Mix Examples-Motion in Plane Questions in English

(A) Given: Linear speed $v = 72 \ km/h = 72 \times \frac{5}{18} = 20 \ m/s$.
Radius $r = 0.25 \ m$.
Initial angular velocity $\omega_0 = \frac{v}{r} = \frac{20}{0.25} = 80 \ rad \ s^{-1}$.
Final angular velocity $\omega = 0 \ rad \ s^{-1}$.
Number of rotations $n = 20$.
Total angular displacement $\theta = 2\pi n = 2 \times \pi \times 20 = 40\pi \ rad$.
Using the kinematic equation $\omega^2 = \omega_0^2 + 2\alpha\theta$:
$0^2 = (80)^2 + 2 \times \alpha \times (40\pi)$
$0 = 6400 + 80\pi\alpha$
$80\pi\alpha = -6400$
$\alpha = -\frac{6400}{80\pi} = -\frac{80}{\pi} \approx -\frac{80}{3.1416} \approx -25.46 \ rad \ s^{-2}$.
Rounding to the nearest value,we get $-25.5 \ rad \ s^{-2}$.

(C) Given: Mass $m = 2 \; kg$,initial velocity $u = 3 \; m/s$ along $OE$,force $F = 4 \; N$ along $OF$,time $t = 4 \; s$.
Along $OE$ (x-direction),there is no force,so acceleration $a_x = 0$. The displacement $x$ is:
$x = u_x t = 3 \times 4 = 12 \; m$
Along $OF$ (y-direction),the initial velocity $u_y = 0$. The acceleration $a_y$ is:
$a_y = \frac{F}{m} = \frac{4}{2} = 2 \; m/s^2$
The displacement $y$ is:
$y = u_y t + \frac{1}{2} a_y t^2 = 0 + \frac{1}{2} \times 2 \times (4)^2 = 16 \; m$
The total displacement $d$ from $O$ is:
$d = \sqrt{x^2 + y^2} = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20 \; m$

(B) Let the initial velocity be $\vec{u} = u \cos \alpha \hat{i} + u \sin \alpha \hat{j}$.
Let the velocity at point $P$ be $\vec{v} = v_x \hat{i} + v_y \hat{j}$.
Since horizontal velocity remains constant,$v_x = u \cos \alpha$.
Given that the velocity $\vec{v}$ is perpendicular to the initial velocity $\vec{u}$,their dot product must be zero:
$\vec{u} \cdot \vec{v} = 0$
$(u \cos \alpha \hat{i} + u \sin \alpha \hat{j}) \cdot (u \cos \alpha \hat{i} + v_y \hat{j}) = 0$
$u^2 \cos^2 \alpha + u \sin \alpha v_y = 0$
$v_y = -\frac{u^2 \cos^2 \alpha}{u \sin \alpha} = -u \frac{\cos^2 \alpha}{\sin \alpha}$
The magnitude of the velocity $v$ is given by $v = \sqrt{v_x^2 + v_y^2}$.
$v = \sqrt{(u \cos \alpha)^2 + (-u \frac{\cos^2 \alpha}{\sin \alpha})^2}$
$v = u \cos \alpha \sqrt{1 + \frac{\cos^2 \alpha}{\sin^2 \alpha}} = u \cos \alpha \sqrt{\frac{\sin^2 \alpha + \cos^2 \alpha}{\sin^2 \alpha}}$
$v = u \cos \alpha \cdot \frac{1}{\sin \alpha} = u \cot \alpha$.

(B) Let the initial velocity be $u$ and the angle of projection be $\theta$.
At the highest point,the vertical component of velocity is zero. The particle reaches the highest point $1 \, s$ after the $2 \, s$ mark,so the total time to reach the highest point is $T = 2 + 1 = 3 \, s$.
Using $T = \frac{u \sin \theta}{g} = 3 \, s$,we get $u \sin \theta = 3g = 30 \, m/s$ ... $(i)$
At $t = 2 \, s$,the velocity makes an angle of $30^\circ$ with the horizontal. The horizontal component of velocity remains constant: $v_x = u \cos \theta$.
The vertical component at $t = 2 \, s$ is $v_y = u \sin \theta - g(2) = 30 - 20 = 10 \, m/s$.
Since $\tan 30^\circ = \frac{v_y}{v_x}$,we have $\frac{1}{\sqrt{3}} = \frac{10}{u \cos \theta}$,so $u \cos \theta = 10\sqrt{3} \, m/s$ ... (ii)
Dividing $(i)$ by (ii): $\tan \theta = \frac{30}{10\sqrt{3}} = \sqrt{3}$,so $\theta = 60^\circ$.
Substituting $\theta = 60^\circ$ in $(i)$: $u \sin 60^\circ = 30 \implies u(\frac{\sqrt{3}}{2}) = 30 \implies u = \frac{60}{\sqrt{3}} = 20\sqrt{3} \, m/s$.

(D) Let $v$ be the velocity of the particle at the top of the incline. The particle performs projectile motion to cross the $40 \, m$ wide well at an angle of $45^o$. The range $R$ of the projectile is given by $R = \frac{v^2 \sin(2\theta)}{g}$.
Given $R = 40 \, m$ and $\theta = 45^o$,we have $40 = \frac{v^2 \sin(90^o)}{g} = \frac{v^2}{g}$.
Assuming $g = 10 \, m/s^2$,we get $v^2 = 400$,so $v = 20 \, m/s$.
Now,consider the motion on the frictionless incline of length $s = 20\sqrt{2} \, m$. The acceleration along the incline is $a = -g \sin(45^o) = -\frac{g}{\sqrt{2}}$.
Using the equation of motion $v^2 = u^2 + 2as$,where $u$ is the initial velocity at point $M$:
$(20)^2 = u^2 + 2(-\frac{g}{\sqrt{2}})(20\sqrt{2})$.
$400 = u^2 - 2g(20) = u^2 - 40(10) = u^2 - 400$.
$u^2 = 800$,which gives $u = \sqrt{800} = 20\sqrt{2} \, m/s$.

