Question

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

Answer 1

Answer:

In first part we are going to take out the final velocity reached when parachutist falls 50m without any friction.

We know that v²=u²+2gs

Since u =0

Therefore v²=0+2×10×50=1000

V=√1000=10√10 m/s

Now coming to the second part of the question

The velocity that we calculated in previous part now serves as the initial velocity for the motion of the parachutist after he falled 50 m

v²=u²-2as

u=10√10 m/s

a=-2m/s²

3²=(10√10)²-2×2×s

s=(9–1000)/-4=247.75meters

Final answer =50+247.75=297.75 m

Explanation: