How does "Professor Splash" work? So there is a daredevil named "Professor Splash" who is known for diving (belly flopping) from over 30 feet into a foot or less of water. You can find more information about him below: http://en.wikipedia.org/wiki/Professor_Splash http://scienceblogs.com/dotphysics/2008/11/the-physics-of-professor-splashs-jump-into-1-foot-of-water.php [Broken] Now the second link got me really intrigued. I have taken a mechanics course, so I mostly understand the discussion of how the work-energy theorem, impulse, etc. work in his jumps, but the one thing I don't understand is what realistically determines the impulse the water exerts on him. Sure, you can take his speed, mass, and the height of the water (1 foot), and divide them to determine the impulse the water would have to exert on him to completely stop his fall, but why does the water provide that needed force? What's to say that the water doesn't just provide half the force and thus not decelerate him completely, leaving him to splatter into the ground, albeit at a reduced velocity? Is there something I'm missing here? I suppose it would have to do with the surface tension of water being able to supply up to so much force per square inch of surface area? And if so, how would you go about calculating it realistically, since realism would logically entail a stronger force for only a portion of the distance (not actually using the entire 1 foot of water in reality). How would professional physicists calculate such a thing?