(D) The initial momentum of the body is $\vec{p}_i = m\vec{v}_i = m(v \cos \theta \hat{i} + v \sin \theta \hat{j})$.
At the maximum height,the vertical component of velocity becomes zero,so the velocity is $\vec{v}_f = v \cos \theta \hat{i}$.
The final momentum is $\vec{p}_f = m(v \cos \theta \hat{i})$.
The change in momentum is $\Delta \vec{p} = \vec{p}_f - \vec{p}_i = -m v \sin \theta \hat{j}$.
The magnitude of the change in momentum is $|\Delta \vec{p}| = m v \sin \theta$.
Given $m = 100 \, g = 0.1 \, kg$,$v = 20 \, m \, s^{-1}$,and $\theta = 30^\circ$.
$|\Delta \vec{p}| = 0.1 \times 20 \times \sin(30^\circ) = 0.1 \times 20 \times 0.5 = 1 \, kg \, m \, s^{-1}$.

(A) The horizontal component of velocity for both objects is $v_x = v \cos\theta$.
Since the objects are moving towards each other,their horizontal velocities are equal in magnitude but opposite in direction.
Let the mass of each object be $m$.
The momentum of the first object is $\vec{p}_1 = m(v \cos\theta) \hat{i}$.
The momentum of the second object is $\vec{p}_2 = m(-v \cos\theta) \hat{i}$.
The total momentum of the system is $\vec{P} = \vec{p}_1 + \vec{p}_2 = m v \cos\theta \hat{i} - m v \cos\theta \hat{i} = 0$.
Therefore,the momentum of the system at the moment of collision is $0$.

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

(D) For a given range $R$,the two angles of projection are $\theta$ and $(90^\circ - \theta)$.
The maximum height for angle $\theta$ is $h_1 = \frac{u^2 \sin^2 \theta}{2g}$.
The maximum height for angle $(90^\circ - \theta)$ is $h_2 = \frac{u^2 \sin^2(90^\circ - \theta)}{2g} = \frac{u^2 \cos^2 \theta}{2g}$.
Multiplying $h_1$ and $h_2$,we get:
$h_1 h_2 = \frac{u^2 \sin^2 \theta}{2g} \cdot \frac{u^2 \cos^2 \theta}{2g} = \frac{u^4 \sin^2 \theta \cos^2 \theta}{4g^2}$.
We know that the range $R = \frac{u^2 \sin 2\theta}{g} = \frac{2u^2 \sin \theta \cos \theta}{g}$.
Squaring the range,$R^2 = \frac{4u^4 \sin^2 \theta \cos^2 \theta}{g^2}$.
Comparing the two expressions,we see that $h_1 h_2 = \frac{R^2}{16}$.
Therefore,$R^2 = 16 h_1 h_2$,which implies $R = 4 \sqrt{h_1 h_2}$.

(B) For the first ball thrown vertically upwards with velocity $u_1$,the time of flight is $T_1 = \frac{2u_1}{g}$.
For the second ball thrown at an angle $(90^\circ - \theta)$ with the horizontal with velocity $u_2$,the time of flight is $T_2 = \frac{2u_2 \sin(90^\circ - \theta)}{g} = \frac{2u_2 \cos \theta}{g}$.
Given that $T_1 = T_2$,we have $\frac{2u_1}{g} = \frac{2u_2 \cos \theta}{g}$,which implies $u_1 = u_2 \cos \theta$.
The maximum height attained by the first ball is $H_1 = \frac{u_1^2}{2g}$.
The maximum height attained by the second ball is $H_2 = \frac{(u_2 \sin(90^\circ - \theta))^2}{2g} = \frac{u_2^2 \cos^2 \theta}{2g}$.
Taking the ratio of the heights: $\frac{H_1}{H_2} = \frac{u_1^2 / 2g}{u_2^2 \cos^2 \theta / 2g} = \frac{u_1^2}{u_2^2 \cos^2 \theta}$.
Substituting $u_1 = u_2 \cos \theta$,we get $\frac{H_1}{H_2} = \frac{(u_2 \cos \theta)^2}{u_2^2 \cos^2 \theta} = 1$.
Thus,the ratio is $1 : 1$.

(B) Let the initial velocity be $v$ and the angle of projection be $\theta = 45^o$. The initial kinetic energy is $K = \frac{1}{2}mv^2$.
At the maximum height,the vertical component of velocity becomes zero,and the projectile only possesses the horizontal component of velocity,which is $v_x = v \cos \theta$.
The kinetic energy at the maximum height is $K' = \frac{1}{2}m(v_x)^2 = \frac{1}{2}m(v \cos \theta)^2$.
Substituting $K = \frac{1}{2}mv^2$,we get $K' = K \cos^2 \theta$.
Given $\theta = 45^o$,we have $\cos 45^o = 1/\sqrt{2}$.
Therefore,$K' = K (1/\sqrt{2})^2 = K/2$.

(C) The potential energy $(PE)$ at maximum height $(H)$ is given by $PE = mgH$.
For a projectile launched with velocity $u$ at an angle $\theta$ with the horizontal,the maximum height is $H = \frac{u^2 \sin^2 \theta}{2g}$.
Thus,$PE = mg \left( \frac{u^2 \sin^2 \theta}{2g} \right) = \frac{1}{2} m u^2 \sin^2 \theta$.
For the first ball thrown vertically upward,the angle with the horizontal is $\theta_1 = 90^\circ$. Therefore,$(PE)_1 = \frac{1}{2} m u^2 \sin^2(90^\circ) = \frac{1}{2} m u^2$.
For the second ball thrown at an angle of $60^\circ$ with the vertical,the angle with the horizontal is $\theta_2 = 90^\circ - 60^\circ = 30^\circ$. Therefore,$(PE)_2 = \frac{1}{2} m u^2 \sin^2(30^\circ) = \frac{1}{2} m u^2 (\frac{1}{2})^2 = \frac{1}{8} m u^2$.
The ratio of their potential energies is $\frac{(PE)_1}{(PE)_2} = \frac{\frac{1}{2} m u^2}{\frac{1}{8} m u^2} = \frac{8}{2} = 4:1$.

(A) The total time given is $2 \,min \,20 \,sec = (2 \times 60) + 20 = 140 \,sec$.
The time period of one revolution is $T = 40 \,sec$.
The number of revolutions $n$ is given by $n = \frac{\text{Total time}}{T} = \frac{140}{40} = 3.5$.
The distance covered in one revolution is the circumference of the circle, $C = 2\pi r$.
Total distance $d = n \times (2\pi r) = 3.5 \times 2 \times \pi \times 100$.
$d = 7 \times \pi \times 100 = 700\pi \,m$.

