A coin is released inside a lift at a height of 2m from the floor of the lift. The height of the is 10 m. The lift is moving with an acceleration of 11m/s downwards.The time after which the coin will strike with the lift is?
Answers
Answered by
10
distance to cover=10-2=8m. acceleration=11-10=1m/s^2. initial velocity=0
s=it+1/2at^2
8=0+1/2(1)t^2
t^2=16
t=+-4=4 as time is always positive.
s=it+1/2at^2
8=0+1/2(1)t^2
t^2=16
t=+-4=4 as time is always positive.
Similar questions