Dot Physics

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/s2. 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.

Comments

  1. #1 agm
    November 15, 2008

    An admirer of Henri LaMothe then?

  2. #2 Jasper Palfree
    November 16, 2008

    That’s neat… but I’m still bothered by something…

    Why should the water act on him with that force? Why doesn’t the water quickly displace itself and leave the rest of the pushing for the hard ground to do? (I guess I’m puzzled by the “fine tuning” element… if you see what I mean).

  3. #3 Alex M
    November 18, 2008

    So the video seems to have been removed, but you’re telling me that an educated man willingly belly-flopped into a 0.305m deep pool of water from a height of 10.8m? Shouldn’t he have been in incredible pain afterward?

  4. #4 rhett
    November 18, 2008

    @Alex,

    Sorry about the video. I found another version. This guy is all over the internet. If this video disappears also, just search for “professor splash”. Also, I am sure he was in pain. Such is the price of fame.

  5. #5 rhett
    November 18, 2008

    @Jasper,

    If you want, you can think of the water as a whole bunch of water particle. When the guy hits the water, he is having a collision with them so that they DO move. Also, when the water moves it collides with other water. The net result can be modeled as some force.

  6. #6 Roy
    January 8, 2009

    You are missing an important fact: The pool with 1 ft. of water is not on solid ground. It is on a several feet thick cushion. Therefore, a great deal of the force is absorbed by the cushion and not the water.

  7. #7 Aza
    November 28, 2009

    Roy, If you read the previous posts you will see that the video currently up is NOT the original video from “Time Warp” on Discovery Chan, the original video has him jumping into the very same kiddy pool with 1ft of water but over solid concrete not a soft cushion. This video is only a replacement video for the one that got taken down.

  8. #8 jiu
    June 1, 2010

    When you went from this line

    http://blog.dotphys.net/wp-content/uploads/2008/11/fwater.jpg

    to this line

    http://blog.dotphys.net/wp-content/uploads/2008/11/avgaccel.jpg

    what happened to the g? (I assuming you meant to divide by m)

    This also demonstrates why it is sometimes hard to follow the working of others

    (Thanks for your website, it is nice to be able to be watching Mythbusters get angry about them messing up, easy calculations or improperly using physics terms then by able to see someone else agrees)

  9. #9 Rhett Allain
    June 1, 2010

    @jiu,

    If I am looking at the right equations, the first one is the force the water exerts on a person. To find the average acceleration, I need to the total force. The total force would be F-water – mg (so the mg cancels).

    Yes – you are correct, maybe this is a little hard to follow. Sometimes I have to choose how much detail to include. Some people like all the details, but others just want to jump to the result. I probably should have included that step.

  10. #10 jiu
    June 1, 2010

    Thanks, for your answer.

    I embarrassed that I didn’t recognise that.

  11. #11 David
    June 14, 2011

    Haha, awesome! I saw that jump on the discover show something like real life superhero’s. They had mentioned a few of the numbers, but it’s nice to see it calculated out. Thank you!

    I’m no math expert, but I was wondering a couple things. When you talk about that insane amount of gforce, is that the total spread across his whole body, or per square something like square inch? The fourth picture down on this site shows that his legs are up (maybe bounced up from impact?), leaving greater force to the rest of his body. Ouch!

    I guess Nasa didn’t research how much of that gforce actually reaches the internal organs, as fat and muscle would play a role there.

    He talked about preferring cold water somewhere as it’s more dense and helps absorb the impact. How might that play in? Would the amount of deceleration change depending on first impact, water density, and how hard he hits the bottom?

    It’s kind of stupid asking all these questions being able to hardly follow your math in the first place, but I guess I’m just wondering what the confounding are, and how much variance the the final number could be. My guess is that it’s not much, unless he lands with only a quarter of the surface area, maybe his side, as opposed to perfectly flat. But even then, I don’t know enough to say that definitively.

    Thanks for putting up with me. Impressive physics/math skills, and I do appreciate the explanation!

    Cheers!

  12. #12 David
    June 14, 2011

    PS. Here’s the super slow motion high def video of it on discover channel: bit.ly/iVgria
    It also shows impact surface area, which I thought his technique was interesting. Kinda cool.

  13. #13 DaveC426913
    April 15, 2012

    Roy is correct. There is more than a foot of cushioning under the pool. Looks like about an 8″ foam pad on top of an 8″ airbag, for a *total* of more than 2 feet.

    If you can produce a video of him doing it on to concrete, no prob. Until then, I saw got a video proving there was a second shooter on the Grassy Knoll, but I can’t find it right now… :)

  14. #14 DaveC426913
    April 16, 2012

    ..and yes, I know how old this article is. It has resurfaced recently on a science forum.

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