A coin is dropped in a lift. It takes time t_(1) to reach the floor when lift is stationary. It takes time t_(2) when lift is moving up with constant acceleration. Then

A coin is dropped in a lift. It takes time t_(1) to reach the floor wh

1.	A coin is dropped in a lift. It takes time t_(1) to reach the floor when lift is stationary. It takes time t_(2) when lift is moving up with constant acceleration. Then
Answer» `t_(1) GT t_(2)` `t_(2) gt t_(1)` `t_(1)=t_(2)` `t_(1) ge t_(2)` Solution :For STATIONARY lift `t_(1)=sqrt((2h)/(G))` and when the lift is moving up with CONSTANT ACCELERATION `t_(2)=sqrt((2h)/(g+a))`