A person standing on the floor of an elevator drops a coin. The...

Let the height from which the coin is dropped be $h$.

When the elevator is stationary, the initial relative velocity of the coin with respect to the elevator is zero. The relative acceleration of the coin is $g$ downwards. Using the equation of motion $S_{rel} = u_{rel} t + \dfrac{1}{2} a_{rel} t^2$, we get $h = \dfrac{1}{2} g t_1^2$, which gives $t_1 = \sqrt{\dfrac{2h}{g}}$.

When the elevator is moving uniformly, its acceleration is zero. The frame of reference of the elevator is an inertial frame. The initial relative velocity of the coin with respect to the elevator is still zero, and the relative acceleration of the coin is $g$ downwards. Using the same equation of motion, we get $h = \dfrac{1}{2} g t_2^2$, which gives $t_2 = \sqrt{\dfrac{2h}{g}}$.

Therefore, $t_1 = t_2$.