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)gxx100=490m`
Then hus velocity =gt=98.0 m/s=u
Velocity in reaching ground =9=`vartheta`
retardation=2
`vartheta^(2)-u^(2)=2as" "8^(2)-(98)^(2)=2xx(-2)S`
`thereforeS=(106xx90)/(4)=2385m`
Total distance =2385+490