A ball whose density is 0.4 x 10³ kg/m³ falls into water from a height of 9 cm . To what depth does the ball sink?

The velocity of ball before entering the water surface:

$v=\sqrt{2gh}=\sqrt{2g\times 9}$

When ball enters into water, due to the upthrust of water, the velocity of ball decreases (or retarded).

$\text{ Retardation, }a=\frac{\text{ Apparent weight }}{\text{ Mass of ball }}$

$=\frac{V(\rho -\sigma )g}{V\rho }=\left(\frac{\rho -\sigma }{\rho }\right)g=\left(\frac{0.4-1}{0.4}\right)\times g=-\frac{3}{2}g$

lf h be the depth up to which the ball sinks. Then

$\begin{array}{rcl}0-v^2& =& 2\times \left(-\frac{3}{2}g\right)\times h\Rightarrow 2g\times 9=3gh\\ \therefore h& =& 6\mathrm{cm}\end{array}$