 # A person standing on the floor of an elevator drops a coin. The coin reaches the floor of the elevator in a time t1 if the elevator is stationary and in time t2 if it is moving uniformly. Then: (a) t1 = t2 (b) t1 t2 or t1   I feel that the answer is d. When the coin is detached from hand of the person, it is on its own and accelerated downward. But at the same elevator is moving up which is now a separate system. So it seems that the coin will have to cover a smaller distance than in the case if the elevator was stationary since both are approaching each other. If I am in the elevator I would definitely see that the coin covers same distance and I would do with answer a. But if I am on ground, I would see the elevator and coin approaching towards each other. What is actually correct?

5 years ago
Let me take both of them.
From lift frame:
when the lift is moving uniformly,inside the lift the condition would be the same as it would have been at rest,becauise both are non-accelerated motion.
So here we can easily say t1=t2.
From ground frame:
Listen in ground frame ,the person in the lift,coin and the lift itself are moving up at constant speed before droping of the coin.But when the coin is dropped by the person actually the moving person gives the coin an initial upward velocity.This will not remain as a simple case of dropping with initial velocity 0m/s( as in the case of lift frame of refrence) but it would be a case of throwing a ball upwards against gravity.So the even though the distance decreases from release point to lift floor,the time remains the same,you can verify mathematically.
Hence A is correct t1=t2.
I have put my best to explain.If you are unable to get please do write I will try to explain again.And please If my answer is correct do approve my answer!!