A body is thrown vertically upwards. If air resistance is to be taken into account, then the time during which the body rises is [assume no air resistance close to earth]

Solution :
Let the initial velocity of ball be $$u$$
$$\therefore $$ Time of rise $${t_1} = \frac{u}{{g + a}}$$  and height reached $$ = \frac{{{u^2}}}{{2\left( {g + a} \right)}}$$
Time of fall $${t_2}$$ is given by
$$\eqalign{ & \frac{1}{2}\left( {g - a} \right)t_2^2 = \frac{{{u^2}}}{{2\left( {g + a} \right)}} \cr & {t_2} = \frac{u}{{\sqrt {\left( {g + a} \right)\left( {g - a} \right)} }} \cr & = \frac{u}{{\left( {g + a} \right)}}\sqrt {\frac{{g + a}}{{g - a}}} \cr & \therefore {t_2} > {t_1}\,{\text{because}}\,\frac{1}{{g + a}} < \frac{1}{{g - a}} \cr} $$