Motion Under Gravity question from JEE Main 2020 (5 Sep, Shift 1)

Answer: (a)

1. At height $h$ the helicopter's speed: $v^2 = 2gh \Rightarrow v = \sqrt{2gh}$ upward; the packet starts with this velocity.
2. $-h = \sqrt{2gh}\,t - \frac{1}{2}gt^2$. Put $t = x\sqrt{h/g}$: $x^2 - 2\sqrt{2}x - 2 = 0$
3. $x = \sqrt{2} + 2 \approx 3.41$
4. So $t \approx 3.4\sqrt{\frac{h}{g}}$, option (a).