JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Kinematics

A helicopter flying horizontally with a speed of

360 km / h

at an altitude of 2 km , drops an object at an instant. The object hits the ground at a point O , 20 s after it is dropped. Displacement of ' O ' from the position of helicopter where the object was released is :
(use acceleration due to gravity

g = 10 m / s^{2}

and neglect air resistance)

(A) $25 km$
(B)4 km
(C)7.2 km
(D) $22 km$ ✓ Correct

Answer: (D)

Solution

The horizontal speed is

u = 360 \times \frac{5}{18} = 100 m/s

, so in

20 s

the object moves horizontally

x = 100 \times 20 = 2000 m = 2 km

. It falls vertically

H = \frac{1}{2} g t^{2} = \frac{1}{2} \times 10 \times 2 0^{2} = 2000 m = 2 km

. The displacement of the landing point from the release position is

x^{2} + H^{2} = 2^{2} + 2^{2} = 22 km

JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Single correct option

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