Few days ago I played with ball(filled with air) in swimming pool. I observed interesting phenomenon.

When I released a ball from 3 meters depth the ball barely jumped above the water surface but when I released it from 50 cm depth it shoot out of the water like nothing.

I observed when released from 3 meter depth the ball goes up in zig-zag trajectory but from 50 cm depth is goes in straight line.

I would be very interested in calculating optimal depth from which the ball would jump the highest above the water surface. And as well I would like to calculate trajectory of the ball under water.

It is obvious that simple drag formula won't help here.

I guess that the zig-zag patterns is happening because there might be something like Karman vortex street behind the ball.

So have anyone idea how to calculate this? Or can you point me to the right literature?

Edit: I forgot one observation I made. It seamed to me that the ball when released from 3m depth was rotating when it hit a surface and that might prevent the jump.

  • $\begingroup$ Nice hypothesis! You even have a formula to test given in the ref you cite. However, you have omitted measurements of some critical observable variables (present in the formula) from the question. Did you try evaluating the formula using the f, V, D, and Re you can observe or estimate from the actual system? Can you eliminate your hypothesis in this way? How might a ball differ from a cylinder? $\endgroup$ Commented Jun 14, 2013 at 23:32
  • $\begingroup$ which formula do you exactly mean? By any formula there I would get straight trajectory and that is definitely not going to predict my observations. One thing I forgot to mention. It seamed like the ball when released from 3m depth was rotating when it hit the surface that might stopped the jump. $\endgroup$
    – Tom
    Commented Jun 15, 2013 at 9:16
  • 3
    $\begingroup$ I am going swimming in an hour or so and will try your experiment. $\endgroup$
    – anna v
    Commented Jun 15, 2013 at 13:30
  • $\begingroup$ Ball was deflated and I could not find the inflating thingamajig . My view is that when you put the ball down with enthusiasm you give it more potential energy ( including the elasticity of the ball) and it bounces out high. When you dive, rather a friend dives, to three meters the ball is released at rest except the potential from the buoyancy. My analysis is that going up the three meters the ball gets a limiting velocity ( viscosity) and this is what is seen which obviously in your case is less than the 50cm one. I would try to have the ball at rest at 50cm and then release it to measure. $\endgroup$
    – anna v
    Commented Jun 16, 2013 at 3:47

4 Answers 4


For the reference of others who might be interested in this here's an article and the peer reviewed publication linked from it that cover this question in a fair amount of detail and has interesting videos demonstrating the different underwater behaviours.

https://phys.org/news/2016-11-pop-up-effect-buoyant-spheres-dont.html https://doi.org/10.1103/PhysRevFluids.1.074501


Try to eliminate your initial hypothesis by using dimensional analysis to make a quantitative estimate based on variables you can observe and measure easily.

You should be able to estimate a Strouhal number for your system from your observations of ball diameter, 'zig-zag' frequency, and the time for the ball to reach the surface from a known depth (i.e. fluid velocity.) That should be near 0.2 for your Karman vortex hypothesis to be a viable hypothesis. If its not you may want to consider an alternative explanation.

  • $\begingroup$ I'll give it a try. $\endgroup$
    – Tom
    Commented Jun 18, 2013 at 20:56
  • $\begingroup$ Can't really say for certain whether this was caused by vortex-shedding, but the math doesn't rule this possibility out if the ball was 10-15cm in diameter and if you saw it make 4-6 zig-zag cycles as it rose from 300 cm depth. $\endgroup$ Commented Jun 18, 2013 at 23:06

At 50cm 'depth' the ball was not actually submerged, because the water did not have sufficient time to refill the 'crater' created by the ball entering the water. The ball can be said to be 'floating' on the base of the crater. As the water returns at very high speed to refill the crater the bottom of the crater with the ball 'floating' on it is effectively uplifted through the air in the crater (not water). Once the crater is refilled the ball is now at the surface level of the water; having traveled at high velocity from its original resting place at 50 cm depth. It now leaves the surface of the water with the momentum gained by its high velocity rise through the crater and flies into the air.. [email protected]

  • 1
    $\begingroup$ Welcome to Physics SE. Look around, and take the tour. In the case of this answer, note that the OP was not dropping the ball in to the water, but releasing it from underneath the water - there is not 'crater' to refill. That kind of invalidates your answer. $\endgroup$
    – Jon Custer
    Commented Jan 20, 2016 at 15:46
  • $\begingroup$ Agreed Jon, apogs. I was speed reading and jumped to a wrong conclusion. I would be interested to understand the physics of the often YouTube'd video of what I describe though. Complex to model accurately I suspect. $\endgroup$
    – sean
    Commented Jan 20, 2016 at 16:01

The answer is quite simple. The pressure exerted by the water column on the ball is more at 3m depth as compared to the pressure exerted on the ball at 50cm depth. As a result, due to the buoyancy of the ball it shoots out of the water with a greater force when it is released from a depth of 50cm.

  • 1
    $\begingroup$ The pressure difference across the height of the ball should be the same though. In both cases, ignoring other factors, as long as both balls are submerged, the buoyant force acting on them is the same. Also, when released from a further depth, even if the exerted pressure was less, it would have the same pressure when it reaches 50 cm, along with all the acceleration it recieved due to pressure going from 3m to 50 cm. It should already have a velocity and the same pressure that the stationary ball had, so by that logic it would go higher; but we see that isn't the case. $\endgroup$
    – JMac
    Commented Dec 20, 2017 at 17:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.