(B) The change in velocity is given by the vector difference $\Delta \vec{v} = \vec{v}_2 - \vec{v}_1$.
Since the speed is constant,let $v = |\vec{v}_1| = |\vec{v}_2| = 5 \, ms^{-1}$.
For a half revolution,the angle between the initial velocity vector $\vec{v}_1$ and the final velocity vector $\vec{v}_2$ is $180^\circ$.
The magnitude of the change in velocity is given by $\Delta v = |\vec{v}_2 - \vec{v}_1| = \sqrt{v^2 + v^2 - 2v^2 \cos(180^\circ)} = \sqrt{2v^2 + 2v^2} = 2v$.
Alternatively,using the formula $\Delta v = 2v \sin(\theta/2)$,where $\theta = 180^\circ$:
$\Delta v = 2 \times 5 \times \sin(180^\circ / 2) = 10 \times \sin(90^\circ) = 10 \times 1 = 10 \, ms^{-1}$.

(B) The net force on the particle is given by the vector sum of the centripetal force $(F_c)$ and the tangential force $(F_t)$:
$F_{Net} = \sqrt{F_c^2 + F_t^2}$ ... $(i)$
Given kinetic energy $K = \frac{1}{2}mv^2 = as^2$,we have $v^2 = \frac{2as^2}{m}$.
The centripetal force is $F_c = \frac{mv^2}{R} = \frac{m}{R} \left( \frac{2as^2}{m} \right) = \frac{2as^2}{R}$ ... (ii)
To find the tangential force,we use $F_t = ma_t$,where $a_t = \frac{dv}{dt} = \frac{dv}{ds} \cdot \frac{ds}{dt} = v \frac{dv}{ds}$.
Since $v = s \sqrt{\frac{2a}{m}}$,we have $\frac{dv}{ds} = \sqrt{\frac{2a}{m}}$.
Thus,$a_t = \left( s \sqrt{\frac{2a}{m}} \right) \left( \sqrt{\frac{2a}{m}} \right) = \frac{2as}{m}$.
Therefore,$F_t = m \left( \frac{2as}{m} \right) = 2as$ ... (iii)
Substituting (ii) and (iii) into $(i)$:
$F_{Net} = \sqrt{\left( \frac{2as^2}{R} \right)^2 + (2as)^2} = \sqrt{(2as)^2 \left( \frac{s^2}{R^2} + 1 \right)} = 2as \sqrt{1 + \frac{s^2}{R^2}}$.

(A) The block has an initial horizontal velocity $u_x = 1.5 \, m/s$ and initial vertical velocity $u_y = 0 \, m/s$.
Horizontal displacement $S_x = u_x \times t = 1.5 \times 4 = 6 \, m$.
Vertical acceleration $a_y = F/m = 5/5 = 1 \, m/s^2$.
Vertical displacement $S_y = u_y t + \frac{1}{2} a_y t^2 = 0 + \frac{1}{2} \times 1 \times (4)^2 = 8 \, m$.
The total distance (displacement) covered by the block is $S = \sqrt{S_x^2 + S_y^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \, m$.

(C) The distance traveled in one complete circular lap is equal to the circumference of the track,which is $2\pi r$.
Given $r = 100\,m$ and time $t = 62.8\,s$.
Average speed is defined as the total distance divided by the total time:
$\text{Average speed} = \frac{2\pi r}{t} = \frac{2 \times 3.14 \times 100}{62.8} = \frac{628}{62.8} = 10\,m/s$.
Since the car completes one full lap,the final position is the same as the initial position.
Therefore,the net displacement is $0\,m$.
Average velocity is defined as the net displacement divided by the total time:
$\text{Average velocity} = \frac{\text{net displacement}}{t} = \frac{0}{62.8} = 0\,m/s$.
Thus,the average velocity is $0\,m/s$ and the average speed is $10\,m/s$.

(A) The initial velocity of the particle is $\vec{u} = v \cos \theta \hat{i} + v \sin \theta \hat{j}$.
The final velocity of the particle when it lands on the ground is $\vec{v}_f = v \cos \theta \hat{i} - v \sin \theta \hat{j}$.
The initial momentum is $\vec{p}_i = m(v \cos \theta \hat{i} + v \sin \theta \hat{j})$.
The final momentum is $\vec{p}_f = m(v \cos \theta \hat{i} - v \sin \theta \hat{j})$.
The change in momentum is $\Delta \vec{p} = \vec{p}_f - \vec{p}_i = -2mv \sin \theta \hat{j}$.
The magnitude of the change in momentum is $|\Delta \vec{p}| = 2mv \sin \theta$.
Given $\theta = 45^\circ$,we have $\sin 45^\circ = \frac{1}{\sqrt{2}}$.
Therefore,$|\Delta \vec{p}| = 2mv \times \frac{1}{\sqrt{2}} = \sqrt{2}mv$.

(B) Given:
Initial velocity,$\vec{u} = (3\hat{i} + 4\hat{j})\,m/s$
Acceleration,$\vec{a} = (0.4\hat{i} + 0.3\hat{j})\,m/s^2$
Time,$t = 10\,s$
Using the first equation of motion for vectors:
$\vec{v} = \vec{u} + \vec{a}t$
$\vec{v} = (3\hat{i} + 4\hat{j}) + (0.4\hat{i} + 0.3\hat{j}) \times 10$
$\vec{v} = (3\hat{i} + 4\hat{j}) + (4\hat{i} + 3\hat{j})$
$\vec{v} = (3+4)\hat{i} + (4+3)\hat{j} = 7\hat{i} + 7\hat{j}$
The speed is the magnitude of the velocity vector $\vec{v}$:
$|\vec{v}| = \sqrt{(7)^2 + (7)^2}$
$|\vec{v}| = \sqrt{49 + 49} = \sqrt{98}$
$|\vec{v}| = 7\sqrt{2}$ units.

