When a body is thrown up in a lift with a velocity `u` relative to the lift, the time of flight is found to be `t`. The acceleration with which the lift is moving up is

Options

A. `("u-gt")/(t)`
B. `("u+gt")/(t)`
C. `("2u-gt")/(t)`
D. `("2u+gt")/(t)`

Answer

C

Explanation

We have , `S_(rel)=u_(rel)t+(1)/(2)a_(rel)t^(2)`
`implies 0 = ut-(1)/(2)(a+g)t^(2)`
`implies a = (2u)/(t)-g=(2u-"gt")/(t)`