A spherical ball of mass m is dropped into a viscous fluid from height ‘h’ and continues moving with constant speed in fluid. The density of ball is three times the density of fluid and viscosity of fluid is assumed to be independent of depth. Let us assume that ball looses 50% of its KE while striking the fluid surface every time. If the ball is dropped from a height ‘4h’ now, then

In the case of dropping from height ‘h’
Speed just before striking $=\sqrt{2\mathrm{gh}}={\mathrm{v}}_0$
Speed just after striking $=\frac{{\mathrm{v}}_0}{\sqrt{2}}=\sqrt{\mathrm{gh}}={\mathrm{v}}_{\mathrm{T}}$
In the case of dropping from height ‘4h’
Just before striking ${\mathrm{v}}_0^1=\sqrt{8\mathrm{gh}}$
Just after striking $\mathrm{v}=\frac{{\mathrm{v}}_0^1}{\sqrt{2}}=\frac{2\sqrt{2\mathrm{gh}}}{\sqrt{2}}=2{\mathrm{v}}_{\mathrm{T}}$
Also viscous force belonging to ${\mathrm{v}}_{\mathrm{T}}\hspace{0.17em}: {\mathrm{F}}_{\mathrm{v}}=\mathrm{W}-\mathrm{B}=\mathrm{mg}-\frac{\mathrm{mg}}{3}=\frac{2\mathrm{mg}}{3}$
$\therefore$Viscous force belong to $2{\mathrm{v}}_{\mathrm{T}}: {\mathrm{F}}_{\mathrm{v}}\text{'}=\frac{4}{3}\mathrm{mg}$

net upward force : $\mathrm{F}{\text{'}}_{\mathrm{up}}=\mathrm{B}+{\mathrm{F}}_{\mathrm{v}}\text{'}=\frac{1}{3}\mathrm{mg}+\frac{4}{3}\mathrm{mg}=\frac{5}{3}\mathrm{mg}$

net force acting on the ball just inside the fluid: ${\mathrm{F}}_{\mathrm{net}}=\mathrm{F}{\text{'}}_{\mathrm{up}}-\mathrm{W}=\frac{5}{3}\mathrm{mg}-\mathrm{mg}=\frac{2}{3}\mathrm{mg}$