A parachutist flying in an aeroplane jumps when it is at a height of 3 km above ground. He opens his parachute when he is about 1 km above ground. Describe his motion.

Explanation

a) Height of fall before opening, h=2 km
=2000 m
`therefore` Velocity at a height of 1 km
`=sqrt(2gh_1)=sqrt(2times10times2000)`
`=sqrt(4000)=200ms^-1`
Time to fail,`t=sqrt((2h)/g)=sqrt((2times2000)/100)`
`=sqrt400 =20sec.`
b) After parachute is opened it touches the ground with almost zero velocity.
`thereforeu=200m//sec,v=0,S=h=1000m`
From `v^2-u^2=2as`
Retardation, `a=(0-200^2)/1000=-40m//sec^2`
Time taken to reach ground , `t=u/a`
`=200/40=5 sec`

The motion is as shown in figure.