Q 11-03-126JEE MainJEE Main 2021 (31 Aug, Shift 1)Easy
A helicopter is flying horizontally with a speed $v$ at an altitude $h$ and has to drop a food packet for a man on the ground. What is the distance of the helicopter from the man when the food packet is dropped?
Answer: (D) $\sqrt{\dfrac{2v^2h}{g} + h^2}$
Time of fall: $t = \sqrt{\dfrac{2h}{g}}$; horizontal distance covered by the packet: $x = v\sqrt{\dfrac{2h}{g}}$.
The man must be at this horizontal distance and $h$ below, so
$$d = \sqrt{x^2 + h^2} = \sqrt{\frac{2v^2h}{g} + h^2}$$
Solution by Sreeraj P, M.Sc Physics