Extreme Jump - Analysis of a Fake Video

So, here is a video (from break.com - so you know it is likely fake).

So, here is a video (from break.com - so you know it is likely fake).

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If for some reason, you can not view this video, here is the plot.

  1. Guy wears parachute and brings a portable thing like a see-saw.
  2. Guy approaches large crane dropping a large mass repeatedly (I assume to flatten a dirt road)
  3. Guy sneaks up an puts the see-saw under the area that the mass drops on and then stands on the other end of the see-saw.
  4. Mass drops, guy shoots up and parachutes down.

So, why is it fake? I think the best thing is to give an analysis of the see-saw. To start this analysis (as usual) I will start with some assumptions. Normally I would use video analysis to get some data from the actual video, but that won't be needed in this case.

  • Person Size: I will assume the
    person is 5'10" (or 1.78 meters). I will use a mass
    of 80 kg.
  • Length of see-saw: Totally an
    estimate from the video - but I choose 2.5
    meters.
  • Mass of huge thing: The giant
    thing that drops, it is about 1 meter x 2 meters x
    2 meters in size (4 m3). What is it made
    of? It could be filled with water, or it could be
    steel. What has a density of 1000 kg/m3.
    Steel has a density around 8000 kg/m3. I
    will randomly use something of density 5000
    kg/m3. This will give it a mass of
    20,000 kg.
  • How high does he go? This one
    is really not possible to tell from the video. If
    he were able to parachute, that would make it at
    least 200 feet? I timed the video, from the time of
    the launch to the time where they show the
    parachute open, it is 5 seconds. During this time
    he went up and down some. Suppose he reached his
    highest point after 3 seconds.

So, how high does he go? I can use the following equation (assuming his acceleration is 9.8 m/s2 downward - a reasonable assumption).

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This equation uses the initial velocity - which I don't have. But, I can cheat. If it is just constant acceleration, the time down is the same as the time up. So, from the highest point down will also take 3 seconds. I know the initial velocity for that motion, 0 meters/sec. If I take his final y-position to be 0 meters, then:

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How fast did he leave the platform? Using the following:

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With a time of 3 seconds, this means his initial velocity was:

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(which is 66 mph - google calculator rocks)
So is this speed the problem? No. Clearly someone can move that fast. But, what would the acceleration be to obtain this speed? Here is my diagram of the apparatus the fake-guy used:

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I estimated the height of the pivot to be around 0.5 meters. Using similar triangles, this means that the guy is accelerated over 0.71 meters. To determine his acceleration, I will use the following kinematic equation:

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Solving for the acceleration (initial velocity is zero and y-y = 0.71 m)

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I put the acceleration in terms of "g's" so it could be easily compared to . They found an acceleration in this position, a person can take 18 g's if it is for less than 0.01 second. The most a person could take was in the prone position and that was 35 g's.

Finally, what kind of force would this guy exert on this board? If his acceleration is 609 m/s2 this would take a net force of

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This is the net force (in non-vector notation - sorry). The force the board exerts would actually be greater because of gravity. Needless to say, it would break unless it was made of something like adamantium.

Clearly, this video was fake. It was still fun to analyze WHY it had to be fake.

Send me your videos and I will attack them - it's what I do best.