A coin is released inside a lift at a ht of 2m from the floor of the lift. The ht of the lift is 10 m . the lift is moving with an acceleration of 9m/s² downwards. The time after which the coin will strike with lift is: (g= 10 m/s)

Answer 1

Both have the same initial velocity. The coin has the greater acceleration. The coin has to travel 2 m further than the bottom of the lift.

distancelift=vi+9t^2/2
distancecoin=vi+10t^2/2
or
vi+9t^2/2-2=vi+10t^2/2
.5t^2=2

t=2 seconds check my math.