A lift is moving up with a constant acceleration a. A man standing on the lift, throws a ball vertically upwards with a velocity v, which returns to the thrower after a time t. show that `v=(a+g) t/2` where g is the acceleration due to gravity.

Explanation

The downward acceleration of the ball with respect to the lift =g(-a)=g+a.
The initial velocity of the ball is v upward. As the ball returns to the thrower, its relative displacement is zero.
Hence, from the equation h=`ut-1/2"gt"^2,` we get ,
`0=vt-1/2(g+a)t^2 or, 0=r{v-1/2(g+a)t}`
Since `tne0`,
`therefore v-1/2(g+a)t=0 or ,v=(a+g)t/2.`