Let an object of mass $m$ and volume $V$ be dropped in water from height $h$, and $a$ be the amplitude of the wave generated. What is the relation between $a$ and $h$. How many waves are generated? What is the relation between the amplitudes of successive waves? Does it depend on the shape of the particle?

Assume the particle is spherical. What would be the shape of the water that rises creating first wave?

  • $\begingroup$ I'm putting the homework tag on this because it seems like it is, but really, experimentX, it would help if you mention why you're interested in this question and what research you've done to try to figure out the answer yourself. $\endgroup$
    – David Z
    Jun 18, 2012 at 21:53
  • $\begingroup$ I'm planning a simulation project .. I'm not in hurry!! We will discuss this later after my exam. $\endgroup$
    – S L
    Jun 18, 2012 at 22:08
  • $\begingroup$ I'd be very curious about the setup and results of such a simulation. $\endgroup$
    – user24901
    Jun 5, 2013 at 10:57

3 Answers 3


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 $$ Y = \propto \sqrt{ \frac{h m}{\rho \pi }} $$

  • $\begingroup$ This sparked my interest to look into scaling laws for impact craters. I found one arbitrary source to give an equation, although for the diameter. Ideally we would look at the depth or the height of ejected material, but the diameter still has everything on the same side of the fraction as you. palaeo.gly.bris.ac.uk/Communication/Brana/equation.html $\endgroup$ Jun 18, 2012 at 20:40
  • $\begingroup$ Interesting, although this equation looks empirical. $\endgroup$ Jun 18, 2012 at 20:41
  • $\begingroup$ The question is empirical only due to the fact that we can't answer it exactly. The link I gave is for an empirical equation, but the point I was trying to establish is that it has basically h, m, rho, and g equivalents present and on the same side of the fraction. $\endgroup$ Jun 18, 2012 at 20:52
  • $\begingroup$ It looks empirical to me because of the fractional powers. $\endgroup$ Jun 19, 2012 at 0:18
  • $\begingroup$ @ja72 What does '∝' stand for in the final equation? $\endgroup$
    – user218127
    Sep 25, 2019 at 12:12

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 $\rho_b/\rho_w$.

  • $\begingroup$ @ja72 so $Y$ doesn't depend on the viscosity of the fluid? Is this reasonable? $\endgroup$ Jun 18, 2012 at 20:07
  • $\begingroup$ I think the last comment was directed to the other answer, but I think the need or the lack of need for viscosity is an interesting question. I would suggest that surface tension might actually matter more than viscosity. The mental image the OP has in mind is probably in a regime where surface tension matters a great deal. Plus, when we are dealing with large scale drops into water where surface tension doesn't matter much, does it make the kind of waves we have in mind in the first place? $\endgroup$ Jun 18, 2012 at 20:25
  • $\begingroup$ Actually it was, but I can't comment the other answer, I don't know why :( $\endgroup$ Jun 18, 2012 at 20:32

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