Here is a video of a guy jumping 35 feet into a pool of water only 1 foot deep.

**UPDATE: Apparently, that video went away. Here is another version.**

How does this work?

I don’t think I even need to do a video analysis of this motion, all the important info is given. I will assume that air resistance did not play a signficant role (and that is a good assumption – or good enough – see this for example: motion of a falling tennis ball). So, here is the situation.

Part 1: guy falls 35 feet 5 inches (10.8 meters).

![Screenshot 16sd](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screenshot-16sd.jpg)

For this part of the motion, it will be easiest to use the Work-Energy theorem to determine his velocity RIGHT before hitting the water. (note that I am assuming the 10.8 meters is the distance to the surface of the water, but really it doesn’t matter much). The work-energy theorem states:

![workenergy](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/workenergy.jpg)

In this example, I will assume just the person as the system. This means that the only change in energy will be the change in his kinetic energy and the gravitational force will do work. The two things to start with are the gravitational force (close to the surface of the Earth):

![fgrav](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/fgrav.jpg)

Here g is the gravitational field (9.8 N/kg) pointing down.

And kinetic energy:

![ke](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/ke.jpg)

When calculating the work done by gravity, the gravitational force and the displacement are both down. This means that the work done by gravity will be a positive quantity. This gives:

![workenergy3](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/workenergy3.jpg)

Putting in some numbers, I get:

![Screenshot 14](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screenshot-14.jpg)

for the speed of the guy right BEFORE he hits the water.

Now I can apply the same idea when he hits the water. The only difference is this time he starts at the above speed and ends at rest – also there is another force acting on him, the water.

![forces1234](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/forces1234.jpg)

I can then use this to find the force the water exerts on him:

![fwater](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/fwater.jpg)

This would be great, but it turns out that a better measure of what a person can handle is in terms of the acceleration. So, solving for the acceleration of the person:

![avgaccel](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/avgaccel.jpg)

Now to get this in terms of “g’s”, where 1 “g” is 9.8 m/s^{2}. This would give an acceleration of 35.4 g’s. Is this ok?

Well, instead of going out and taking human g-force tolerance data, I will use [NASA's data as listed on wikipedia](http://en.wikipedia.org/wiki/G_force#NASA_g-tolerance_data). This says a human can take 35 g’s “eyeballs in” if it is for less than 0.01 minutes. (eyeballs in means the acceleration is in the opposite direction that your eyes look)

So, how long was this guy accelerating? If I assume a constant acceleration, I can use the definition of average velocity where his average velocity while stopping would be 7.275 m/s.

![dtsdf](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/dtsdf.jpg)

So, it looks like this is within the range of what NASA recommends. No wonder this guy is a professor, his jump is NASA-approved.