An open elevator is ascending with constant speed `v=10m//s.` A ball is thrown vertically up by a boy on the lift when he is at a height `h=10m` from the ground. The velocity of projection is `v=30 m//s` with respect to elevator. Find
(a) the maximum height attained by the ball.
(b) the time taken by the ball to meet the elevator again.
(c) time taken by the ball to reach the ground after crossing the elevator.

Explanation

(i) Absolute velocity of ball = 40 m/s (upwards)
`therefore h_(max)=h_(i)+h_(f)`
Here, `h_(i)` = initial height = 10 m
and `h_(f)` = further height attained by ball
`=(u^(2))/(2g)=((40)^(2))/(2xx10)=80 m`
`therefore h_(max)=(10+80)m=90 m`
(ii) The ball will meet the elevator again when displacement of lift = displacement of ball
or `10xxt=40xxt-(1)/(2)xx10xxt^(2)` or t = 6 s
(iii) Let `t_(0)` be the total time taken by the ball to reach the ground. then,
`-10=40xxt_(0)-(1)/(2)xx10xxt_(0)^(2)`
Therefore, time taken by the ball to reach the ground after crossing the elevator `=(t_(0)-t)=2.24 s`