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

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Question

(A) $1$. Free fall phase (from point $A$ to $B$): The parachutist falls freely under gravity for a distance $s_1 = 50\, m$. Initial velocity $u_1 = 0$

Vedclass · Accepted Answer

$293$

Vedclass · Answer

(A) $1$. Free fall phase (from point $A$ to $B$): The parachutist falls freely under gravity for a distance $s_1 = 50\, m$. Initial velocity $u_1 = 0$,acceleration $a_1 = 9.8\, m/s^2$.The velocity $v$ at point $B$ is given by $v^2 = u_1^2 + 2a_1s_1 = 0 + 2 	imes 9.8 	imes 50 = 980$.So,$v = \sqrt{980}\, m/s$.$2$. Deceleration phase (from point $B$ to $C$): The parachute opens,and the parachutist decelerates at $a_2 = -2\, m/s^2$. The final velocity at the ground is $v_f = 3\, m/s$. Let the distance covered be $h$.Using $v_f^2 = v^2 + 2a_2h$:$(3)^2 = 980 + 2(-2)h$$9 = 980 - 4h$$4h = 980 - 9 = 971$$h = 971 / 4 = 242.75\, m$.$3$. Total height: The total height from which he bailed out is $H = s_1 + h = 50 + 242.75 = 292.75\, m \approx 293\, m$.

$A$ 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 (in $, m$)?

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