I just saw a video on youtube, where an excavator drops a load of water on a car, destroying it completely. I also saw a video with a fire-fighting plane dropping even more water on a burning truck. I guess the firefighters do not intend to squish everything on their way, so how do they know how to drop water without damaging stuff under the plane?
How to determine height of dropping without destroying drenched object or killing people in bombed area?
In an earlier life, I studied liquid impact for a living (albeit on the scale of very small drops). There are a few things to note:
very high speed impact: when a small drop hits a surface with sufficient velocity, the pressure buildup at the impact point can be very large. See this earlier answer
large object impact traveling slowly: when the liquid travels slowly, and the time scale of interest is large, the "brief initial pressure" described above doesn't matter so much. Instead you have to ask about the pressure of the body of water that is slowing down. Now in the limiting case of a very large body of water, the force per unit area will just be the weight of the water above that point: regardless of the height of the drop, that weight can be sufficient to crush an object. Imagine all that water in a big bag: drop a large enough bag of water on anything, and you will crush it. The inertia of the water at the periphery of your "big drop" in essence acts like that bag.
object traveling quickly: if you have an object traveling quickly (but not so quickly that the phenomena under the first bullet play a role) then you need to stop a certain amount of liquid per second: if the density is $\rho$ and the velocity $v$, then the pressure needed is $P=\rho v$. From this we conclude that we control the pressure by either controlling $v$ (the velocity - or the height from which you drop the liquid) or $\rho$ - the density.
If you drop the liquid from a great height, two things happen: the liquid disperses into drops, and these drops are slowed down by drag in the air. If the liquid disperses, it decreases the effective $\rho$. The slowing down of the individual drops also helps. In essence, you spread out the time available to slow all the water down. If the water has the same total momentum $p$, then in the equation $p = F\Delta t$ you are increasing $\Delta t$ and therefore decreasing $F$.
A perfect example of the difference between Laminar and Turbulence. Even the difference in optical appearance is present.
It should be noted that the excavators shovel has a radius, which means that the water is released with a minimal disturbance. It has also most propably been filled with a small hose, so it's undisturbed. In hydrostatic equilibrium. See this question and answer; How quickly should a fluid come to hydrostatic equilibrium?
As I claim (against mainstram physics) that Turbulence is surface or crack inside the fluid, and this leads to a situation where the collision and friction defines the flow, and not viscous forces and continuity. So with this idea, you can understand that the main difference here is that, at the first picture the water mass falls in one piece, and on the other picture it falls in many pieces.
The one piece drop is self explaining. But the many drops can been best understood through newtons cradle. the first drop hits the object and bounces back, and this causes the following drop's to hit first to this bouncing back drop, and loosing energy energy before even hitting the object. Turbulence is friction and collision, not continuity. The premises of the Navier-Stokes existence and Smoothness problem is wrong; It's not a homogenous, the vector velocity and pressure fields simply dont go smooth through surfaces. It's collision and friction. And thus You actually can derive the fluid dynamics throught the first principles of physics. (not NS-equations, but fluid dynamics) I have an answer to this, but it's not "mainstream-physics", though it's simply Newtons laws explained. Why can't the Navier Stokes equations be derived from first principle physics?
The answer; hight is not relevant. The drop size is. The answer of Floris is dealing with single drops. If such a mass of drops are falling, the drag of air is not so relevant. It's back and for bouncing drops which are cancelling each others energies which makes the difference.
