A parachutist after bailing out falls `80 m` without friction. When the parachute opens, he decelerates downward with `5m//s^(2)` and reaches the ground with a speed of `10 m//s`. (a) How long is the parachutist in air? (b) At what height did bail out?

Explanation

Here, motion is in two parts. In the first part, the parachutist falls freely and after opening of parachute, he falls with retardation.

`O` to `A`(free fall)
`v_(0)^(2)=0+2g h_(1)=2xx10xx80impliesv_(0)=40 m//s`
`h_(1)=0+1/2g t_(1)^(2)`
`80=5t_(1)^(2)impliest_(1)=4 s`
`A` to `B` (motion under constant retardation)
`v=v_(0)+a_(2)t_(2)`
`10=40-5t_(2)impliest_(2)=6s`
`v^(2)=v_(0)^(2)+2a_(2)h_(2)`
`(10)^(2)=(40)^(2)-2xx5xxh_(2)`
`h_(2)=((40)^(2)-(10)^(2))/10=150 m`
Total time: `t_(1)+t_(2)=4+6=10 s`
Total height: `h_(1)+h_(2)=80+150=230m`