Time taken by coin to reach the floor is given by$:$

$h=1 / 2 \mathrm{gt}^{2}$            $(\because u=0)$

or $t=\sqrt{\frac{2 h}{g}}$

In stationary lift, $\mathrm{t}_{1}=\sqrt{\frac{2 \mathrm{h}}{\mathrm{g}}}$

In upward moving lift with constant acceleration

a,

$\mathrm{g}^{\prime}=\mathrm{g}+\mathrm{a}$

$\therefore t_{2}=\sqrt{\frac{2 h}{(g+a)}}$

Clearly, $\quad \mathrm{g}^{\prime}>\mathrm{g}$

$\text { Thus, } \quad \mathrm{t}_{2}<\mathrm{t}_{1}$