Question

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

Answer 1

Hello Dear !!
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ATQ potential energy = kinetic energy 
        mgh                    = 1/2mv²
        v                         = √2gh
        v                         = √2X10 X 50
        v                         = √1000 
        v                         = 10√10 m/sec
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Also , 
it decelerates at 2 m/s2. He reaches the ground with a speed of 3 m/s
here,
v = 3 m/s
u = 10√10 m/s
a = - 2 m/s²
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We know 
v²- u² = 2as 
3²- (10√10)² = -4s 
-991            = -4s 
s                 =   999/4
s                 = 247.75 m
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