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Question

A parachutist after bailing out falls 50 m without friction. When parachute opens, it decelerates at 2 m/s2. He reaches the ground with a speed of 3 m/s. At what height, did he bail out?


A

293 m

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B

111 m

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C

91 m

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D

182 m

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Solution

The correct option is A

293 m


After bailing out from point A parachutist falls freely under gravity. The velocity acquired by it at the end of free fall is v.

From, v2=u2+2as=0+2×9.8×50=980
[As u=0, a=9.8 m/s2, s=50 m]
So, v2=980


At point B, parachute opens, and it moves with retardation of 2 m/s2 and reach at ground (Point C) with velocity of 3 m/s.

For the part 'BC' by applying the equation we get:
v2=u2+2as
v=3 m/s,u=980,a=2 m/s2,s=h

(3)2=(980)2+2×(2)×h9=9804h

h=98094=9714=242.7243 m

So, the total height by which parachutist bail out
=50+243=293 m


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