A wooden ball of density $\rho$ is immersed in water of density ${\mathrm{\rho }}_0$ to depth h and then released. The height H above the surface of water upto which the ball jump out of water is

Let V be the volume of wooden ball. The mass of ball is m = V$\rho$.

Upward acceleration,

$\begin{array}{l} \mathrm{a}=\frac{\text{ upward thrust-weight of ball }}{\text{ mass of ball }}\\ \Rightarrow \mathrm{a}=\frac{{\mathrm{V\rho }}_0\mathrm{g}-\mathrm{V\rho g}}{\mathrm{V\rho }}=\frac{\left({\mathrm{\rho }}_0-\mathrm{\rho }\right)\mathrm{g}}{\mathrm{\rho }}\end{array}$

If v' is the velocity of ball on reaching the surface after being released at depth his

$\mathrm{v}=\sqrt{2\mathrm{as}}={\left[2\left(\frac{{\mathrm{p}}_0-\mathrm{\rho }}{\mathrm{\rho }}\right)\mathrm{gh}\right]}^{1/2}$

If H is the vertical distance reached by ball above the surface of water, then

$\begin{array}{c}\mathrm{H}=\frac{{\mathrm{v}}^2}{2\mathrm{g}}=\frac{2\left({\mathrm{\rho }}_0-\mathrm{\rho }\right)}{\mathrm{\rho }}\mathrm{gh}\times \frac{1}{2\mathrm{g}}\\ =\left(\frac{{\mathrm{\rho }}_0-\mathrm{\rho }}{\mathrm{\rho }}\right)\mathrm{h}=\left(\frac{{\mathrm{\rho }}_0}{\mathrm{\rho }}-1\right)\mathrm{h}\end{array}$