I was reading on this post on Height of Water 'Splashing'. One of the answers given by user Martin Gales derived a simple correlation which I could logically understand:

From the law of momentum conservation we get $$m_{s}v_{0}=(m_{s}+m_{w})v$$ or


So $v$ above is velocity with which water (with volume of $V$) bursts up. To what height of $H$ ?

$$H=\frac{v^{2}}{2g}=\frac{v_{0}^{2}}{2g}\frac{1}{\left (1+\frac{\rho_{w}}{\rho_{s}}\right )^{2} }=\frac{H_{0}}{\left (1+\frac{\rho_{w}}{\rho_{s}}\right )^{2} }$$

or $$\frac{H}{H_{0}}=\frac{1}{\left (1+\frac{\rho_{w}}{\rho_{s}}\right )^{2} }$$

Because $\rho_{w}=1\frac{g}{cm^{3}}$ and $\rho_{s}=3\frac{g}{cm^{3}}$ (roughly) we get an estimation:

$$\frac{H}{H_{0}}=\left ( \frac{3}{4} \right )^{2}\approx 0.6$$

I wanted to use this formula in my experiment to test if that equation really holds. However my professor asked me - although the derivation seems ok as it removes complex details - to find an actual report that supports this derivation and tests if it actually works. Now the problem is that I am not sure if there are any sources which back this derivation so I was wondering if anyone in this forum is aware of any studies conducted to verify this relationship?

I would appreciate any sources and pdf file links.

  • $\begingroup$ The approach is interesting in theory but I agree with your professor: There are many assumptions in there which could and should be questioned. The water and the object are treated as point-particles (?) participating in a completely inelastic collision (?) with a water volume of equal size as the sphere (?) that results in the same amount of water (?) splashing. Other than that it neglects any kind of friction (?) including the fluid's viscosity. As a result it would predict the same height for any kind of object shape and type of liquid (e.g. water and honey). $\endgroup$
    – 2b-t
    Dec 22, 2019 at 13:27
  • $\begingroup$ Take a small solid steel sphere (something like a computer mouse's ball), which has a density of approximately 8 times the one of water, drop it from the $10 \, m$ board into the pool. The formula predicts an average (not even maximum) splash height of $0.8 H_0$ ergo $8 \, m$. This is clearly unrealistic! Thus, I think it is unlikely that there exist any serious publications considering this approach. $\endgroup$
    – 2b-t
    Dec 22, 2019 at 13:49
  • $\begingroup$ Thanks for your feedback, then what source could I use to understand a more accurate relationship between initial drop height and the splash height? Although i believe it is very difficult to come to an accurate estimation, whatever sources which aims to do this and establish some sort of relationship will be appreciated. $\endgroup$ Dec 22, 2019 at 16:25
  • $\begingroup$ Please don't post screenshots of text. It breaks search functionality, and it doesn't work for blind users. The size is also much too large. $\endgroup$
    – user4552
    Dec 22, 2019 at 16:36
  • 1
    $\begingroup$ Hmm so would it be ok to use this equation as my model and later when experiment results give otherwise results I can just state the limitations of my experiment in my evaluation on why it did not work very well? And also thank you for the edit. $\endgroup$ Dec 23, 2019 at 3:42

1 Answer 1


The use of conservation of momentum in the calculation which you quote does not make any sense to me. So I would not use it. A better idea which is closer to the description below is found in the unanswered question How does a ball cause a splash? (With the relevant math).

The splash phenomenon is called a Worthington Jet.

The object falling into water pulls down a column of air behind it. This column collapses as the water at the sides rushes inwards, causing a "pinch" effect at the middle of the air column. Two columns or jets of water then rush upwards and downwards from the pinch. The upper jet jumps out of the water with a speed which can be several times that of the object at impact. In some circumstances this jet can break up, leaving a droplet at the tip which continues rising as the remainder collapses back into the water.

The above explanation can be found in Worthington jets explanation: fluid phenomenon. This contains a link to useful videos on YouTube which show that spectacular jets can also be achieved by dropping heavy objects into fine sand or powder. Thus surface tension does not have much effect in liquid jets.

Also included is pre-print of the article Generation and Breakup of Worthington Jets After Cavity Collapse by S Gekle and J M Gordillo (2009), from which the following diagrams are taken. These show simulations of the formation of the jets, which are the spikes in the right-hand diagrams.

enter image description here

Gekle and Gordillo's paper develops a numerical model to identify how the formation and break up of the jet depend on the dimensionless Froude Number $Fr=V^2/gR$ and Weber Number $We=\rho R V^2/\sigma$ which characterise the experiment. They not give a complete analytical derivation of the relation between the jet ejection velocity $U$ and the radius $R$ and velocity of impact $V$ of the object (which is modelled as a circular disk) and the density $\rho$ and surface tension $\sigma$ of the liquid. When $U$ is known the maximum height of the splash is $U^2/2g$.

Another factor which has a significant effect on jet velocity $U$ is the wettability of the surface material of the object, which is measured by contact angle. If the surface is hydrophilic, meaning that contact angle is small, then the jet is almost eliminated. See the video Dynamics of Water Entry by TT Truscott et al, MIT. A review article for the whole phenomenon was published by the same team in 2014; it can be accessed from ResearchGate.

Also useful is the Masters Thesis An Experimental Study of Worthington Jet Formation After Impact of Solid Spheres by Jenna Marie McKown, MIT (2011).

  • $\begingroup$ Thank you very much this is exactly what i needed. $\endgroup$ Jan 1, 2020 at 14:14

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