A coin is thrown vertically upward in a lift which is ascending with acceleration $a$. The coin takes time $t_1$ to reach the maximum height $h$. In the later instant, the coin is dropped from the same height in a lift which is descending with acceleration $a$. The time taken by the coin to reach the floor of the lift is $t_2$, then

In the frame of the ascending lift, the effective acceleration is $g_{eff} = g + a$.

For the coin thrown upward to reach a maximum height $h$ relative to the lift, the final velocity is zero.

Using $v^2 = u^2 - 2g_{eff}h$, we get $u = \sqrt{2(g+a)h}$.

The time taken to reach the maximum height is $t_1 = \dfrac{u}{g_{eff}} = \sqrt{\dfrac{2h}{g+a}}$.

In the frame of the descending lift, the effective acceleration is $g'_{eff} = g - a$.

For the coin dropped from height $h$, the initial velocity relative to the lift is zero.

Using $h = \dfrac{1}{2}g'_{eff}t_2^2$, we get $t_2 = \sqrt{\dfrac{2h}{g-a}}$.

Since $g - a < g + a$, it follows that $\sqrt{\dfrac{2h}{g-a}} > \sqrt{\dfrac{2h}{g+a}}$.

Therefore, $t_2 > t_1$.

Answer: $t_2 > t_1$