Motion Under Gravity question from JEE Main 2024 (6 Apr, Shift 2)

Answer: (b)

1. Up throw: $-h = ut_1 - \frac{1}{2}gt_1^2$; down throw: $h = ut_2 + \frac{1}{2}gt_2^2$
2. Eliminate $u$ (multiply first by $t_2$, second by $t_1$, subtract): $h(t_1 + t_2) = \frac{1}{2}g t_1 t_2 (t_1 + t_2) \Rightarrow h = \frac{1}{2}gt_1t_2$
3. Dropped: $h = \frac{1}{2}gt^2 \Rightarrow t = \sqrt{t_1t_2}$
4. So the answer is (b) $\sqrt{t_1t_2}$.