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. ` 2.5 ms ^(-1) `
B. ` 7.5 ms ^(-1) `
C. ` 5 ms ^(-1) `
D. ` 10 ms ^(-1) `

Answer

C

Explanation

Velocity at the time of opening the parachute ` = 0+ 10 ( 10 ) =100 m//s `
Distance traveled = ` 0 +( 1)/(2) (10 ) (100) =500 cm " " `Height remaining ` = 2495 -500 = 1995`
Acceleration ` = - 2.5m//s^(2) `
` v^(2) =u^(2) +2as =(100 )^(2) + 2(2.5) (1995) " "v^(2) = 25 " "rArr v= 5m//s `