Dot Physics

Check this out.

So, the guy jumps from 150 feet into some cardboard boxes. Why are the boxes important? You want something that can stop you in the largest distance to make your acceleration the smallest. Here is my Dangerous Jumping Calculator. Basically, you put in how high you will jump from and how much distance you will take to land and it tells you your acceleration.

You will probably need this G-force tolerance info from wikipedia.

One problem – this calculator doesn’t really work for this case. It doesn’t take into account air resistance. Does air resistance even matter in this case? How about a plot of a person falling from 150 feet with and without air resistance? (I just used my code from the Red Bull Stratos Jump calculation)


This is a difference in speed of just about 3 m/s (or almost 7 mph). I guess every little bit helps.


  1. #1 Alex Besogonov
    April 24, 2010

    Your terminal speed will be about 200 km/h, which is 55 m/s. So after about 5 seconds it won’t really matter how much more you’re going to fall :)

  2. #2 Omega Centauri
    April 25, 2010

    The force the jumper feels is actually one G worse than your calculation. The jumper must feel os force of 1G just to counteract gravity, i.e. if the boxes gave him a force of 11G his speed would remain constant. So instead of feeling 14G, he is actually feel 15.

    It is also pretty unlikely that the decleration is constant.

    From your graph, the wind resistance slows his speed by about 10%. But that is a twenty percent drop in kinetic energy, so it does help considerably. Perhaps he could also try holding an umbrella as he falls as well, that would increase wind drag considerably.

  3. #3 Anonymous Coward
    April 27, 2010

    Re: Omega Centauri at #2

    As one nit-picker to another, I agree with what I think is your general idea, but I’m pretty sure when you say

    > i.e. if the boxes gave him a force of 11G his speed would remain constant.

    you’re wrong.

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