lekh2003 said:
This question is quite interesting, I'm not well versed in drag and fluid movement, what would the terminal velocity of something like a regular sized basketball look like?
If I use the equation for turbulent drag and ignore the mass of the ball i get:
\frac{1}{2} \rho v^2 C_d \pi r^2 = \frac {4}{3} g \pi \rho r^3
ρ is the density of the water
Cd = 0.47 for a spherical shape.
r is the radius of the ball. (12 cm for a full-sized basketball, volume is 0.0072 m
3 weight 0.62 kg)
The result is.
v^2 = \frac {8 r g} {3 C_d}
v is proportional to the square root of the ball radius.
I get v = 2.6 m/s from this. This would result in a height of only 0.33 m, and I've seen balls that were smaller than a basketball go higher.
It seems it is very important what happens at the surface. If the ball is half out of the water, it would still have 30 N of buoyancy, there would be no more drag, and it would accelerate at 30/0.62 = 49 ms
-2. The drag formula is likely not valid close to the surface. A very light ball such as a ping-pong ball might accelerate much faster near the surface and go higher than a basket ball, even if its speed through the water at larger depths is slower.
What happens to the water in the wake of the ball is also critical.
According to this link, the ball often goes higher if released at a moderate depth (a few ball diameters ) than at larger depths.
At larger depths sideways movement is observed under water, probably due to vortex shedding, and the ball sometimes doesn't clear the water at all.
Water exit dynamics of buoyant spheres
Tadd T. Truscott, Brenden P. Epps, and Randy H. Munns
Phys. Rev. Fluids 1, 074501 – Published 1 November 2016
https://journals.aps.org/prfluids/abstract/10.1103/PhysRevFluids.1.074501
Behind a paywall, but you can still see graphs of the results and pictures of the ball popping out with the turbulence behind it.