Explanation

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