Q 11-03-178JEE MainJEE Main 2025 (7 Apr, Shift 2)Easy
A helicopter flying horizontally with a speed of $360\ \text{km/h}$ at an altitude of $2\ \text{km}$ drops an object at an instant. The object hits the ground at a point O, $20\ \text{s}$ after it is dropped. The displacement of O from the position of the helicopter where the object was released is: (use $g = 10\ \text{m/s}^2$ and neglect air resistance)
Answer: (D) $2\sqrt2\ \text{km}$
The object keeps the helicopter's horizontal velocity, $u = 360\times\dfrac{5}{18} = 100\ \text{m/s}$.
Horizontal distance: $x = 100\times20 = 2000\ \text{m}$. Vertical drop: $\tfrac12\times10\times20^2 = 2000\ \text{m}$ (matches the altitude).
$$D = \sqrt{2000^2 + 2000^2} = 2000\sqrt2\ \text{m} = 2\sqrt2\ \text{km}$$
Solution by Sreeraj P, M.Sc Physics