How does a ball cause a splash? (With the relevant math)

Problem Statement:

Imagine a spherical ball is dropped from a height $h$, into a liquid. What is the maximum average height of the displaced water? For instance, although one particular drop of water might travel a high distance, the average displacement height of all the displaced water might be very low.

Approximation:

Let's make some assumptions. Let's assume that on impact, a fraction $\lambda$ of the ball's Kinetic Energy is transferred to the liquid. Let's also assume that all of the energy transferred will go into shooting water up. This should give the upper bound on how high the water "tower" will be.

The Kinetic Energy of the ball at impact with the surface of the liquid is,

$$KE=m \cdot g \cdot h$$

The fraction that is transferred into making the water tower is,

$$KE_t=m \cdot g \cdot \lambda \cdot h$$

When the formation of the water tower reaches its maximum height, the potential energy $PE$ must be equal to $KE_t$. We'll assume that the tower has cross-sectional area $A$, and integrate the $PE$ of each slice of the tower at a particular height to get the total $PE$.

$$PE=\int_0^{h_t} g \cdot l \cdot \rho \cdot A \ dl={{g \cdot \rho \cdot A \cdot {h_t}^2} \over 2}$$

Where $h_t$ is the height of the water tower, $\rho$ is the density of the liquid, and $g$ is gravity.

Set $KE_t=PE$ and get,

$$m \cdot g \cdot \lambda \cdot h={{g \cdot \rho \cdot A \cdot {h_t}^2} \over 2}$$

We'll use $A=\pi \cdot r^2$, which assumes the water tower has a radius equal to the ball. Solving for $h_t$, we note that the $g$ terms cancel, and we get.

$$h_t={1 \over r} \cdot \sqrt{{{2 \cdot \lambda \cdot m \cdot h} \over {\pi \cdot \rho}}}$$

Let's take $\lambda={1}$. let's use a water droplet. It has a radius of $0.5 \ cm$, a mass of $0.125 \ g$. Let's drop it into water from a height of $20 \ cm$.

We get,

$$h_t = 2.52... \ cm$$

Referring back to the original problem statement, we need only divide $h_t$ by two to get the average height of the displaced water.

Thus,

$$\mu_{h_t}=1.26... \ cm$$

Questions

First, I don't have access to materials other than water and a sink. So I can only test my approximation for water droplets dropped into water. Are there any results that point to flaws in my model?

Second, given this initial approximation, how does one go about adding in more complex phenomena? For instance, I've assumed a very ideal energy and water tower mass distribution. Is there any way you can go from this original approximation to a more useful one? Perhaps, there is an efficient way to conduct an experiment to get useful data?

Edit: This question inspired from another answer I wrote. I deleted the answer, and put the material here instead.

• +1. I have that problem about once or twice a day. – Rikki-Tikki-Tavi Sep 15 '15 at 15:51
• I would suggest this model: When the ball enters the water, it acts like a newtonian penetrator, so it goes dept=d*(density of ball/density of water) deep and forms a cyllinder of air in the water. Water then enters the cyllinder from the sides, with increasing speed towards the middle (Bernoulli). The potential splash height is 1/2v^2/g. v is infinite in the middle, but the minimum radius is capped by surface tension, though how exactly I haven't worked out. – Rikki-Tikki-Tavi Sep 15 '15 at 16:09
• This appears to be much the answer provided to this question: physics.stackexchange.com/q/203753, but now turned into a question. Same member too. – Gert Sep 15 '15 at 16:39
• Well the integration part is not necessary you know ... You only have to consider the center of gravity of the water pillar when calculating the GPE as the volume is uniform – slhulk Sep 15 '15 at 17:45
• Also a great deal of energy will be given out as sound rather than being used to raise water – slhulk Sep 15 '15 at 17:49