A parachutist after bailing out falls 50, m w | vedclass

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Question

(D) Step $1$: Calculate the velocity after falling $50\, m$ freely.Using $v^2 = u^2 + 2as$,where $u = 0$,$a = g = 9.8\, m/s^2$,and $s = 50\, m$:$v_1^2

Vedclass · Accepted Answer

$293$

Vedclass · Answer

(D) Step $1$: Calculate the velocity after falling $50\, m$ freely.Using $v^2 = u^2 + 2as$,where $u = 0$,$a = g = 9.8\, m/s^2$,and $s = 50\, m$:$v_1^2 = 0 + 2 	imes 9.8 	imes 50 = 980\, m^2/s^2$.Step $2$: Calculate the distance covered during deceleration.Let $h_2$ be the distance covered after the parachute opens. The initial velocity is $v_1$,final velocity $v_f = 3\, m/s$,and deceleration $a = -2\, m/s^2$.Using $v_f^2 = v_1^2 + 2ah_2$:$3^2 = 980 + 2(-2)h_2$$9 = 980 - 4h_2$$4h_2 = 980 - 9 = 971$$h_2 = 971 / 4 = 242.75\, m \approx 243\, m$.Step $3$: Calculate the total height.Total height $H = h_1 + h_2 = 50 + 243 = 293\, m$.

$A$ parachutist after bailing out falls $50\, m$ without friction. When the 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?

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