A parachutist drops freely from an aeroplane for 10 seconds before the parachute opens out. Then he descends with a net retardation of `2 m//s^(2)`. His velocity when he reaches the ground is `8 m//s` find the height at which he gets out of the aeroplane?

Explanation

Distance he falls before parachute opens is
`S_(1)=(1)/(2)gxx100=490m`
Then his velocity `,u=g t=98.0 m//s`
Velocity on reaching ground `=v=8 m//s`
Retardation `2= m/s^(2)`
`v^(2)u^(2)=2aS_(2) impliesS_(2)=2385m`
Total distance `S=S_(1)+S_(2)=2385+490`
`=2875m` (height of aeroplane)