Explanation

(b) Let the initial velocity of ball be $u$
Time of rise $t_{1} = \frac{u}{g + a}$ and height reached $= \frac{u^{2}}{2(g + a)_{\ }}$
Time of fall $t_{2}$ is given by
$\frac{1}{2}(g - a)t_{2}^{2} = \frac{u^{2}}{2(g + a)}$
$\Rightarrow t_{2} = \frac{u}{\sqrt{(g + a)(g - a)}} = \frac{u}{(g + a)}\sqrt{\frac{g + a}{g - a}}$
$\therefore t_{2} > t_{1}$ because $\frac{1}{g + a} < \frac{1}{g - a}$