Q 11-02-061JEE MainJEE Main 2024 (6 Apr, Shift 2)Medium
A body projected vertically upwards with a certain speed from the top of a tower reaches the ground in $t_1$. If it is projected vertically downwards from the same point with the same speed, it reaches the ground in $t_2$. The time required to reach the ground if it is dropped from the top of the tower is
Answer: (A) $\sqrt{t_1t_2}$
Taking downward as positive, with tower height $h$:
$$h = -ut_1 + \tfrac12gt_1^2,\qquad h = ut_2 + \tfrac12gt_2^2$$
Multiply the first by $t_2$, the second by $t_1$ and add: $h(t_1 + t_2) = \tfrac12gt_1t_2(t_1 + t_2)$, so $h = \tfrac12gt_1t_2$.
Dropped from rest: $h = \tfrac12gt^2 \Rightarrow t = \sqrt{t_1t_2}$.
Solution by Sreeraj P, M.Sc Physics