A wooden ball of density D is immersed in water of density d to a depth h and then released . The height  above the surface of water up to which the ball will jump out of water is ( neglect viscous effect )

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{ball} \mathrm{be} V. \mathrm{Force} \mathrm{on} \mathrm{the} \mathrm{ball} \mathrm{due} \mathrm{to} \\ \mathrm{upthrust} =V \mathrm{d} \mathrm{g} \\ \mathrm{Net} \mathrm{upward} \mathrm{force} =\mathrm{V} \mathrm{d} \mathrm{g} -\mathrm{V} \mathrm{D} \mathrm{g} \\ .\text{'}. \mathrm{Upward} \mathrm{accelemtion} \mathrm{is} \mathrm{given} \mathrm{bY} \\ \mathrm{VDa}=\mathrm{Vdg}-\mathrm{VDg} \\ \mathrm{a}=\left(\frac{\mathrm{d}-\mathrm{D}}{\mathrm{D}}\right)\mathrm{g} \\ \text{ velocity on reaching the surface, u}=\sqrt{2\mathrm{ah}} \\ \text{ Further u}=\sqrt{2gH} \\ \therefore 2\mathrm{ah}=2\mathrm{gH}\text{ or }\mathrm{H}=\frac{\mathrm{ah}}{\mathrm{g}} \\ \text{ So, }\mathrm{H}=\left(\frac{\mathrm{d}-\mathrm{D}}{\mathrm{D}}\right)\mathrm{h}=\left(\frac{\mathrm{d}}{\mathrm{D}}-1\right)\mathrm{h} \end{gathered}$$