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If I jump from height of 5ft to ground -what is g-force of impact

  1. Apr 9, 2009 #1
    I jump off a high bench (5ft high) to the ground. What is the g-force of the impact and how do you calculate it. It seems mass is irrelevant for this

    Joe
     
  2. jcsd
  3. Apr 9, 2009 #2

    mgb_phys

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    You need to know the rate that you decelerate. This is difficult to calculate
    This depend son the surface, your shoes and how you fall - you can feel this if you jump onto a gym mat in sneakers compared to jumping onto concrete in shoes
     
  4. Apr 9, 2009 #3

    HallsofIvy

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    We keep getting this kind of question. This ought to be in a "faq": You can calculate the speed and kinetic energy at impact, but what force is required to stop you depends upon how fast you decelerate which, in turn, depends upon the surface on which you land. If you land on a very hard surface, say metal or very hard wood, you will decelerate quickly and the force is comparatively large. If you land on soft earth you will decelerate slowly and the force will be less.
     
  5. Apr 9, 2009 #4

    jtbell

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    IN the case of a person landing on the ground feet first, it also depends on how much his knees and other joints flex during impact, which determines how far his center of mass moves as he comes to a stop.
     
  6. Apr 9, 2009 #5

    Nabeshin

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    Everyone is exactly correct this is not the kind of question that has a straightforward answer. You need to know a lot more than just the height you jump at, but here are some values:

    Regardless of anything else your velocity prior to impact will be:
    [tex]\frac{mv^2}{2}=mgh[/tex]
    [tex]v=\sqrt{2gh}\approx 6 m/s[/tex] For [tex]h\approx 2m[/tex]
    Corresponding to a momentum of:
    [tex]p=mv=6m[/tex]
    I don't think it's a terrible assumption to say force is constant, so:
    [tex]\Delta p=F*t[/tex]
    So we can arrive at a basic equation for the force since we know final momentum is zero:
    [tex]\frac{6m}{t}=F[/tex]

    Notes: From jtbell's post, it is obvious that this answer assumes no flexibility in the person.

    If we consider a deceleration time of jumping on concrete or some other hard surface to be perhaps .01~.05s,
    [tex]F=\frac{6m}{.01~.05}[/tex] For an average sized dude, (m=70kg),
    [tex]F=8400-42000N=12-60g's[/tex]

    These numbers seem huge, but they're actually not that outrageous. Even assuming no flexibility, the numbers are less than those achieved in some football hits.

    On a softer surface, with a deceleration time of perhaps .1~.2s,
    [tex]F=2100-4200N=3-6g[/tex]
    Which is around the roller coaster range. (A little higher).

    Consider that flexing your knees will give lower g forces, and also that I made up these deceleration times, but they are probably typical of certain impacts. Of course, the combination of shoe type and ground type will give an approximation for the deceleration time in any given circumstance. These are merely two ends of the spectrum.

    Cheers!
     
  7. Apr 10, 2009 #6
    Assume I have shoes on and jump down onto concrete, how do you compute the
    g-force. Assume I weight 70kg if that matters. G(accelerationDueToGravity)=9.8m/sec;
    I think it should be a simple equation, i.e. you start at v=0 and then hit v=final
    which is when you hit the concrete. So Vfinal-Vinitial=Vfinal/timeToStop=g-force??

    Joe
     
  8. Apr 11, 2009 #7

    Nabeshin

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    Did you read anyone's posts?

    It is not a simple equation for a complete solution and depends in an extremely complex way on the anatomy of the human body, the type of shoes you are wearing, and the surface you are jumping down onto.

    My post gave a ballpark calculation given some made up values of stopping time, which is the easiest and most basic way to get an estimate of the force.
     
  9. Apr 12, 2009 #8
    with the help of many and if I assume I jump down from 2ft and come to a stop from final velocity to 0 over the course of 0.1 second:
    given D=1/2at^2
    by solving for t: t=(2d/a)^0.5
    and given Vf=at (final velocity just before impact)

    given that my G factor is given by: (Vf/0.1)/a
    by substitution for Vf
    ((a(2d/a)^0.5)/.1)/a

    by simplifying
    ((2d/a)^0.5)/.1

    Thus from a height of 2 ft, I calculate
    ((2x2/32)^0.5)/.1 or a g-factor of 3.5355

    Does this make sense?

    Joe
     
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