Thanks to Nick for showing me this video (Check out his blog – Fine Structure):

Wow. That was my first reaction. My second reaction was: no way. Is this real life? I just don’t know. How hard would it be to find out exactly where to place that pool and where did they get the water from? Obviously, this one requires some analysis. First, on the VAS for this video: 4/8. Not too good. Oh here are the questions I would like to answer:

- What is the guy’s acceleration after he leaves the ramp?
- What was his initial velocity leaving the ramp?
- How high above this would he have to start?

Ok. Time for Tracker Video Analysis. Here is the y-motion for the ‘flight’:

Notes:

- The unit of scale is the height of the ramp.
- There is an obvious perspective problem. As the camera pans, the angle between the motion and the camera is not constant and not perpendicular. I have no simple way to correct for this (yet).
- From this, the acceleration is about 4 ramps/s
^{2}. - At this point, I can’t really tell if it is fake or not. The motion does not fit a parabola very well, but that could be because of perspective issues.

Now, if I assume that the acceleration in real life is 9.8 m/s^{2}, then the ramp would be 2.45 meters tall. Using this new scale, I can look at the horizontal motion:

This looks linear-ish. From this, the horizontal velocity is mostly constant at a value of around 16 m/s (which is about 36 mph for the metric-challenged). At least I have enough info to make some calculations. Note that 16 m/s is the guy’s horizontal velocity, not the total initial velocity. The initial vertical speed can be determined by looking at the time in the air. (here is a review of projectile motion) If I assume that the guy starts and lands at the same height, then I can use:

Since y and y_{0} are the same, I can solve for the initial velocity:

From the video, ?t = 2.1 seconds. This gives an initial y-velocity of 10.3 m/s. This will give a total initial speed of:

Putting in the values for the x- and y-velocity, this gives a magnitude of the initial velocity of 19 m/s. Why do I care about this velocity? Two reasons. First, I can estimate how high up the hill the guy would need to start to get this speed. Second, this is the same speed the guy will hit the pool. So, I can estimate the acceleration when he lands and see how deadly it would be (I already suspect he should be ok – think about professor splash)

How high up the hill would he have to start? If I ignore friction (always a good place to start), then I can use the work-energy principle to calculate this. Let me make a sketch.

The work-energy principle is great to use here because it essentially deals with change in position. I will start with the Earth-guy as my system (this means that there will be a gravitational potential energy and NOT work done by the gravitational force). When working with the work-energy principle, you need two positions. In this case, that will be at the top of the hill and at the top of the ramp. During this motion, there are only two forces acting on the guy: the normal force from the ground and the gravitational force. The normal force does no work since it is always perpendicular to the direction the guy is moving. Gravity doesn’t do any work because I am using the gravitational potential energy. If the guy starts from rest at the top of the hill, and I set the zero gravitational potential at the top of the ramp, then:

I didn’t want to be too confusing about the velocity in the above expressions. That is the velocity at the top of the ramp. If I wanted to be consistent with the stuff from before, this would be v_{0}. Using this stuff and solving the for the height above the ramp, I get:

Notice that this solution does not depend on the mass of the guy nor does it depend on the angle the hill is inclined. If I plug in the value for the speed at the top of the ramp, then the starting point must be at least 18 meters higher than the top of the ramp. If there is significant friction it would need to be even higher.

It is very difficult to estimate the height of the starting point because of the angle the camera is viewing from. There is one thing that does not change with perspective though – time. I can get the time it takes the guy to get from the top of the hill to the bottom of the hill and calculate how steep the hill would have to be (again assuming no friction). From the video, this is about 3 seconds. The ramp looks pretty big, but I am going to use the velocity at the top of the ramp as though it were the velocity at the bottom of the ramp just to get an estimate of the angle of the ramp. Ok, so if he goes from 0 to 19 m/s in 3 seconds, then his acceleration (average) would be:

So, if this were a hill at a constant slope with no friction, how steep would it be? Here is a free body diagram of an object sliding down a slope.

I want to find the acceleration down the plane as a function of the angle of the plane. In this case, the only force acting in the direction of acceleration would be a component of the gravitational force. This gives:

If I put in 6 m/s^{2} in for a, then I get an angle of 40 degrees. Pretty steep – but it is a mountain I guess. I guess this is real. But there are still some things to investigate. I will leave the following questions for homework:

- Suppose you are planning this “stunt” and your initial velocity is off by dv (some small amount). What would the resulting change in range be? If dv = 0.5 m/s, would the guy still land in the pool?
- Suppose the coefficient of kinetic friction was 0.1. What would be the new velocity at the top of the ramp? You can assume that the hill is straight.
- Estimate the acceleration of the guy when he hits the water. Look up the NASA g-force tolerance tables and see if he is ok.
- Where did they get all the water to fill up the pool?
- Who inflated the pool and how long did it take if they just used their lungs?

Homework hint. If you look at that Professor Splash jumping into a foot of water, it will really help. In that analysis, Prof Splash is going about 15 m/s before hitting the water. Yes, that is slower than this guy, but this guy lands in much deeper water (maybe 3 feet?) and at a non-perpendicular angle (which means he has a greater distance to slow down).

### Update:

As pointed out in the comments, this is indeed a fake video. I lost. Anyway, there are some points that are still true.

- This is not impossible. Even for a computer. We used to bullseye wamprats back home and they are not much bigger than 2 meters. No really, it could be done even if it would be stupid to do so.
- The pool should be plenty deep enough to land in. Look at professor Splash jumping into 1 foot of water.