A ball is thrown upwards with a speed u from a height h above the...

Time taken to reach the highest point from the height h is obtained from equation, $\mathrm{v}=\mathrm{u}-\mathrm{gt}$ $\therefore \quad 0=\mathrm{u}-\mathrm{gt}$ or $\mathrm{t}=\frac{\mathrm{u}}{\mathrm{g}}$ Height attained above h is obtained from $\mathrm{v}^2-\mathrm{u}^2=2(-\mathrm{g}) \mathrm{h}_1$ or $\quad 0-\mathrm{u}^2=2(-\mathrm{g}) \mathrm{h}_1$ or $\mathrm{h}_1=\frac{\mathrm{u}^2}{2 \mathrm{~g}}$ Total height, $\mathrm{h}_2=\left[\mathrm{h}_1+\mathrm{h}=\frac{\mathrm{u}^2}{2 \mathrm{~g}}+\mathrm{h}\right]$ Time taken to hit the ground is obtained from, $$\begin{aligned} \mathrm{h}_2 & =\mathrm{ut}+\frac{1}{2} \mathrm{at}^2 \\ \frac{\mathrm{u}^2}{2 \mathrm{~g}}+\mathrm{h} & =0+\frac{1}{2} \mathrm{gt}^2\end{aligned}$$ $\therefore$ Total time taken, $\mathrm{t}=\sqrt{\frac{\left(\mathrm{u}^2+2 \mathrm{~g} \mathrm{~h}\right)}{\mathrm{g}^2}}$