Q 11-02-120JEE MainJEE Main 2020 (5 Sep, Shift 1)Medium
A helicopter rises from rest on the ground vertically upwards with a constant acceleration $g$. A food packet is dropped from the helicopter when it is at a height $h$. The time taken by the packet to reach the ground is close to [$g$ is the acceleration due to gravity]:
Answer: (C) $t = 3.4\sqrt{\dfrac hg}$
At height $h$ the helicopter (and packet) has upward speed $u = \sqrt{2gh}$.
Taking up as positive: $-h = ut - \tfrac12gt^{2}$, i.e. $gt^{2} - 2\sqrt{2gh}\,t - 2h = 0$.
$$t = \frac{2\sqrt{2gh} + \sqrt{8gh + 8gh}}{2g} = (\sqrt2 + 2)\sqrt{\frac hg} \approx 3.4\sqrt{\frac hg}$$
Solution by Sreeraj P, M.Sc Physics