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 )

Physics · Motion in one Dimension

JEE Main 2026 — 23 January, Evening Shift — Question 29

A paratrooper jumps from an aeroplane and opens a parachute after 2 s of free fall and starts deaccelerating with $3 \mathrm{~m} / \mathrm{s}^{2}$ . At 10 m height from ground, while descending with the help of parachute, the speed of paratrooper is $5 \mathrm{~m} / \mathrm{s}$ . The initial height of the aeroplane is $\_\_\_\_$ m. $\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)$

Option A:
62.5
Option B:
92.5
Correct
Option C:
20
Option D:
82.5

Answer: B

Step-by-step solution

A to B $x_{1}=\frac{1}{2} \times 10 \times 2^{2}=20 \mathrm{~m}$ $\mathrm{V}=0+10 \times 2$ B to C $5^{2}=20^{2}-2(3) \mathrm{x}_{2}$ $\mathrm{x}_{2}=\frac{375}{6}$ $\mathrm{x}_{2}=62.5$ C to D $\mathrm{x}_{3}=10 \mathrm{~m}$ $\mathrm{H}=\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}=92.5$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 23 January, Evening Shift
Subject: Physics
Chapter: Motion in one Dimension
Topic: Motion Under Gravity