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`.A. `5ms^(-1)`B. `10ms^(-1)`C. `15ms^(-1)`D. `20ms^(-1)`

A parachutist drops first freely from a plane for `10 s` and then his

1.	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`.A. `5ms^(-1)`B. `10ms^(-1)`C. `15ms^(-1)`D. `20ms^(-1)`
Answer» Correct Answer - A The velocity v acquired by the parachutist after 10 s. `v=u+g t=9+10xx10=100ms^(-1)` Then, `s_(1)=ut+(1)/(2)g t^(2)=0+(1)/(2)xx10xx10^(2)=500m` The distance travelled by the parachutist under retardation is `s_(2)=2495-500=1995m` Let `v_(g)` be the velocity on reaching the ground then `v_(g)^(2)-v^(2)=2as_(2)` or `v_(g)^(2)-(100)^(2)=2xx(-2.5)xx1995` or `v_(g)=5ms^(-1)`

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