Q 11-02-091JEE MainJEE Main 2022 (24 Jun, Shift 1)Medium
From the top of a tower, a ball is thrown vertically upward which reaches the ground in $6\ \text{s}$. A second ball thrown vertically downward from the same position with the same speed reaches the ground in $1.5\ \text{s}$. A third ball released from rest from the same location will reach the ground in ______ s.
Numerical value type. Enter your answer.
Answer: 3
With height $h$ and speed $u$ (downward positive):
$$h = -6u + \tfrac12 g(36),\qquad h = 1.5u + \tfrac12 g(2.25)$$
Eliminating $u$ gives $h = \tfrac12 g\,(6\times1.5)$. For the ball released from rest, $h = \tfrac12 g t^2$, so
$$t = \sqrt{t_1t_2} = \sqrt{6\times1.5} = 3\ \text{s}$$
Solution by Sreeraj P, M.Sc Physics