JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Kinematics

A paratrooper jumps from an aeroplane and opens a parachute after 2 s of free fall and starts deaccelerating with

3 m / s^{2}

. At 10 m height from ground, while descending with the help of parachute, the speed of paratrooper is

5 m / s

. The initial height of the aeroplane is ____ m.

(g = 10 m / s^{2})

Answer: (B)

Solution

During the

2 s

of free fall the paratrooper descends

x_{1} = \frac{1}{2} g t^{2} = \frac{1}{2} \times 10 \times 2^{2} = 20 m

and reaches a speed

v = g t = 20 m/s

. The parachute then decelerates him at

3 m/s^{2}

until his speed is

5 m/s

; from

v^{2} = u^{2} - 2 a x_{2}

5^{2} = 2 0^{2} - 2 (3) x_{2}

, so

x_{2} = \frac{400 - 25}{6} = 62.5 m

. A further

10 m

remains to the ground. The initial height of the aeroplane is

H = 20 + 62.5 + 10 = 92.5 m

JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Single correct option