An open elevator is ascending with zero acceleration. The speed v=10m/s. A ball is thrown vertically up by a boy when he is at a height h=10 m from the ground. The velocity of projection is v= 30 m/s with respect to elevator. Find 1. The maximum height attained by the ball. 2. The time taken by the ball to reach the ground after crossing the elevator.
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Answer:
Explanation:
VELOCITY OF BALL = VELOCITY OF BALL WITH RESPECT TO ELEVATOR + VELOCITY OF ELEVATOR
V=40 m/s
max height = u^2/2g
=1600/20 = 80m
elevator displacement when ball is at max height
s=ut+1/2at^2
80 = 40t -5t^2
t^2-8t+16 = 0
t=4 sec
s=vt
s=40 m
time taken by ball to cover 40 m
40=5t^2
t=2.828
time taken to reach ground after crossing the elevator = 4-2.828 =1.172 sec
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