A parachutist drops first freely form an areophone for `10 s` and then his parachut opens out. Now he descends with a net retardtion of `2.5 ms^(-2)` If the balil 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^(-1)`
B. `10 ms^(-1)`
C. `15 ms^-1)`
D. `20 ms^(-1)`

Answer

A

Explanation

The velocity `v` acquired by the parachutist after `10 s`
`v=u+g t=0+ 10 xx 10 =100 ms^(-1)`
Then, `s_(1)=ut +(1)/(2) g t^(2) =0+(1)/(2) xx 10^(2) =500 m`
The distance travelled by the parachutist under retardation is
`s_(2) =2495 -500 =1995 m`
Let `v_(g)` be the velocity on reaching the ground. Then
`v_(g)^(2)-v^(2) =2 as_(2)`
or ` v_(g)^(2) -(100)^(2) =2 xx (-2.5) xx 1995` or `v_(g) ms^(-1)`.