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 the 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 the height at which he gets out of the aeroplane?
Answer» Solution :Distance he FALLS before the parachute OPENS is `(1)/(2)gxx100=490m` Then his velocity = `g t=98.0m//s=u` Velocity on reaching ground = `8=vartheta` retardation = 2 `vartheta^(2)-u^(2)=2AS` `8^(2)-(98)^(2)=2xx(-2)SthereforeS=(106xx90)/(4)=2385m` TOTAL distance = `2385+490` = `2875m` = height of aeroplane