Physics Asked on June 12, 2021
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.
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
Correct answer by Sparky on June 12, 2021
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.
Answered by Mark Rovetta on June 12, 2021
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]
Answered by sean on June 12, 2021
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.
Answered by Amy on June 12, 2021
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