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How much force or speed is needed to jump

  1. Jun 9, 2009 #1

    inh

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    How fast does something need to accelerate in order to jump? Say a weight of 100 lbs, or kg's, etc.
     
  2. jcsd
  3. Jun 9, 2009 #2

    Dale

    Staff: Mentor

    How high?

    The equation is:

    mgh = 1/2 mv²

    Which is assuming that you are jumping straight up.
     
  4. Jun 9, 2009 #3

    inh

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    I don't have a set height, or weight really, just wondering how to determine how much force it takes to make X amont of weight move Y feet vertically.

    I get the rest of the equation, but what does mgh stand for?
     
  5. Jun 9, 2009 #4

    Nabeshin

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    m: mass
    g: gravitational acceleration near the surface of the earth. (9.8m/s^2 in SI units, about 32 feet per second^2 in imperial units).
    h: the height of the jump.
     
  6. Jun 9, 2009 #5

    inh

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    exactly what i was looking for

    thank you very much :)
     
  7. Jun 9, 2009 #6

    inh

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    how off am i? using a mass of 10 lbs, height of 1 foot, and g = 32.15 feet, i get 10336.23 for mgh.

    v^2 = .5 * 10 * 10336.23

    v = 227.33 feet per second

    so to move a 10 weight 1 foot in the air it needs to have a velocity of 227 feet per second? seems a bit high....
     
  8. Jun 9, 2009 #7

    Dale

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    Uhh, 10*32.15*1 = 321.5 < 10336.23
     
  9. Jun 9, 2009 #8

    Nabeshin

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    Uhh your algebra is a bit off. Solve for v in the equation DaleSpam posted earlier, you'll find the mass doesn't even matter.

    +DaleSpam's calculation correction.
     
  10. Jun 9, 2009 #9

    inh

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    isnt it 10 * 32.15^2 * 1 and not 10 * 32.15 * 1 ?

    also, i was simplifying the equation earlier, and i believe i got it down to v^2 = .5gh which netted 4.009 for v for a 1 foot jump. that looks better to me, how about you?
     
  11. Jun 9, 2009 #10

    diazona

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    4.009 what?

    The formula you got is almost right though - you just put your factor of 2 in the wrong place ;-)
     
  12. Jun 9, 2009 #11
    [itex]\frac{1}{2}mv^2=mgh[/itex]

    rearranges to: [itex]h=\frac{v^2}{2g}[/itex] of [itex]v=\sqrt{2gh}[/itex] so to get 1 foot (which i'm going to say is approximately a third of a meter since I'm not american) and g is approximately 10m/s%2 gets me [itex]v=\sqrt{2(10)(1/3)}\approx 2.6 m/s[/itex]
     
  13. Jun 9, 2009 #12

    inh

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    4.0 fps^2

    looks like i was off by half :) thanks for the help guys
     
  14. Jun 9, 2009 #13

    Nabeshin

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    Note the answer for velocity has units of fps, not fps^2.
     
  15. Jun 9, 2009 #14

    inh

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    ah yea, sorry about that. v = 4.0 fps, the actual acceleration is fps^2, sorry about the confusion
     
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