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JEE Main 2026 (23 Jan, Shift 2) · Motion Under Gravity

Motion Under Gravity question from JEE Main 2026 (23 Jan, Shift 2)

Q5JEE Main 2026 · 23 JAN · SHIFT 2Medium

Motion Under Gravity

A paratrooper jumps from an aeroplane and opens a parachute after 2 s of free fall and starts decelerating with 3 m/s23\text{ m/s}^2. At 10 m height from ground, while descending with the help of parachute, the speed of paratrooper is 5 m/s5\text{ m/s}. The initial height of the airplane is ______ m.

(g=10 m/s2)(g = 10\text{ m/s}^2)

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Answer: (a)

  1. Free fall for 2 s: v=20v = 20 m/s, h1=12×10×22=20h_1 = \frac{1}{2} \times 10 \times 2^2 = 20 m
  2. Decelerating at 3 m/s2^2 from 20 m/s to 5 m/s: h2=202−522×3=3756=62.5h_2 = \frac{20^2 - 5^2}{2 \times 3} = \frac{375}{6} = 62.5 m
  3. Initial height =20+62.5+10=92.5= 20 + 62.5 + 10 = 92.5 m
  4. So the answer is (a) 92.5 m.
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