Waves on water generated by a falling object

This is crude.

Maybe there can be an energy approach. Initially the mass has potential energy $T=m g h$. At the point of peak splash-back lets assume all the energy has been transferred to the water with peak potential energy related to the radial wave height function $y(r)=?$. A small volume of water a distance $r$ from impact has differential volume ${\rm d}V = y(r) 2\pi r {\rm d}r$ .

The potential energy of the small volume of water is ${\rm d}T = \rho g \frac{y}{2} {\rm d}V$ where $\rho$ is density of water. The total energy is thus:

$$ T = \int_0^\infty \rho g \frac{y(r)^2}{2} 2\pi r {\rm d} r $$

Putting a nice smooth wave height function of $$y(r) = Y \exp(-\beta\, r) \left(\cos(\kappa\, r) +\frac{\beta}{\kappa} \sin(\kappa\, r)\right)$$ with $Y$ a height coefficient. This has the properties of ${\rm d}y/{\rm d}r=0$ at $r=0$ with $y(0)=Y$.

$$ T = \frac{\pi Y^2 g \rho \left(9 \beta^4+2 \beta^2 \kappa^2+\kappa^4\right)}{8 \beta^2 \left( \beta^2+\kappa^2 \right)^2 } = m g h $$

So wave height should be $$ Y = \propto \sqrt{ \frac{h m}{\rho \pi }} $$

There isn't any simple relationship between the size of the splash and the impact velocity. Modelling splashes turns out to be a surprisingly hard thing to do. The problem is that the response of the water to the falling object is described by the Navier-Stokes equations, and apart from a few simple cases these are fiendishly difficult to solve.

I had a quick Google and found various school experiments where pupils measured splash height as a function of impact velocity. The nearest I found to a comprehensive description is this PhD thesis.

If you have a look on Youtube there are loads of slow motion videos of splashes.

I want just to add a thing. I think that the only reasoning we can do without solving the Navier-Stokes equations, nor doing experiments, is a dimensional analysis calculation.

(I suggest these notes sec 3.6 pag.83 for a complete treatment of this approach.)

The physical parameters we have are the density of water, the density of the ball (we have its mass and volume), the speed of the ball hitting the surface of the flow (via the conservation of kinetic energy, assuming no air resistance), and the viscosity of water. So:

$$\rho_b=m/V,\ \ mgh=\frac{1}{2}mv_b^2 \Longrightarrow v_b=\sqrt{2gh}$$

The only group of these parameters having the dimensions of a lenght is:

$$ \left(\frac{\rho_b}{\rho_w}\right)^\alpha\frac{\mu}{v_b}=\left( \frac{m}{\rho_wV}\right)^\alpha \frac{\mu}{\sqrt{2gh}}$$

with arbitrary $\alpha$. So we can't estimate neither the value of the amplitude, nor the frequency of the wave with only this argument, and we should solve the Navier-Stokes equations in this particular case (as said in the previous answer, this is possible only in a few special cases). Therefore, we can just state the dependence on some physical parameters, saving the arbitrariness about the power of the non-dimensional number $\rho_b/\rho_w$.