A ball is thrown vertically up in air where the resistive force can be considered to be constant. If time of ascent is $t_1$ and time of descent is $t_2$, then the ratio of speed of projection to the speed just before it hits the ground is

The acceleration of ball during upward motion is $g+a,$ where a is the acceleration due to resistive force. The time of accent, $t_1=\frac{u}{g+a}\left(1\right)$

The acceleration during descent g-a and time taken is $t_2=\frac{v}{g-a}\mathrm{......}\left(2\right)$ 

The time taken can also be written as $t_2=\sqrt{\frac{2h}{g-a}}=\sqrt{\frac{\frac{u^2}{g+a}}{g-a}}\Rightarrow t_2^2=\frac{u^2}{(g+a)(g-a)}\mathrm{.......}\left(3\right)$

Dividing equation (3) by equation (1) and (2) $\frac{t_2}{t_1}=\frac{u}{v}$