(A) Let $u$ be the initial speed of the projectile and $\theta$ be the angle of projection.
At the maximum height,the vertical component of velocity becomes zero,and the velocity of the projectile is only due to the horizontal component.
Thus,the speed at maximum height is $v = u \cos \theta$.
According to the problem,the speed at maximum height is half of its initial speed:
$v = \frac{u}{2}$
Equating the two expressions for $v$:
$u \cos \theta = \frac{u}{2}$
$\cos \theta = \frac{1}{2}$
$\theta = \cos^{-1}(\frac{1}{2}) = 60^o$
Therefore,the angle of projection is $60^o$.

(B) Given:
Initial velocity,$\vec{u} = 2\hat{i} + 3\hat{j}$
Acceleration,$\vec{a} = 0.3\hat{i} + 0.2\hat{j}$
Time,$t = 10\,s$
Using the first equation of motion for vectors:
$\vec{v} = \vec{u} + \vec{a}t$
$\vec{v} = (2\hat{i} + 3\hat{j}) + (0.3\hat{i} + 0.2\hat{j})(10)$
$\vec{v} = 2\hat{i} + 3\hat{j} + 3\hat{i} + 2\hat{j}$
$\vec{v} = 5\hat{i} + 5\hat{j}$
Speed is the magnitude of the velocity vector:
$|\vec{v}| = \sqrt{(5)^2 + (5)^2}$
$|\vec{v}| = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}\,\text{units}$

(A) The equation of the trajectory of a projectile is given by: $y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta}$.
For two trajectories to be identical with the same angle of projection $\theta$,the term $\frac{g}{u^2}$ must be constant.
Therefore,$\frac{g_{earth}}{u_{earth}^2} = \frac{g_{planet}}{u_{planet}^2}$.
Given $g_{earth} = 9.8 \, m s^{-2}$,$u_{earth} = 5 \, m s^{-1}$,and $u_{planet} = 3 \, m s^{-1}$.
Substituting the values: $\frac{9.8}{5^2} = \frac{g_{planet}}{3^2}$.
$\frac{9.8}{25} = \frac{g_{planet}}{9}$.
$g_{planet} = \frac{9.8 \times 9}{25} = \frac{88.2}{25} = 3.528 \, m s^{-2}$.
Rounding to the nearest provided option,the value is $3.5 \, m s^{-2}$.

(D) At time $t_1 = 0 \, s$,the position vector of the particle is $\vec{r}_1 = 2\hat{i} + 3\hat{j}$.
At time $t_2 = 5 \, s$,the position vector of the particle is $\vec{r}_2 = 13\hat{i} + 14\hat{j}$.
The displacement $\Delta \vec{r}$ from $t = 0$ to $t = 5 \, s$ is given by $\Delta \vec{r} = \vec{r}_2 - \vec{r}_1$.
$\Delta \vec{r} = (13\hat{i} + 14\hat{j}) - (2\hat{i} + 3\hat{j}) = 11\hat{i} + 11\hat{j}$.
The average velocity $\vec{v}_{av}$ is defined as the total displacement divided by the total time interval $\Delta t = t_2 - t_1 = 5 - 0 = 5 \, s$.
$\vec{v}_{av} = \frac{\Delta \vec{r}}{\Delta t} = \frac{11\hat{i} + 11\hat{j}}{5} = \frac{11}{5}(\hat{i} + \hat{j}) \, m/s$.

(D) Given position vector: $\vec{R} = 4\sin(2\pi t)\hat{i} + 4\cos(2\pi t)\hat{j}$.
Since $x = 4\sin(2\pi t)$ and $y = 4\cos(2\pi t)$,then $x^2 + y^2 = 16(\sin^2(2\pi t) + \cos^2(2\pi t)) = 16$. This represents a circle of radius $4 \ m$.
Velocity $\vec{v} = \frac{d\vec{R}}{dt} = 8\pi\cos(2\pi t)\hat{i} - 8\pi\sin(2\pi t)\hat{j}$.
Magnitude of velocity $|\vec{v}| = \sqrt{(8\pi\cos(2\pi t))^2 + (-8\pi\sin(2\pi t))^2} = 8\pi \ m/s$.
Acceleration $\vec{a} = \frac{d\vec{v}}{dt} = -16\pi^2\sin(2\pi t)\hat{i} - 16\pi^2\cos(2\pi t)\hat{j} = -4\pi^2\vec{R}$.
Since $\vec{a} = -4\pi^2\vec{R}$,the acceleration is directed towards the origin (along $-\vec{R}$).
For uniform circular motion,$a = \frac{V^2}{R}$. Here $V = 8\pi$ and $R = 4$,so $a = \frac{(8\pi)^2}{4} = 16\pi^2$. This matches the magnitude of $\vec{a}$.
Statement $(D)$ claims the magnitude of velocity is $8 \ m/s$,but it is $8\pi \ m/s$. Thus,$(D)$ is the wrong statement.

(B) For the first ball thrown vertically upwards,the time of flight is $T_1 = \frac{2u_1}{g} = 3 \; s$. Thus,$u_1 = \frac{3g}{2}$. The maximum height is $H_1 = \frac{u_1^2}{2g} = \frac{(3g/2)^2}{2g} = \frac{9g}{8}$.
For the second ball thrown at an angle $\theta = 60^{\circ}$ with the vertical,the angle with the horizontal is $\alpha = 90^{\circ} - 60^{\circ} = 30^{\circ}$. The time of flight is $T_2 = \frac{2u_2 \sin \alpha}{g} = 3 \; s$. Thus,$u_2 \sin \alpha = \frac{3g}{2}$.
The maximum height is $H_2 = \frac{(u_2 \sin \alpha)^2}{2g} = \frac{(3g/2)^2}{2g} = \frac{9g}{8}$.
Comparing the two heights,$\frac{H_1}{H_2} = \frac{9g/8}{9g/8} = 1:1$.

