Explanation

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