Which should be the correct option? Question--

A person standing on the floor of an elevator drops a coin. The coin reaches the floor of the elevator in time t1 if the elevator is stationary and in time t2 if it is moving uniformly. Then
(a) t1 = t2 
(b) t1 < t2 
(c) t1 > t2 
(d) t1 < t2
  or
t1 > t2 depending on whether the lift is going up or down.

Explanation-
As the elevator is moving at uniform speed, so it's acceleration is zero, so, no pseudo force. Thus it can not affect the motion of the coin. Thus in both cases, the coin takes the same time. i.e, t1=t2.Therefore correct option is (a).
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But I am not satisfied with the above explanation and the answer.
I think that option (d) is correct.
Let the coin be dropped from a height h from the floor of the elevator.
Suppose, if the elevator is moving down, then the coin will have to travel a distance h+H(>h), where H is the distance travelled by the elevator in the time the coin reaches it's floor. Since the coin has to travel a greater distance than previous, with the same acceleration, so it should take more time than t1.
Similarly, if the lift moves up the coin will have to travel lesser distance to reach the bottom. And thus it should take less time.
Correct me, if I am wrong.
(This question is from HC Verma's "Concepts of Physics".)
 A: The thing you're missing is that if the elevator is travelling uniformly at some velocity $v$, the coin starts with that same velocity.
So taking our SUVAT equation - in the case where the elevator isn't moving, you have:
$$\tfrac12 g t_1^2  = h$$
and when it is, $H$ is clearly equal to $v t_2$ so:
$$v t_2 + \tfrac12 g t_2^2 = h + H = h + v t_2$$
$\implies \tfrac12 gt_2^2 = h$ after cancellation.
The constant motion of the elevator cancels out!
This is actually a very general principle of physics: the speed you're moving at doesn't affect the outcome of experiments,  if it's a constant speed. Look up the strong equivalence principle if you're interested!
A: you are missing one small thing that initial velocity is not the same in both cases if you look from the outside frame of reference. in the second case, the initial velocity of the ball will be the same as that of elevator. 
also let's look it from the inside frame of reference. inital velocity will be zero. distance to be covered will be same as well. so value of everything is same as in the previous rest case.
