A parachute after bailing out falls 50 m without friction. When a parachute opens it decelerates at 2 ${m s}^{- 2}$ . He reaches the ground with a speed of 3 ${m s}^{- 1}$ . At what height did he bail out?

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Solution

Given,

$h = 50 m$

$g = 9.8 m / s^{2}$

$u = 0 m / s$

3rd equation of motion,

$v^{2} - u^{2} = 2 g h$

$v^{2} - 0^{2} = 2 \times 50 \times 9.8$

$v^{2} = 980$

When a parachute opens, $a = - 2 m / s^{2}$ , $u = 3 m / s$

3rd equation of motion,

$v^{2} - u^{2} = 2 a h^{'}$

$980 - 9 = - 4 h^{'}$

$h^{'} = 243 m$

Total height, $H = 50 + 243 = 293 m$

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A parachutist after bailing out falls

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. He reaches the ground with speed

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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?

Q. A parachutist after bailing out falls 50m without friction. When parachute opens, it decelerates at 2 m/

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A parachute after bailing out falls 50 m without friction. When a parachute opens it decelerates at 2ms−2. He reaches the ground with a speed of 3ms−1. At what height did he bail out?

Given,h=50mg=9.8m/s2u=0m/s3rd equation of motion,v2−u2=2ghv2−02=2×50×9.8v2=980When a parachute opens, a=−2m/s2, u=3m/s3rd equation of motion,v2−u2=2ah′980−9=−4h′h′=243mTotal height, H=50+243=293m

A parachute after bailing out falls 50 m without friction. When a parachute opens it decelerates at 2 ${m s}^{- 2}$ . He reaches the ground with a speed of 3 ${m s}^{- 1}$ . At what height did he bail out?