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.

An open elevator is ascending with constant speed `v=10m//s.` A ball i

1.	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.
Answer» Correct Answer - A (a) Absolute velocity of ball=`40 m//s` (upwards) `:. h_(max)=h_i+h_f` Here, `h_i`=initial height =`10m` and `h_f`=further height attained by ball `=u^2/2g=(40)^2/(2xx10)=80m` `:. h_(max) =(10+80)m=90m` (b) The ball will meet the elevator again when displacement of lift=displacement of ball or `10xxt=40xxt-1/2xx10xxt^2` or `t=6s` (c) Let `t_0` be the total time taken by the ball to each the ground. Then, `-10=40xxt_0-1/2xx10xxt_0^2` Solving this equation we get, `t_0=8.24s` Therefore, time taken by the ball to reach the ground after crossing the elevator, `=(t_0-t)=2.24s`

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