This is such an incredible stunt that the very first question that comes up is - fake? or not fake? From my previous analysis, I can say:

• Even though there is some slight perspective issues, the vertical acceleration seems to be constant and has a reasonable scale to give the acceleration of -9.8 m/s²
• The horizontal motion is essentially constant (as a real jump would be)
• The launch speed is about 19 m/s
• The launch angle is 32.8 degrees
• If the guy had no friction on the slide, he would have to go down a 40 degree slope.
• Landing in the pool should not be a problem, Professor Splash did it in much less water.

So, from all that, there is nothing that says it has to be fake. I know what you are thinking, still no way. I agree. Here is what bothers me. What if he had been off in his calculations or initial start? He would be dead if he collided with the ground? It just seems uber-risky. If you could assure the angle and the velocity at launch, and you could accurately measure the location of the calculated landing spot, then ok. Let me just say that I wouldn't do this (obviously).

One way to look at this a little more is to do an error analysis of the landing spot. What if the initial velocity was off by 1 m/s, how much would that change the landing location? First, let me go through the calculation of the landing spot with the following assumptions:

• No air resistance (or negligible)
• Guy starts and lands at the same height (if this is not true, the equations could be adjusted, but this will give me a good idea)

If I assume no air resistance, then the motion of the guy after he leaves the ramp is standard projectile motion stuff. In projectile motion, the key is that the horizontal and vertical motions are independent except for the time. So, the vertical motion would be:

With the initial and final y being the same value. The initial y-velocity is just the vertical component of the total initial velocity, so:

From this, I can solve for the time of flight. I get:

Now to look at the horizontal motion. Since there are no forces in the horizontal direction, the horizontal acceleration is zero. This means that:

It doesn't really matter where the origin is, so let me say that x₀ = 0 meters. Putting in my expression for the change in time, I get:

This does have the correct units (velocity squared over acceleration gives units of distance). Also, for angles between 0 and pi, this gives a positive range.

Ok, I could stop here and just play around with the initial values for the velocity and angle of launch, but I can do more. Suppose that I know that the initial velocity is somewhere between 18 and 20 m/s and the angle of launch is from 32.0 to 33.0 degrees. What would my uncertainty in range be? (note that I am assuming there is no uncertainty in the "lining up" of the ramp). To do this, I would need to look at propagation of error.

There are two ways to do this, the non-calculus way and the calculus way. Really, there won't be a huge difference between these two, but I will use calculus. The non-calculus way is pretty simple though. You just use the numbers for initial velocity and angle that give you the max and min range. But for this calculation, I will use the method I describe in the uncertainty in the calculation to the distance of the Sun. Basically, if you have some function (f) that depends on two measured variables (x and y), then the uncertainty in that function is:

Where delta y and delta x are the uncertainties in those values. So, I have my function x(v, theta). Do you want me to break this into small pieces? Well, I won't. You can do that as a homework exercise. Ok. Here is what I get so you can check my answer:

Two things to check. Do the two terms inside the square root have the same units? (if they don't you can't add them together) In this case, they both have units of m²/s² (theta is in radians). Also, is the units for delta x in meters? Yes.

Now to put this stuff into a spread sheet so you and I can play around with the numbers quite easily.

So, if I put in a completely made up uncertainty of 0.5 m/s for the velocity and 0.005 radians for the angle, then the range is 34 meters plus or minus 2 meters. I am going to completely guess that the pool is a 10 foot diameter pool (maybe it is this 10 foot Intex pool). So, the pool goes from, let's say, x = 32.5 meters to 35.5 meters and he will likely land from 32 to 36 meters. A little too close for me. If you increase the uncertainty in the initial velocity to 1 m/s, then the uncertainty in the range goes up to almost +/- 4 meters.

So, does this mean it is fake? I still see nothing that says it HAS to be fake other than it seems very dangerous.

Update:

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

1. 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.
2. The pool should be plenty deep enough to land in. Look at professor Splash jumping into 1 foot of water
3.  
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5. VIEW HERE
6. http://scienceblogs.com/dotphysics/2009/08/more-on-the-giant-water-slide-jump/