A lift is movingup with a constant acceleration a.A man standingon 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.

A lift is movingup with a constant acceleration a.A man standingon the

1.	A lift is movingup with a constant acceleration a.A man standingon 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.
Answer» Solution :The downward acceleration of theball with respect to the LIFT =g(-a)=g+a. The initialvelocity 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.`