# Height of Water 'Splashing'

Suppose from a height $$H$$, I throw a ball of mass $$M$$ and radius $$R$$ with initial velocity $$u$$ into a pool of depth $$x$$ having a liquid with density $$\rho$$ and coefficient of viscosity $$\eta$$.

Upto what height will the Water Splash?

This may be a silly question! (Source : my brain)

Assume every condition to be identical with no air resistance.

Feel free to assume any other parameters if needed.

P.S. $$\rightarrow$$ I've see similar questions but some of them want an exact answer and some are not so direct while some are incomplete questions...

Small Note : We do not have to consider the highest 'single' drop... Just assuming almost every water droplet jumped upto a height $$h$$. Ideal conditions...

Feel free to assume the liquid as WATER if required!

• This is a famous problem Commented Sep 1, 2015 at 19:43
• This is an incredibly complex, nonlinear (and turbulent) two phase problem. An exact solution is certainly analytically out of reach, although you might be able to make some bulk scale arguments. Or you can employ some robust numerical techniques. Commented Sep 5, 2015 at 12:49
• @NickP yeah..I am aware of that...I want to get some approximate answers tho...Not asking for the exact real one Commented Sep 5, 2015 at 12:51
• @NeilRoy Okay. The mechanism for the jet formation is likely the creation of a standing wave, due to the work done by the ball to deform the free surface. See, for example, here: royalsocietypublishing.org/content/royprsa/457/2006/… rspa.royalsocietypublishing.org/content/royprsa/457/2013/… My approach would then be to try to come up with a scaling between the initial geometry and the height of the wave. But this would require numerical/lab experiments. Commented Sep 5, 2015 at 12:59
• Notice in this video of a golf ball dropped into water that the greatest vertical height is attained by the column of water ejected upon collapse of the cavity created by passage of the ball through the water. This phenomenon is described in the two links provided by Nick P in his comment. Commented Sep 5, 2015 at 14:06

Basically, the whole kinetic energy is transferred to pressure, and then this pressure will be transferred to kinetic energy again; this time only the direction is as defined by hydrostatic pressure; perpendicular to surface.

This above gives a following basis;

Kinetic energy of the ball is also its potential energy (no friction on fall) $$E_{\textrm{kin}} = m g H$$. This is then transferred to pressure through the ball surface; $$A = 4 \pi r^2$$.

This pressure then splashes the fluid up;

In optimal case the diameter of the ball is almost zero, and viscosity of the fluid is such, that the ball would stop in a distance of slightly more than $$r$$. This would lead to a situation where the vertical velocity of the water is very low, and thus the water would jump almost directly upwards. This doesn't actually matter too much, if the air friction is not considered.

Ok, so an answer, if the density of the ball is same as the density of the fluid. Then the fluid would jump to the same height as the ball was dropped, if we also consider that there are no viscous losses. This is never true, and thus the ball drops deeper in the fluid and the losses reduce the available energy.

This all could be calculated. But the interesting thing is that there is a hole in the water when it goes deeper; And this means that the fluid which has maximum pressure has now a surface with no pressure. And therefore the fluid goes with even higher velocity back to fill this hole; as the velocity come from pressure difference, what happens;

It collides in the middle of the hole, but this time there is many velocities reaching the same point at the same time. Again all these velocities are transferred to pressure and the fluid takes new direction.

In 2-dimensional world this new velocity component would be 2-times the original. In 3-D reality it's more, and in true reality it's limited by viscous losses, surface tensions, etc. etc.

So to conclude this all; The splash height can be anything.

• The "round splash" could theoretically reach the Height $$H$$, but it can never be more.
• The "middle-splash" can be even higher than the Height $$H$$.

At this video found from comments, there is a golf ball used to do the splash. And such a golf ball makes a higher middle-splash than a round ball, because the boundary layer of the ball makes less losses, but also disturbs the fluid less. And therefore the returning middle splash is so big in this video; the collision happens with minimal disturbances; and the velocity vectors really hits against each others.

• The density of the ball plays a very important role which you neglect. If density is much greater than that of water, the ball will barely slow down as it hits the water, resulting in significantly volume of water being forced up. This will affect the splash height. Commented Nov 1, 2015 at 19:55
• @Floris "if the density of the ball is same as the density of the fluid.",, -no I didn't forget it. But as this can be anything, so can also the splash. Give me your case with all the details and I calculate you a real world answer. But please don't neglect even the fluid temperature and the balls contact angle. (surface tension contact angle) Commented Nov 1, 2015 at 20:02

Indeed, this seems to be very complicated problem.

But let's throw away all this complexity and concentrate in the core of the phenomenon.

So consider an ordinary stone with volume $$V$$ that falls into the water from a height of $$H_{0}$$. The shape of the stone may be arbitrary. Ignore air resistance.

Velocity of the stone before collision with water surface is $$v_{0}=\sqrt{2gH_{0}}$$ It is reasonable to consider the collision as inelastic. The volume of the water in which the stone is interacting at the moment of the collision we will take equal with the volume of the stone $$V$$.

So two masses that interact are $$m_{s}=\rho_{s}V$$ and $$m_{w}=\rho_{w}V$$ where $$\rho_{s}$$ is the density of the stone and $$\rho_{w}$$ is the density of water.

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

$$v=\frac{\rho_{s}v_{0}}{\rho_{s}+\rho_{w}}=\frac{v_{0}}{1+\frac{\rho_{w}}{\rho_{s}}}$$

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 found this derivation easy to understand and actually wanted to use this as a basis for my experiment in University. But my professor kind of prohibited it as he said I need to find sources. Are there any sources or real life studies conducted on this derivation you made? I would appreciate it if you can share if you have any. Commented Dec 22, 2019 at 8:10
• Since the bulk of water, after a series of complex interactions which are left out here, should gain upwards momentum in order to splash into the air, should the momentum conservation equation be $m_s v_0 = \left( m_s - m_w\right)v$, with a minus sign? But then the numbers don't feel plausible, so please correct me if I am wrong. Commented May 25, 2023 at 16:16
• @jphys $m_s v_0 = \left( m_s - m_w\right)v$ doesn't make sense. For example if we take $m_s =m_w$ then $v_0=0$. This is against the terms of the problem. In reality, we have case $m_s =m_w$ if instead of a stone we take not the freshest egg, barely floating. I agree that it is difficult to imagine this collision process, what is actually happening there. But that's what conservation laws are good for. You don't need to know exactly how this whole process works. Commented May 28, 2023 at 7:37