A body is thrown vertically upwards. If air resistance is to be taken into account, then the time during which the body rises is

Let the initial velocity of ball be u.

Time of rise ${\mathrm{t}}_1 = \frac{\mathrm{u}}{\mathrm{g}+\mathrm{a}} \mathrm{and} \mathrm{height} \mathrm{reached} \mathrm{h} =\frac{{\mathrm{u}}^2}{2(\mathrm{g}+\mathrm{a})}$

Time of fall ${\mathrm{t}}_2 \mathrm{is} \mathrm{given} \mathrm{by}$

$\frac{1}{2}(\mathrm{g}-\mathrm{a}){{\mathrm{t}}_2}^2 = \frac{{\mathrm{u}}^2}{2(\mathrm{g}+\mathrm{a})}$

$\Rightarrow {\mathrm{t}}_2 = \frac{\mathrm{u}}{\sqrt{(\mathrm{g}+\mathrm{a})(\mathrm{g}-\mathrm{a})}} = \frac{\mathrm{u}}{(\mathrm{g}+\mathrm{a})}\sqrt{\frac{\mathrm{g}+\mathrm{a}}{\mathrm{g}-\mathrm{a}}}$

$\therefore {\mathrm{t}}_2 > {\mathrm{t}}_1 \mathrm{because} \frac{1}{\mathrm{g}+\mathrm{a}} < \frac{1}{\mathrm{g}-\mathrm{a}}$