A parachutist drops first freely from a plane for `10 s` and then his parachute opens out. Now he descends with a net retardation of `2.5 ms^(-2)` If he bail out of the plane at a height of `2495 m` and `g=10 ms^(-2)`, his velocity on reaching the ground will be`.

Options

A. `5 m//s`
B. `10 m//s`
C. `15 m//s`
D. `20 m//s`

Answer

A

Explanation

Velocity after `10` sec is equal to
`0+(10)(10)=100 m//s`
Distance covered in `10` sec is equal to
`1/2 (10)(10)^(2)=50m`
Now from `v^(2)=u^(2)+2as`.
`rArr v^(2)=(100)^(2)-2(2.5)(2495-400)=25rArrv=5ms^(-1)`