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. `8 m//s`
B. `5 m//s`
C. `6 m//s`
D. `2.5 m//s`

Answer

B

Explanation

Suppose the man drops at `A`m from `A` to `B` he is falling freely and then at `B` parachute opens out and he falls with a retardation of `2.5 m//s^2`.
`:. AB = (1)/(2) xx 10 xx 10^2 = 500 m`
`:. BC = AC - AB = 2495 - 500 = 1995 m`.
Velocity at `B`,
`V_B = g t = 10 xx 10 = 100 m//s darr`
Velocity at `C`, `V_C = sqrt(V_B^2 + 2 ay)`
=`sqrt(100^2 + 2 xx 2.5 xx (-1995))`
=`sqrt(25) = 5 m//s darr`.