# Kinetic energy loss when a rigid body falls into water

Assume you have a rigid body falling into the ocean at terminal velocity. Also assume that the rigid body does not break on impact. How could you figure out how much kinetic energy would be lost in the resulting splash? I promise this isn't a homework question, but an engineering thought experiment. I began to reason that water is incompressible so the body has to displace the water. I know the volume of my object so I could say I know how much water gets displaced. I also have a figure that the splash (radial jet) leaves at 20-30 times the impact speed. From there I can calculate a KE, but I don't believe it because this doesn't account for the surface tension. Any guidance would really be appreciated.

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Thanks, very helpful. Falling into molasses is going to result in a different final velocity than falling into water. MY point is just that there is another parameter that is involved that I am not accounting for. Perhaps I meant viscosity. – user3166 Apr 17 '11 at 19:02
It is sufficient to know the final velocity in the liquid. All the difference $\frac{1}{2} m (v_1^2-v_2^2)$ is lost somehow. – Vladimir Kalitvianski Apr 17 '11 at 19:06
Thanks, I agree, but I'm trying to determine the final velocity in the liquid via kinetic energy. – user3166 Apr 17 '11 at 19:08
I "know" that when falling with terminal velocity on water the impact is the same as falling on cement.Quotation marks because this is a statement circulating widely, not a physics one. Though if you dive from high up and fall on your stomach it really hurts. – anna v Apr 17 '11 at 19:18
Continued: I suspect that the delta(t) of the impact at the high velocity has something to do with it. There is not enough time for liquid to behave as liquid. I have to think about this. – anna v Apr 17 '11 at 19:27