A body is thrown vertically upwards. If ai

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 $\frac{1}{{g + a}} < \frac{1}{{g - a}}$