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

Explanation

Distance he falls before the parachute opens is `1/2 g xx 100` =490 m
Then his velocity = gt = 98.0 m/s = u
Velocity on reaching ground =8=v
retardation =2
`v^2-u^2=2as`
`8^2-(98)^2=2xx(-2)S therefore S=(106xx90)/4` =2385 m
Total distance =2385+490
=2875 m = height of aeroplane