(B) Let $v$ be the tangential speed of the heavier stone.
Given that the centripetal force experienced by both stones is the same.
The centripetal force for the lighter stone (mass $m$,radius $r$,speed $nv$) is given by:
$F_{lighter} = \frac{m(nv)^2}{r} = \frac{mn^2v^2}{r}$
The centripetal force for the heavier stone (mass $2m$,radius $r/2$,speed $v$) is given by:
$F_{heavier} = \frac{(2m)v^2}{r/2} = \frac{4mv^2}{r}$
Equating the two forces:
$\frac{mn^2v^2}{r} = \frac{4mv^2}{r}$
$n^2 = 4$
$n = 2$

(D) Given: Mass $m = 10\, g = 10^{-2}\, kg$,Radius $R = 6.4\, cm = 6.4 \times 10^{-2}\, m$,Final kinetic energy $K_f = 8 \times 10^{-4}\, J$,Initial kinetic energy $K_i = 0$.
The work done by the tangential force is equal to the change in kinetic energy,as the centripetal force does no work.
Work done $W = F_t \times d$,where $F_t = m a_t$ and $d$ is the distance covered in two revolutions.
Distance $d = 2 \times (2\pi R) = 4\pi R$.
Using the work-energy theorem: $W = K_f - K_i = K_f$.
$m a_t (4\pi R) = K_f$.
$a_t = \frac{K_f}{4\pi R m}$.
Substituting the values: $a_t = \frac{8 \times 10^{-4}}{4 \times 3.14159 \times 6.4 \times 10^{-2} \times 10^{-2}}$.
$a_t = \frac{8 \times 10^{-4}}{4 \times 3.14159 \times 6.4 \times 10^{-4}} = \frac{8}{25.6 \times 3.14159} \approx \frac{8}{80.42} \approx 0.0995\, m/s^2$.
Rounding to the nearest value,$a_t \approx 0.1\, m/s^2$.

(D) The centripetal force $F_c$ acting on a particle in uniform circular motion is given by $F_c = \frac{mv^2}{r}$.
Given,$F_c = 10 \, N$ and $r = 20 \, cm = 0.2 \, m$.
So,$\frac{mv^2}{0.2} = 10$,which implies $mv^2 = 10 \times 0.2 = 2 \, J$.
The kinetic energy $K$ of the particle is given by $K = \frac{1}{2}mv^2$.
Substituting the value of $mv^2$,we get $K = \frac{1}{2} \times 2 = 1 \, J$.
Therefore,the correct option is $D$.

(C) For the particle thrown vertically downwards:
Using the equation of motion $v^2 = u^2 + 2gh$,the final velocity $v_1$ at the ground is $v_1 = \sqrt{u^2 + 2gh}$.
For the particle thrown horizontally:
The horizontal component of velocity remains $v_x = u$.
The vertical component of velocity at the ground is $v_y = \sqrt{0^2 + 2gh} = \sqrt{2gh}$.
The resultant velocity $v_2$ at the ground is $v_2 = \sqrt{v_x^2 + v_y^2} = \sqrt{u^2 + 2gh}$.
Comparing the two,$v_1 = v_2 = \sqrt{u^2 + 2gh}$.
Therefore,the ratio of their velocities is $1:1$.

(A) The horizontal range $R$ of a projectile is given by the formula: $R = \frac{u^2 \sin(2\theta)}{g}$,where $u$ is the initial velocity,$\theta$ is the angle of projection,and $g$ is the acceleration due to gravity.
For the first case,$\theta_1 = 30^{\circ}$,so $R = \frac{u^2 \sin(2 \times 30^{\circ})}{g} = \frac{u^2 \sin(60^{\circ})}{g}$.
For the second case,$\theta_2 = 60^{\circ}$,so $R' = \frac{u^2 \sin(2 \times 60^{\circ})}{g} = \frac{u^2 \sin(120^{\circ})}{g}$.
Since $\sin(120^{\circ}) = \sin(180^{\circ} - 60^{\circ}) = \sin(60^{\circ})$,we have $R' = \frac{u^2 \sin(60^{\circ})}{g} = R$.
Thus,for complementary angles (where $\theta_1 + \theta_2 = 90^{\circ}$),the horizontal range is the same.

(B) The initial velocity of the body is $v$ at an angle $\theta = 45^{\circ}$ with the horizontal.
The initial momentum vector is $\vec{p}_i = m(v \cos \theta \hat{i} + v \sin \theta \hat{j})$.
When the body strikes the ground,the horizontal component of velocity remains $v \cos \theta$ and the vertical component becomes $-v \sin \theta$.
The final momentum vector is $\vec{p}_f = m(v \cos \theta \hat{i} - v \sin \theta \hat{j})$.
The change in momentum $\Delta \vec{p} = \vec{p}_f - \vec{p}_i = -2mv \sin \theta \hat{j}$.
The magnitude of the change in momentum is $|\Delta \vec{p}| = 2mv \sin \theta$.
Substituting $\theta = 45^{\circ}$,we get $|\Delta \vec{p}| = 2mv \sin(45^{\circ}) = 2mv \times \frac{1}{\sqrt{2}} = \sqrt{2}mv$.

(C) The time of flight $T$ for a projectile is given by $T = \frac{2u_y}{g}$,where $u_y$ is the vertical component of the initial velocity.
Since both balls stay in the air for the same period of time,their vertical components of initial velocity $(u_y)$ must be equal.
The maximum height attained by a projectile is given by $H = \frac{u_y^2}{2g}$.
Since $u_y$ and $g$ are the same for both balls,the maximum height $H$ attained by both balls will be identical.
Therefore,the ratio of the heights attained is $H_1 : H_2 = 1:1$.

(D) Let the radius of the wheel be $R = 5 \, m$. After half a revolution,the wheel moves forward by a horizontal distance equal to half its circumference,which is $\pi R$. The vertical displacement of point $P$ is equal to the diameter of the wheel,which is $2R$. The total displacement $d$ is the straight-line distance between the initial and final positions of point $P$,given by the Pythagorean theorem: $d = \sqrt{(\pi R)^2 + (2R)^2} = R\sqrt{\pi^2 + 4}$. Substituting $R = 5 \, m$,we get $d = 5\sqrt{\pi^2 + 4} \, m$.

(D) The kinetic energy $E$ of a particle in uniform circular motion is given by $E = \frac{1}{2} L \omega$,where $\omega = 2\pi n$ and $n$ is the frequency of motion.
Thus,$E = \frac{1}{2} L (2\pi n) = \pi L n$.
This implies $E \propto L \times n$,or $L \propto \frac{E}{n}$.
Given the initial state $(E_1, n_1, L_1)$ and final state $(E_2, n_2, L_2)$ where $E_2 = \frac{E_1}{2}$ and $n_2 = 2n_1$,we have:
$\frac{L_2}{L_1} = \frac{E_2}{E_1} \times \frac{n_1}{n_2} = \frac{E_1/2}{E_1} \times \frac{n_1}{2n_1} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.
Therefore,$L_2 = \frac{L_1}{4} = \frac{L}{4}$.

