A parachutist after bailing out falls 50 m without friction. When parachute opens, it decelerates at 2 m/s^2 . He reaches the ground with a speed of 3 - Physics MCQ AIEEE 2026 | ExamTest.live

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A parachutist after bailing out falls 50 m without friction. When parachute opens, it decelerates at 2 m/s^{2} . He reaches the ground with a speed of 3 m/s . At what height, did he bail out?

Solution & Step-by-step Explanation

1. Free fall phase (s_{1} = 50 m): v_{1}^{2} = 2 g s_{1} = 2 (9.8) (50) = 980 (Taking g \approx 10 m/s^{2} ⟹ v_{1}^{2} = 1000).2. Deceleration phase (a = - 2 m/s^{2}, v_{f} = 3 m/s): v_{f}^{2} = v_{1}^{2} + 2 a s_{2} 3^{2} = 1000 + 2 (- 2) s_{2} 9 = 1000 - 4 s_{2} ⟹ 4 s_{2} = 991 ⟹ s_{2} = 247.75 m .Total height = s_{1} + s_{2} = 50 + 247.75 = 297.75 m \approx 293 m (depending on g value used).

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A parachutist after bailing out falls 50 m without friction. When parachute opens, it decelerates at 2 m/s^{2} . He reaches the ground with a speed of 3 m/s . At what height, did he bail out?

91 m

182 m

293 m

111 m

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