A ball is thrown upwards with a speed u from a height h above the ground. The time taken by the ball to hit the ground is 

Time taken to reach the highest point from the height h is obtained from equation, $\mathrm{v} = \mathrm{u} - \mathrm{gt}$ 

$\therefore$                           $0=u-g{t}_1\text{ or }{t}_1=\frac{u}{g}$

                                           ($\because$ At highest point, final velocity of ball = 0)

Height attained above h is obtained from ${\mathrm{v}}^2 - {\mathrm{u}}^2 = 2 (-\mathrm{g}){\mathrm{h}}_1$

or               $0-u^2=2(-g)h_1\text{ or }{h}_1=\frac{u^2}{2g}$

Total height, $h_2=\left(h_1+h=\frac{u^2}{2g}+h\right)$

Time taken to hit the ground is obtained from 

                    $h_2=ut+\frac{1}{2}a{t}^2\text{ or }\frac{u^2}{2g}+h=0+\frac{1}{2}g{t_2}^2$

$\therefore$   , $t_2=\sqrt{\frac{\left(u^2+2gh\right)}{g^2}}$

Therefore , total time , T = $t_1+t_2 =\frac{u}{g}[1+\sqrt{1+\frac{2gh}{u^2}}]$