{2g}$.
By substituting the value of $R_{\max}$ into the expression for $H_{\max}$,we get:
$H_{\max} = \frac{1}{2} \left( \frac{u^2}{g} \right) = \frac{1}{2} \times 80 \, m = 40 \, m$.
Therefore,the maximum height to which the ball can be thrown is $40 \, m$.

(C) For a projectile,the range is the same for complementary angles $\theta$ and $90^o - \theta$.
The maximum heights attained for these angles are given by:
$y_1 = \frac{u^2 \sin^2 \theta}{2g}$
$y_2 = \frac{u^2 \sin^2(90^o - \theta)}{2g} = \frac{u^2 \cos^2 \theta}{2g}$
Adding the two heights:
$y_1 + y_2 = \frac{u^2 \sin^2 \theta}{2g} + \frac{u^2 \cos^2 \theta}{2g}$
$y_1 + y_2 = \frac{u^2}{2g} (\sin^2 \theta + \cos^2 \theta)$
Since $\sin^2 \theta + \cos^2 \theta = 1$,we get:
$y_1 + y_2 = \frac{u^2}{2g}$

(B) The kinetic energy of a projectile is given by $K = \frac{1}{2} m v^2$.
Since the horizontal component of velocity $v_x = u \cos \theta$ remains constant throughout the motion,the kinetic energy is minimum when the vertical component of velocity $v_y$ is zero.
This occurs at the highest point of the trajectory.
The horizontal distance covered to reach the highest point is half of the total horizontal range $R$.
Therefore,the horizontal distance is $R/2$ or $0.5 \, R$.

(D) The kinetic energy of a projectile at its maximum height is given by $KE = \frac{1}{2} m (u \cos \theta)^2$,where $u$ is the initial velocity and $\theta$ is the angle of projection.
For a ball thrown vertically upwards,the angle of projection $\theta = 90^o$.
At the maximum height,the vertical component of velocity becomes zero.
Since $\cos 90^o = 0$,the kinetic energy at the maximum height for a ball thrown vertically is $KE = \frac{1}{2} m (u \cos 90^o)^2 = 0$.
Therefore,the kinetic energy of the first ball at its maximum height is $0$.

(D) The initial kinetic energy $K$ is given by $K = \frac{1}{2}mv^2$.
At the maximum height,the vertical component of velocity is zero,so the velocity is only the horizontal component $v_x = v \cos \theta$.
The kinetic energy at the maximum height $K'$ is given by $K' = \frac{1}{2}m(v \cos \theta)^2 = K \cos^2 \theta$.
Given $K = 100 \, J$ and $K' = 30 \, J$,we have:
$30 = 100 \cos^2 \theta$
$\cos^2 \theta = \frac{30}{100} = \frac{3}{10}$
$\cos \theta = \sqrt{\frac{3}{10}}$
$\theta = \cos^{-1} \left( \sqrt{\frac{3}{10}} \right)$.

(C) The equation of the trajectory of a projectile launched horizontally from the top of a cliff is given by $y = \frac{g x^2}{2 u^2}$.
For the ball to hit the edge of the $n^{th}$ step,the horizontal distance covered is $x = n b$ and the vertical distance covered is $y = n h$.
Substituting these values into the trajectory equation:
$n h = \frac{g (n b)^2}{2 u^2}$
$n h = \frac{g n^2 b^2}{2 u^2}$
Dividing both sides by $n$ (assuming $n \neq 0$):
$h = \frac{g n b^2}{2 u^2}$
Solving for $n$:
$n = \frac{2 h u^2}{g b^2}$