The velocity of ball just before striking the surface of water is given by using kinematical equation of motion under the gravity
v2=u2−2gy
⇒v2=−2g(−h)
⇒v=2gh...(1)
Now, the terminal velocity of spherical ball inside the water tank is given by,
vT=92ηr2(ρ−σ)g...(2)
After falling through h the velocity of ball should be equal to its terminal velocity.
From equation (1) and (2) we get,
⇒2gh=92ηr2(ρ−σ)g
Squaring the above equation, we get
⇒h=812η2r4g(ρ−σ)2
⇒h∝r4