A parachutist after bailing out falls 50 m without friction. When parachute opens, it decelerates at 2 m/s². He reaches the ground with a speed of 3 m/s. At what height, did he bail out? (a) 182 m (b) 91 m (c) 111 m (d) 293 m

Answer: (d). Speed after 50 m free fall: v² = 2gh = 2 × 9.8 × 50 = 980 With deceleration 2 m/s ² down to 3 m/s: s = (980 - 9)/(2 × 2) ≈ 243 m Total height = 50 + 243 = 293 m So the answer is (d) 293 m.

A parachutist after bailing out falls 50 m without friction. When… | JEE Main 2005 (AIEEE)

Q54JEE Main 2005 · AIEEEMedium

Motion Under Gravity

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

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Answer: (d)

Speed after 50 m free fall: $v^2 = 2gh = 2 \times 9.8 \times 50 = 980$
With deceleration 2 m/s $^2$ down to 3 m/s: $s = \frac{980 - 9}{2 \times 2} \approx 243$ m
Total height $= 50 + 243 = 293$ m
So the answer is (d) 293 m.

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