Question

An elevator with a boy in it is moving upwards with a constant acceleration of 1 m/s2 at some intant when speed of elevator is 10 m/s the time taken by the stone to reach the ground of the elevator

Answer 1

Component of velocity of ball relative to lift are

ux=4cos30∘=23–√m/sux=4cos⁡30∘=23m/s

uy=4sin30∘=2m/suy=4sin⁡30∘=2m/s

and acceleration of ball relative to lift is a=10+2=12m/s2a=10+2=12m/s2 in negative y direction acting downwards.

Therefore Time of flight T=2uyaT=2uya

=2×212=2×212

=13s=13s