A body is dropped from a height of `40 m`. After it crosses half distance, the acceleration due to gravity ceases to act. The body will hit ground with velocity `("Take" g = 10 m//s^2)` :

Options

A. `20 m//s`
B. `10 m//s`
C. `2 m//s`
D. `15 m//s`

Answer

A

Explanation

Suppose `v` be the velocity of the body after falling through half the distance. Then
`s =(40)/(2) = 20 m, u = 0 and g = 10 m//s^2`
`v^2 = u^2 + 2gh = 0^2 + 2 xx 10 xx 20`
`rArr v = 20 m//s`.
When the acceleration due to gravity ceases to act, the body travels with the uniform velocity of `20 m//s`. So it hits the ground with velocity `20 m//s`.