A parachutist,after bailing out,falls 20,m wi | vedclass

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(D) Step $1$: Calculate the velocity of the parachutist after falling $20\,m$ under gravity $(g = 10\,m/s^2)$.Using $v^2 = u^2 + 2as$,where $u = 0$,$a

Vedclass · Accepted Answer

$116$

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(D) Step $1$: Calculate the velocity of the parachutist after falling $20\,m$ under gravity $(g = 10\,m/s^2)$.Using $v^2 = u^2 + 2as$,where $u = 0$,$a = 10\,m/s^2$,and $s = 20\,m$:$v^2 = 0 + 2 	imes 10 	imes 20 = 400$$v = 20\,m/s$.Step $2$: Consider the motion after the parachute opens.The initial velocity for this phase is $u' = 20\,m/s$,the final velocity is $v' = 4\,m/s$,and the deceleration is $a' = -2\,m/s^2$.Using $v'^2 = u'^2 + 2a'h$,where $h$ is the distance covered after the parachute opens:$4^2 = 20^2 + 2(-2)h$$16 = 400 - 4h$$4h = 400 - 16 = 384$$h = 96\,m$.Step $3$: Calculate the total height $H$.$H = 20\,m + 96\,m = 116\,m$.

$A$ parachutist,after bailing out,falls $20\,m$ without friction. When the parachute opens,it decelerates at $2\,m/s^2$. He reaches the ground with a speed of $4\,m/s$. At what height did he bail out (in $,m$)?

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