A ball of relative density 0.8 falls into water from a height of 2 m. The depth to which the ball will sink is

$a$ = deceleration of the ball = $\frac{\left({\rho }_w\frac{m}{\sigma }g-mg\right)}{m}=\left(\frac{{\rho }_w}{\sigma }-1\right)g$

$=\left(\frac{1}{0.8}-1\right)\times 10m/s^2=2.5m/s^2$

Now u = velocity of the ball at water surface

$\begin{array}{l}=\sqrt{2\times 10\times 2}m/s=2\sqrt{10}m/s\\ \therefore h=\frac{u^2}{2a}=\frac{40}{2\times 2.5}m=8m\end{array}$