JEE Advanced 2024 Paper 2 · Q09 · Projectile Motion

Question

A ball is thrown from the location $(x_0, y_0) = (0, 0)$ of a horizontal playground with an initial speed $v_0$ at an angle $	heta_0$ from the $+x$-direction. The ball is to be hit by a stone, which is thrown at the same time from the location $(x_1, y_1) = (L, 0)$. The stone is thrown at an angle $(180^\circ - 	heta_1)$ from the $+x$-direction with a suitable initial speed. For a fixed $v_0$, when $(	heta_0, 	heta_1) = (45^\circ, 45^\circ)$, the stone hits the ball after time $T_1$, and when $(	heta_0, 	heta_1) = (60^\circ, 30^\circ)$, it hits the ball after time $T_2$. In such a case, $(T_1/T_2)^2$ is _____.

SolveKar AI · Accepted Answer

For two projectiles aimed at each other (one from origin at angle $	heta_0$, the other from $(L,0)$ at angle $180^\circ - 	heta_1$), the relative motion is along the line joining them initially because both share gravity. Their meeting time is $T = \dfrac{L}{v_{0x} + v_{1x}} = \dfrac{L}{v_0\cos	heta_0 + v_1\cos	heta_1}$.

For them to meet at the same height: vertical component of meeting requires $v_0\sin	heta_0 = v_1\sin	heta_1$ (so they hit each other in the air). Hence $v_1 = v_0\sin	heta_0/\sin	heta_1$.

Meeting time: $T = \dfrac{L}{v_0\cos	heta_0 + (v_0\sin	heta_0/\sin	heta_1)\cos	heta_1} = \dfrac{L\sin	heta_1}{v_0(\sin	heta_1\cos	heta_0 + \sin	heta_0\cos	heta_1)} = \dfrac{L\sin	heta_1}{v_0\sin(	heta_0+	heta_1)}$.

For $(	heta_0,	heta_1) = (45°,45°)$: $T_1 = \dfrac{L\sin 45°}{v_0\sin 90°} = \dfrac{L/\sqrt{2}}{v_0} = \dfrac{L}{v_0\sqrt{2}}$.

For $(	heta_0,	heta_1) = (60°,30°)$: $T_2 = \dfrac{L\sin 30°}{v_0\sin 90°} = \dfrac{L/2}{v_0} = \dfrac{L}{2v_0}$.

$\dfrac{T_1}{T_2} = \dfrac{L/(v_0\sqrt{2})}{L/(2v_0)} = \dfrac{2}{\sqrt{2}} = \sqrt{2}$.

$(T_1/T_2)^2 = 2$. Units: $T$ in seconds, ratio dimensionless. Answer: $2$.

JEE Advanced 2024 Paper 2 · Q09 · Projectile Motion

Solution