JEE Advanced 2024 — Paper 2 — PHY-III solution

JEE Advanced2024Paper 2PHY-III

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 $θ_{0}$ 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 $(180 - θ_{1})$ from the $+ x$ -direction with a suitable initial speed. For a fixed $v_{0}$ , when $(θ_{0}, θ_{1}) = (4 5^{\circ}, 4 5^{\circ})$ , the stone hits the ball after time $T_{1}$ , and when $(θ_{0}, θ_{1}) = (6 0^{\circ}, 3 0^{\circ})$ , it hits the ball after time $T_{2}$ . In such a case, $(T_{1} / T_{2})^{2}$ is ________.

Your answer

Solution

Both projectiles experience the same gravitational acceleration. In the frame of the ball, the stone moves in a straight line at constant relative velocity. For the stone to meet the ball, the vertical component of relative velocity must vanish:

$v_{0} sin θ_{0} = u_{s} sin θ_{1},$ where $u_{s}$ is the stone's initial speed.

The horizontal closing speed is $v_{0} cos θ_{0} + u_{s} cos θ_{1}$ , and the meeting time is

$T = \frac{L}{v _{0} cos θ _{0} + u _{s} cos θ _{1}} .$

Case 1: $θ_{0} = θ_{1} = 4 5^{\circ}$ , so $u_{s} = v_{0}$ and

$T_{1} = \frac{L}{v _{0} ( \frac{1}{2} + \frac{1}{2} )} = \frac{L}{v _{0} 2} .$

Case 2: $θ_{0} = 6 0^{\circ}, θ_{1} = 3 0^{\circ}$ . From the vertical condition $v_{0} sin 6 0^{\circ} = u_{s} sin 3 0^{\circ}$ , $u_{s} = 3 v_{0}$ . Then

$T_{2} = \frac{L}{v _{0} \cdot \frac{1}{2} + 3 v _{0} \cdot \frac{3}{2}} = \frac{L}{2 v _{0}} .$

$∴ (\frac{T _{1}}{T _{2}})^{2} = (\frac{2 v _{0}}{v _{0} 2})^{2} = (2)^{2} = 2.$

← Back to full paper

Practice more PHY-III

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →