A ball whose density is $0.4\times {10}^3\mathrm{kg}/{\mathrm{m}}^3$ falls into water from a height of 9m. To what maximum depth in metre does the ball sink?

Velocity of the ball as it reaches water surface: 

$u=\sqrt{2gh}=\sqrt{18g} m/s$

ρ = density of ball,  σ = density of water

Net upward force acting on the body: $\mathrm{F}=\sigma \mathrm{Vg}-\rho \mathrm{Vg}$

$\begin{array}{r}\Rightarrow \mathrm{ma}=\sigma \mathrm{Vg}-\rho \mathrm{Vg}\\ \Rightarrow \left(\rho \mathrm{V}\right)\mathrm{a}=\mathrm{Vg}(\sigma -\rho )\end{array}$

$\Rightarrow \mathrm{a}=\mathrm{g}\left(\frac{\sigma }{\rho }-1\right)=\mathrm{g}\left(\frac{1}{0.4}-1\right)=\frac{3\mathrm{g}}{2} \left(upward\right)$

at the lowest depth, final velocity: v=0,

$\Rightarrow 0=(\sqrt{18g}{)}^2-2\times \frac{3g}{2}\times H$

$\Rightarrow 18g=3gH$

$\Rightarrow \mathrm{H}=6\mathrm{m}$