The velocity of ball before entering the water surface

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

when ball enters into water, due to up thrust of water the velocity of ball decreases (or retarded)

The retardation,

             $\text{a}=\frac{\text{apparent weight}}{\text{mass of ball}}$

             $=\frac{\text{V}\left(\rho -\sigma \right)\text{g}}{\text{V}\rho }=\frac{\left(\rho -\sigma \right)\text{g}}{\rho }$

              $=\frac{0\text{.}4-1}{0\text{.}4}\text{ g}=-\frac{3}{2}\text{g}$

If h be the depth upto which ball sin k, then

            $0-{\text{v}}^2=2\times \frac{-3}{2}\text{g}\times \text{h}$

  $\Rightarrow 2\text{g}\times 9=3\text{gh}$

   $\therefore \text{h}=6\text{ cm.}$