Imagine a breaking wave 10m high. The wave breaks onto a beach. What is the maximum pressure created by the wave on the beach, ignoring effects such as might be cause by air?
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The pressure of falling water can vary by a LOT - it depends in detail on the shape of the interface between the water and the surface it hits.
When a perfectly spherical drop of water hits a hard surface, there will actually be a short moment in time when the contact point between the water and the surface travels faster than the speed of sound in water - this means that the pressure "can't escape" and it builds up to something greater than the water hammer pressure $\rho c v$ where $c$ is the speed of sound in water and $v$ is the velocity of the drop (image source).
If a drop lands from a height of 10 m, it has a speed of about 14 m/s; the speed of sound in water is about 1500 m/s, and density is 1000 kg/m$^3$. That makes the peak pressure 21 MPa - almost 200 atmospheres. But that only happens on very small local points (incidentally, this is the phenomenon that gives rise to erosion by liquid impact - see for example http://blog.kmt-waterjet.com/tag/waterjet-impact/).. In general, when water hits a surface with a velocity $v$, the pressure generated will be
$$P = \frac12 \rho v^2$$
Since $v = \sqrt{2 g h}$ we can rewrite this as
$$P = \rho g h$$
that is, the pressure you would have if you supported a column of water of that height. Which would make the answer, for your wave of 10 m, approximately 10$^5$ Pa above atmopheric pressure - 1 atmosphere. Much less than 200...
Floris
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Each kilogram of water that falls from 10 m will have the kinetic energy of about 100J. I can't tell you the pressure since you didn't specify the area. Shorter falls will have less energy.
LDC3
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