(B) The velocity components are obtained by differentiating the position coordinates with respect to time $t$.
$v_{x} = \frac{dx}{dt} = \frac{d}{dt}(5 \sin 10t) = 50 \cos 10t$
$v_{y} = \frac{dy}{dt} = \frac{d}{dt}(5 \cos 10t) = -50 \sin 10t$
The speed of the particle is the magnitude of the velocity vector:
$\text{Speed} = \sqrt{v_{x}^{2} + v_{y}^{2}}$
$\text{Speed} = \sqrt{(50 \cos 10t)^{2} + (-50 \sin 10t)^{2}}$
$\text{Speed} = \sqrt{2500 \cos^{2} 10t + 2500 \sin^{2} 10t}$
$\text{Speed} = \sqrt{2500(\cos^{2} 10t + \sin^{2} 10t)}$
Since $\sin^{2} \theta + \cos^{2} \theta = 1$,we get:
$\text{Speed} = \sqrt{2500 \times 1} = 50$

(B) The initial velocity of the ball is $u = 15 \ m/s$ and the angle of projection is $\theta = 37^o$. Taking $g = 10 \ m/s^2$,we have $\sin(37^o) \approx 0.6$.
The total time of flight $T$ is given by:
$T = \frac{2u \sin(\theta)}{g} = \frac{2 \times 15 \times 0.6}{10} = \frac{18}{10} = 1.8 \ s$.
The man is at point $P$,which is $9 \ m$ away from the landing point $B$. To catch the ball,the man must cover this distance $d = 9 \ m$ within the time of flight $T = 1.8 \ s$.
The minimum velocity $v$ required is:
$v = \frac{d}{T} = \frac{9}{1.8} = 5 \ m/s$.

(D) Let the position of the first particle projected from $O$ be $\vec{r}_1(t) = (0, vt - \frac{1}{2}gt^2)$.
Let the position of the second particle projected from $P(0, h)$ be $\vec{r}_2(t) = (vt \cos \theta, h + vt \sin \theta - \frac{1}{2}gt^2)$.
The relative position vector $\vec{r}_{rel} = \vec{r}_2 - \vec{r}_1 = (vt \cos \theta, h + vt \sin \theta - vt) = (vt \cos \theta, h - vt(1 - \sin \theta))$.
The square of the distance $D^2 = |\vec{r}_{rel}|^2 = (vt \cos \theta)^2 + (h - vt(1 - \sin \theta))^2$.
To minimize $D^2$,we differentiate with respect to $t$ and set to zero:
$\frac{d(D^2)}{dt} = 2(vt \cos \theta)(v \cos \theta) + 2(h - vt(1 - \sin \theta))(-v(1 - \sin \theta)) = 0$.
$v^2 t \cos^2 \theta - hv(1 - \sin \theta) + v^2 t(1 - \sin \theta)^2 = 0$.
$v^2 t [\cos^2 \theta + (1 - \sin \theta)^2] = hv(1 - \sin \theta)$.
$v^2 t [\cos^2 \theta + 1 + \sin^2 \theta - 2 \sin \theta] = hv(1 - \sin \theta)$.
$v^2 t [2 - 2 \sin \theta] = hv(1 - \sin \theta)$.
$2v^2 t (1 - \sin \theta) = hv(1 - \sin \theta)$.
$t = \frac{h}{2v}$.

(C) Let the origin be at the left projectile $A$ and the $x$-axis be along the line connecting $A$ and $B$. The initial position of $A$ is $(0, 0)$ and $B$ is $(20, 0)$.
The velocity of $A$ is $\vec{v}_A = 20\sqrt{3}(\cos 60^{\circ} \hat{i} + \sin 60^{\circ} \hat{j}) = 20\sqrt{3}(\frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j}) = 10\sqrt{3} \hat{i} + 30 \hat{j}$.
The velocity of $B$ is $\vec{v}_B = 20(\cos 150^{\circ} \hat{i} + \sin 150^{\circ} \hat{j}) = 20(-\frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j}) = -10\sqrt{3} \hat{i} + 10 \hat{j}$.
The relative velocity of $A$ with respect to $B$ is $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B = (10\sqrt{3} - (-10\sqrt{3})) \hat{i} + (30 - 10) \hat{j} = 20\sqrt{3} \hat{i} + 20 \hat{j}$.
The relative initial position is $\vec{r}_{AB} = \vec{r}_A - \vec{r}_B = (0 - 20) \hat{i} + (0 - 0) \hat{j} = -20 \hat{i}$.
The minimum distance is the perpendicular distance from the origin to the line of relative motion,given by $d_{\min} = \frac{|\vec{r}_{AB} \times \vec{v}_{AB}|}{|\vec{v}_{AB}|}$.
$|\vec{r}_{AB} \times \vec{v}_{AB}| = |(-20 \hat{i}) \times (20\sqrt{3} \hat{i} + 20 \hat{j})| = |-400 \hat{k}| = 400$.
$|\vec{v}_{AB}| = \sqrt{(20\sqrt{3})^2 + 20^2} = \sqrt{1200 + 400} = \sqrt{1600} = 40$.
$d_{\min} = \frac{400}{40} = 10 \ m$.

(D) Let the speed of projection be $u$. For projectile $A$,the angle with the horizontal is $\theta_A = \theta$. For projectile $B$,the angle with the vertical is $\theta$,so the angle with the horizontal is $\theta_B = 90^{\circ} - \theta$.
Since $\theta_B$ is the complement of $\theta_A$,these are complementary angles.
For complementary angles,the horizontal range is the same $(R_A = R_B)$.
However,the time of flight $T = \frac{2u \sin \theta}{g}$ and maximum height $H = \frac{u^2 \sin^2 \theta}{2g}$ depend on the specific angle $\theta$.
If $\theta = 45^{\circ}$,then $\theta_A = \theta_B = 45^{\circ}$,and all parameters (time of flight,maximum height,and range) will be identical.
If $\theta \neq 45^{\circ}$,the time of flight and maximum height will be different for $A$ and $B$.
Therefore,the statement that they 'may' have the same time of flight is correct,as it holds true for the case $\theta = 45^{\circ}$.

