Saved Bookmarks
| 1. |
A parachutist drops from an aeroplane for `10s` before the parachute opens out. Then he descends with a net retardation of `2.5ms^(-2)`. If he bails out of plane at a height of `2495m` and `g=10ms^(-2)`, his velocity on reaching the ground will veA. `2.5ms^(-1)`B. `7.5ms^(-1)`C. `5ms^(-1)`D. `10ms^(-1)` |
|
Answer» Correct Answer - C Distance covered in first `10 sec` `S_(i)=(1)/(2)(10)(10)^(2)=500m` Remaining height from ground `=2495-500=1995m` `u=gt=10xx10=100m//s` velocity on reaching the ground `v^(2)=(100)^(2)+2(-2.5)xx1995` `v^(2)=10000-9975=25` `v=5m//s`. |
|