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Question

# A parachutist after bailing out falls 80 m without air resistance. Then the parachute is opened, and he decelerates at a uniform rate of 2m/s2. If he reaches the ground with a speed of 4m/s, then the height at which he bailed out is [g=10m/s2]

A
576m
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B
676m
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C
476m
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D
376m
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Solution

## The correct option is C 476mFinal velocity after falling 80m,V2s=0+2×10×80(vc=0)Vs=√1600Vs1=40m/sFinal velocity when reaching ground =4 m/s(Vs2)2=V2s1−2ah42=402−4hh=(40−4)(40+4)4h=9×44=396Total height of bailing =396+80=476m

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