(D) Let the initial positions of particles $A$ and $B$ be $(-x_0, 0)$ and $(0, -y_0)$ respectively,where $x_0, y_0 > 0$.
Let their constant velocities be $\vec{v}_A = v_A \hat{i}$ and $\vec{v}_B = v_B \hat{j}$.
At any time $t$,the positions are $\vec{r}_A(t) = (-x_0 + v_A t) \hat{i}$ and $\vec{r}_B(t) = (-y_0 + v_B t) \hat{j}$.
The separation $S$ between them is given by $S = |\vec{r}_B - \vec{r}_A| = \sqrt{(v_B t - y_0)^2 + (x_0 - v_A t)^2}$.
Squaring both sides,$S^2 = (v_B^2 + v_A^2) t^2 - 2(v_B y_0 + v_A x_0) t + (y_0^2 + x_0^2)$.
This is a quadratic equation of the form $S^2 = at^2 + bt + c$,which represents a hyperbola. The graph of $S$ versus $t$ is a hyperbola that decreases to a minimum value and then increases,which corresponds to the curve shown in option $D$.

(B) Let $\vec{u}_{1}$ and $\vec{u}_{2}$ be the initial velocities of the two particles and $\theta_{1}$ and $\theta_{2}$ be their angles of projection with the horizontal.
The velocities of the two particles at any time $t$ are given by:
$\vec{v}_{1} = (u \cos \theta_{1}) \hat{i} + (u \sin \theta_{1} - gt) \hat{j}$
$\vec{v}_{2} = (u \cos \theta_{2}) \hat{i} + (u \sin \theta_{2} - gt) \hat{j}$
The relative velocity of one particle with respect to the other is:
$\vec{v}_{12} = \vec{v}_{1} - \vec{v}_{2} = u(\cos \theta_{1} - \cos \theta_{2}) \hat{i} + u(\sin \theta_{1} - \sin \theta_{2}) \hat{j}$
Since the relative velocity $\vec{v}_{12}$ is independent of time $t$,the relative acceleration is zero $(\vec{a}_{12} = \vec{a}_{1} - \vec{a}_{2} = -g\hat{j} - (-g\hat{j}) = 0)$.
Because the relative acceleration is zero and the relative initial position is zero (both start from the same point),the relative motion is a straight line passing through the origin. Since the relative velocity vector is constant,the path is a straight line making a constant angle with the horizontal.

(B) Let the velocities of the two projectiles be $\vec{v}_1 = v_1 \cos \theta_1 \hat{i} + v_1 \sin \theta_1 \hat{j}$ and $\vec{v}_2 = v_2 \cos \theta_2 \hat{i} + v_2 \sin \theta_2 \hat{j}$.
Both projectiles experience the same acceleration due to gravity,$\vec{a}_1 = \vec{a}_2 = -g \hat{j}$.
The relative acceleration of particle $1$ with respect to particle $2$ is $\vec{a}_{12} = \vec{a}_1 - \vec{a}_2 = 0$.
Since the relative acceleration is zero,the relative velocity $\vec{v}_{12} = \vec{v}_1 - \vec{v}_2$ is constant.
The relative velocity is $\vec{v}_{12} = (v_1 \cos \theta_1 - v_2 \cos \theta_2) \hat{i} + (v_1 \sin \theta_1 - v_2 \sin \theta_2) \hat{j}$.
Since the relative acceleration is zero and the relative velocity is constant,the relative position vector $\vec{r}_{12} = \vec{v}_{12} t$ represents a straight line equation in the $xy$-plane.

(C) The horizontal components of the velocities are given as $u_{x1} = v_1 \cos \theta_1$ and $u_{x2} = v_2 \cos \theta_2$. Given $v_1 \cos \theta_1 = v_2 \cos \theta_2$,the horizontal velocities are equal.
Since the horizontal acceleration for both projectiles is zero,their horizontal positions at any time $t$ are $x_1 = (v_1 \cos \theta_1)t$ and $x_2 = (v_2 \cos \theta_2)t$.
Because $v_1 \cos \theta_1 = v_2 \cos \theta_2$,it follows that $x_1 = x_2$ for all $t$.
This means one particle will always remain exactly below or above the other particle (Option $A$ is correct).
The relative horizontal displacement is $x_{rel} = x_1 - x_2 = 0$. Thus,the trajectory of one with respect to the other is a vertical straight line (Option $B$ is correct).
The range of a projectile is $R = \frac{v^2 \sin(2\theta)}{g} = \frac{2(v \cos \theta)(v \sin \theta)}{g}$. Since $v_1 \cos \theta_1 = v_2 \cos \theta_2$,the ranges are equal only if $v_1 \sin \theta_1 = v_2 \sin \theta_2$,which is not necessarily true. Therefore,the statement that both will have the same range is not necessarily true. Thus,Option $C$ is the incorrect statement.

(D) The time of flight for a projectile is given by $T = \frac{2v \sin \theta}{g}$. Since $v_1 \sin \theta_1 = v_2 \sin \theta_2$,it follows that $T_1 = T_2$. Thus,option $A$ is correct.
The maximum height is given by $H = \frac{v^2 \sin^2 \theta}{2g}$. Since $v_1 \sin \theta_1 = v_2 \sin \theta_2$,it follows that $H_1 = H_2$. Thus,option $B$ is correct.
The relative acceleration between the two projectiles is $\vec{a}_{rel} = \vec{g} - \vec{g} = 0$. Since the relative acceleration is zero,the relative velocity $\vec{v}_{rel}$ remains constant. The relative vertical velocity is $v_{1y} - v_{2y} = v_1 \sin \theta_1 - v_2 \sin \theta_2 = 0$. Since the relative vertical velocity is zero,the relative motion is purely horizontal. Thus,the trajectory of one with respect to the other is a horizontal straight line. Option $C$ is correct.
Since all statements $A$,$B$,and $C$ are correct,the incorrect statement is none of these.

