A stone is dropped from a height h. Simultaneously, another stone is thrown up from the ground which reaches a height 4h. The two stones cross each other after time

$\text{ For first stone }u=0\text{ and }$

$\text{ For second stone }\frac{{\mathrm{u}}^2}{2\mathrm{g}}=4\mathrm{h}\Rightarrow {\mathrm{u}}^2=8\mathrm{gh}$

$\therefore \mathrm{u}=\sqrt{8\mathrm{gh}}$

${\mathrm{Now}}_{2,}{\mathrm{h}}_1=\frac{1}{2}{\mathrm{gt}}^2;{\mathrm{h}}_2=\sqrt{8\mathrm{gh}}\mathrm{t}-\frac{1}{2}{\mathrm{gt}}^2$

Where   t is the time to cross each other.$::h_1+h_2=h$

$$\begin{gathered} \Rightarrow \frac{1}{2}{\mathrm{gt}}^2+\sqrt{8\mathrm{gh}}\mathrm{t} -\frac{1}{2}{\mathrm{gt}}^2=\mathrm{h} \\ \mathrm{t}=\frac{\mathrm{h}}{\sqrt{8\mathrm{gh}}}=\sqrt{\frac{\mathrm{h}}{8\mathrm{g}}} \end{gathered}$$