Maybe this could fall under my “physics of parkour”. It could also apply to the MythBusters “dumpster diving” episode. In both of these cases, the question is: how far can you jump off of something and not severely hurt yourself. They do this a lot in parkour. Here are some examples:

There are a ton of these things on youtube. Let me go ahead and say it. I would not recommend trying any of this stuff. Even reading this blog won’t adequately prepare you. So, if you go ahead and try to do some cool jump, don’t blame me for your injuries.

Now that the warning is out there – let me get on to the physics. Here is the general situation. A person jumps off a building and lands. The two important parameters are: how high was the building and how far did the person take to stop. Here is a diagram:

I am going to make the choice to have *h* the distance from where the person starts the jump to where the person ends the jump (vertical motion only). The distance the person stops will be *s*. Since this deals with distances and forces, but not forces and time, I should use the work-energy principle. Basically, the work-energy principle says:

So, what is the energy and what does work? This really depends on what I choose as my system. In this case, I will choose the jumper plus the Earth as the system. This way, the only external force that does work on the system will be from the force of the thing that stops the jumper (let me pretend for now that it is a mat or something). I will use the jumper at the top of the building and at zero velocity to start. The end will be the jumper at the bottom (after stopping) with a zero velocity again. The work-energy principle will look like this:

First for the work. It is the dot product between the force stopping the jumper and the displacement vector (I called s). I can re-write this as:

Where theta is the angle between the stopping force and the direction the object is moving. For this case, theta = 180 degrees. This means the work is negative.

Kinetic energy is easy to deal with. Since the jumper starts and ends at the same speed (zero m/s) the change in kinetic energy is zero Joules.

When an object is near the surface of the Earth, the gravitational potential energy can be written as:

It doesn’t really matter where *y* is measured from since the only thing I need to deal with is the change in potential. So, putting all of this together:

The change in gravitational potential energy is negative because the final *y* position is lower than the initial position. Ok, what next? Well, what matters when you land? Yes, the force matters, but what if I jump off the roof (which is unlikely) and then a child jumps off the roof. There would be different forces on us as we land, but probably a similar effect. One way to characterize the landing is with the acceleration. There is probably some maximum acceleration that you could handle without injury. I could divide both sides of the above equation by –*m*, but that would not give me the acceleration. The force on the left side is just one of the forces acting on the jumper while stopping. The net force on the jumper while landing is:

From the work expression, I can substitute in for the stopping force:

There you have it. That is the acceleration for stopping an object (or person) after jumping off of something into something. The nice thing about this form of a solution is that:

- It doesn’t matter what units
*h*and*s*are in as long as they are the same units of distance - If you want the acceleration in g’s, just leave the g off

Well, what is the maximum acceleration you could handle? Who knows? NASA has some data, but everyone is different. The wikipedia page on g-tolerance used to have a cool g-force tolerance table from NASA data. Don’t know why they took it off. Here it is:

Here is your handy jumping calculator:

In the above calculator, the first calculation is to determine the acceleration for a given case. The second calculation is if you want to design what you are going to land in.

Now, let me briefly look at two cases. First, jumping off a two story building and just landing in the grass. Here you see that there is a trick you could use. If you let yourself roll as you hit the ground, you can increase the distance over which you are stopping. This will decrease the acceleration. Suppose the ground does not really “squish” at all. What would be the stopping distance for a human jumper? Here I will look at the center of mass.

I will estimate y_{1} as about 1 meter and y_{2} as about 0.15 meters. Now, what kind of acceleration could someone take like this? Not sure. The NASA data is for either standing or laying down. I will go with the standing values since they have smaller tolerances. From that table, a NASA-trained astronaut could handle 18 g’s for less than 1 second. If I use this value for acceleration and 0.85 meters for s, then a person could jump from:

Wow, that really seems too high. Maybe my estimates for the stopping distance are off. You get the idea though.

Now for the crazy Russian guy jumping from one building to the next. First, let’s just not worry about his horizontal motion (although this would be a good homework assignment: how far could the two buildings be apart). Looking at the comments online, it seems like the guy jumped from a 8 story building to a 5 story building. What is that change in height? I guess 1 story is like 10 feet, so this change in height (*h*) would be about 10 meters. If he stops in 0.85 meters, what is his acceleration? Using the above calculator, I get 10.7 g’s. I guess that is reasonable-ish. I am still not going to do that. (and neither should you)