(D) For two projectiles with the same initial velocity $v$ and different angles $\theta_1$ and $\theta_2$,the equations of motion are $x_1 = v \cos \theta_1 t$,$y_1 = v \sin \theta_1 t - \frac{1}{2}gt^2$ and $x_2 = v \cos \theta_2 t$,$y_2 = v \sin \theta_2 t - \frac{1}{2}gt^2$.
Since $\theta_1 > \theta_2$,we have $\cos \theta_1 < \cos \theta_2$,so $x_1 < x_2$ for any time $t > 0$.
Also,the relative vertical position is $y_1 - y_2 = v t (\sin \theta_1 - \sin \theta_2)$. Since $\theta_1 > \theta_2$,$\sin \theta_1 > \sin \theta_2$,so $y_1 > y_2$.
This means particle $1$ is always above particle $2$,so particle $2$ moves under particle $1$.
The slope of the relative trajectory is $\frac{dy_1 - dy_2}{dx_1 - dx_2}$,which remains positive under these conditions.
Option $C$ is a general property of complementary angles,but the question asks for the incorrect statement regarding the specific condition $\theta_1 > \theta_2$. Since $A$ and $B$ are correct,and $C$ is a true statement about range,$D$ is the correct choice as there is no incorrect statement among the given options.

(A) The projectile crosses the two walls of equal height $H$ symmetrically.
Let the time taken to reach the first wall be $t_1 = 2 \text{ s}$ and the time taken to reach the second wall be $t_2 = 6 \text{ s}$.
Due to the symmetry of the projectile motion,the peak of the trajectory occurs exactly halfway between the two walls.
The time at which the projectile reaches its maximum height is $t_{peak} = \frac{t_1 + t_2}{2} = \frac{2 + 6}{2} = 4 \text{ s}$.
For a projectile launched from the ground,the total time of flight $T$ is twice the time taken to reach the maximum height.
Therefore,$T = 2 \times t_{peak} = 2 \times 4 = 8 \text{ s}$.

(C) The projectile crosses the two walls at times $t_1 = 2 \ s$ and $t_2 = 6 \ s$.
Since the motion is symmetric,the time taken to reach the peak of the trajectory is $t_{peak} = \frac{t_1 + t_2}{2} = \frac{2 + 6}{2} = 4 \ s$.
The vertical displacement $H$ at time $t = 2 \ s$ is given by $H = (v_0 \sin \theta) t - \frac{1}{2} g t^2$.
At the peak,the vertical velocity is zero,so $v_y = v_0 \sin \theta - g t_{peak} = 0$,which gives $v_0 \sin \theta = g t_{peak} = 10 \times 4 = 40 \ m/s$ (taking $g = 10 \ m/s^2$).
Substituting this into the equation for $H$:
$H = (40)(2) - \frac{1}{2} (10)(2)^2$
$H = 80 - 20 = 60 \ m$.

(A) The horizontal velocity $v_x$ of the projectile is constant. The horizontal distance between the two walls is $d = 120\, m$,and the time taken to travel this distance is $\Delta t = t_2 - t_1 = 6\, s - 2\, s = 4\, s$.
Thus,the horizontal velocity is $v_x = \frac{d}{\Delta t} = \frac{120\, m}{4\, s} = 30\, m/s$.
Due to the symmetry of the projectile motion,the total time of flight $T$ is the sum of the times at which it crosses the walls symmetrically relative to the peak. The peak occurs at the midpoint of the time interval $[t_1, t_2]$,which is $t_{peak} = \frac{t_1 + t_2}{2} = \frac{2 + 6}{2} = 4\, s$.
The total time of flight $T$ is $2 \times t_{peak} = 2 \times 4\, s = 8\, s$.
The range $R$ of the projectile is given by $R = v_x \times T$.
Substituting the values,$R = 30\, m/s \times 8\, s = 240\, m$.

(B) The vertical displacement of the projectile is given by $y = u \sin \theta t - \frac{1}{2} g t^2$.
At $t_1 = 2 \, s$ and $t_2 = 6 \, s$,the projectile is at the same height $H$.
Thus,$H = u \sin \theta (2) - \frac{1}{2} g (2)^2 = 2 u \sin \theta - 2g$ and $H = u \sin \theta (6) - \frac{1}{2} g (6)^2 = 6 u \sin \theta - 18g$.
Equating the two expressions for $H$:
$2 u \sin \theta - 2g = 6 u \sin \theta - 18g$
$4 u \sin \theta = 16g$
$u \sin \theta = 4g = 40 \, m/s$ (assuming $g = 10 \, m/s^2$).
The horizontal distance between the two walls is $d = 120 \, m$. The time interval between crossing the two walls is $\Delta t = 6 - 2 = 4 \, s$.
Since horizontal velocity $u \cos \theta$ is constant:
$u \cos \theta = \frac{d}{\Delta t} = \frac{120}{4} = 30 \, m/s$.
Now,$\tan \theta = \frac{u \sin \theta}{u \cos \theta} = \frac{40}{30} = \frac{4}{3}$.
Therefore,$\theta = tan^{-1}(4/3)$.

(B) Given: Initial speed $u = 50 \ m/s$,angle of projection $\theta = 53^{\circ}$,and height $y = 75 \ m$.
Using the trajectory equation of a projectile: $y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta}$.
Taking $g = 10 \ m/s^2$,$\tan 53^{\circ} = \frac{4}{3}$,and $\cos 53^{\circ} = \frac{3}{5}$.
Substituting the values: $75 = x \left(\frac{4}{3}\right) - \frac{10 x^2}{2 \times (50)^2 \times (3/5)^2}$.
$75 = \frac{4x}{3} - \frac{10 x^2}{2 \times 2500 \times (9/25)}$.
$75 = \frac{4x}{3} - \frac{10 x^2}{1800} \times 25 = \frac{4x}{3} - \frac{x^2}{180}$.
Multiplying by $180$: $13500 = 240x - x^2$.
Rearranging into a quadratic equation: $x^2 - 240x + 13500 = 0$.
Solving for $x$ using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{240 \pm \sqrt{57600 - 4(13500)}}{2} = \frac{240 \pm \sqrt{57600 - 54000}}{2} = \frac{240 \pm \sqrt{3600}}{2} = \frac{240 \pm 60}{2}$.
$x_1 = \frac{300}{2} = 150 \ m$ and $x_2 = \frac{180}{2} = 90 \ m$.
The horizontal separation is $\Delta x = |x_1 - x_2| = 150 - 90 = 60 \ m$.