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 when he reaches the ground is 8m/sec. Find te height at which he gets out of the aeroplane?

A parachutist drops freely from an aeroplane for 10 seconds before the

1.	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 when he reaches the ground is 8m/sec. Find te height at which he gets out of the aeroplane?
Answer» SOLUTION :Distance he falls before the parachute OPENS is `1/2g xx 100=490m` 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)=2 xx (-2)S` `S=(106 xx 90)/(4)=2385m` Total distance =2385+490 =2875m=height of aeroplane.