The velocity of ball before entering the water surface $v =\overline{2 gh }=\overline{2 g } X$ When ball enters into water, due to upthrust of water, the velocity of ball decreases (or retarded) The retardationm $a =\frac{\text { apparent weight }}{\text { mass of ball }} $ $=\frac{ v (\rho-\sigma) g }{ V \rho}=\frac{(\rho-\sigma}{=} \frac{(\rho-\sigma) g }{\rho}$ $=\left(\frac{(0.4-1)}{0.4}\right) g =-\overline{2} g$ If $h$ be the depth upto which ball $\sin x$, then $0- v ^{2}=2 \times h \left(\frac{-3}{2} g \right) X$ $\Rightarrow 2 g \times 9=3 \,gh $ $\therefore h =6\, cm$