Question

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?

Answer 1

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`