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Question

A parachute after bailing out falls 50 m without friction. When a parachute opens it decelerates at 2ms2. He reaches the ground with a speed of 3ms1. At what height did he bail out?

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Solution

Given,

h=50m

g=9.8m/s2

u=0m/s

3rd equation of motion,

v2u2=2gh

v202=2×50×9.8

v2=980

When a parachute opens, a=2m/s2, u=3m/s

3rd equation of motion,

v2u2=2ah

9809=4h

h=243m

Total height, H=50+243=293m

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