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Water splash depth

  1. May 20, 2006 #1

    I have a problem.

    Say I have a wooden ball, 5 cm in diameter, 40 grams, I drop it from a height of 1 meter into some still water. Obviously, it will splash in and dip a little before buoyancy forces take over and bring it to the surface,

    I have never heard of a way, but maybe the pros here know a formula or something,

    Is their any way to calculate how deep it will go after the initial splash before the buoyancy forces take over??

  2. jcsd
  3. May 20, 2006 #2


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    I suppose you could consider a simplified verson using conservation of energy. The kinetic energy of the ball before it impacts the water plus the change in gravitational potential energy, must equal the work done against the buoyancy force.

  4. May 20, 2006 #3


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    There is also the complication of the huge drag factor while the ball moves through the water, and the flow will probably be turbulent.

    Also the impact at the surface of the water will involve a combination of displacement and drag.
  5. May 21, 2006 #4
    Thanks guys!

    I tried using the energy converting from potential into work done, but the figure comes out obviously wrong...I think you're right about the drag and other factors which are hard to account for...

    is their a single complete formula for working this out? or could someone please post a quick and dirty formula which may work?
  6. May 21, 2006 #5


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    There is one located here http://hyperphysics.phy-astr.gsu.edu/hbase/lindrg.html#c2, which would probably give a good approximation. Note here, that the initial velocity (V0) would be the velocity before the ball impacted the water, you would have to calculate this yourself.

    Last edited: May 21, 2006
  7. May 21, 2006 #6
    wow! This is getting way over my head, I was expecting something simple like f=ma... :(

    I also found this - http://www.owlnet.rice.edu/~phys111/Lab/expt02.pdf (Its a PDF!)

    Any chance someone who knows the high level maths of double differentials and cosh functions could solve this for the displacement?

    As I have said, the ball, of mass m and diameter r, dropped from rest at a height h, top of water can be at h=0,...I guess at the deepest dip point (h=d), the vertical velocity will be 0 (?) sort of like a vertical pendulum experiment we did in school with a spring and weight at the end...


    Thanks in advance!
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