JEE Main 2026 — Jan 23, Shift 2 — Physics solution

JEE Main2026Jan 23, Shift 2Physics

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 $^{2}$ )

A
62.5
B
92.5
C
20
D
82.5

Solution

Phase A → B (free fall, 2 s):

$x_{1} = \frac{1}{2} (10) (2)^{2} = 20$ m. Speed at B: $v_{B} = 10 \cdot 2 = 20$ m/s (downward).

Phase B → C (deceleration at 3 m/s $^{2}$ until speed reaches 5 m/s):

$5^{2} = 2 0^{2} - 2 (3) x_{2} \Rightarrow x_{2} = \frac{400 - 25}{6} = \frac{375}{6} = 62.5$ m.

Phase C → D (last 10 m to ground): $x_{3} = 10$ m.

Total initial height $H = x_{1} + x_{2} + x_{3} = 20 + 62.5 + 10 = 92.5$ m.

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