A ball of relative density 0.8 falls into water from a height of 2m. The depth upto which the ball will sink is, (in m),  (neglect viscous forces).

Let ‘a’ be the retardation of the body in side the water Fig
$\therefore$$a=\frac{v\rho g-mg}{m}$$=\frac{v{\rho }_{w}g-v{\rho }_{o}g}{V{\rho }_b}$                              ${\rho }_r=\frac{{\rho }_b}{{\rho }_w}$

$=\frac{{\rho }_w}{{\rho }_b}g-g$

$=\frac{{\rho }_w}{{\rho }_r{\rho }_w}g-g=g\left(\frac{1}{{\rho }_r}-1\right)$$=g\left(\frac{1}{0.8}-1\right)$

$a=g\left(\frac{10}{8}-1\right)$$=g\left(\frac{\cancel{2}}{\cancel{8}4}\right)=\frac{g}{4}$

From $v^2-u^2=2ag$

$v^2-u^2=\cancel{2}\left(-\frac{g}{\cancel{4}}\right)h_1$   $0\cancel{-}39.2=\cancel{L}\frac{g{h}_1}{2}$

$h_1=\frac{{\cancel{39.2}}^4\times 2}{\cancel{9.8}}$

$h_1=8\text{m}$