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If I lost 50lbs how much higher could I jump?

  1. Jun 11, 2007 #1
    I'm 230lbs and I jump about 36 inches, if I were to lose 50lbs and get down to 180lbs and lets assume I dont lose any strength in my legs.. how would I calculate how much higher I would be able to jump. I wrote some stuff down and I got the jump would increase by the same percentage of % weight loss, but that doesn't seem right. Anyone got any ideas? I use f=ma and w=mxd but now that I think about it, that wouldn't work
     
  2. jcsd
  3. Jun 11, 2007 #2

    Danger

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    Well, let's see... you'll have to factor in the aerodynamics of your sveldt new shape... :biggrin:
    Sorry, dude... I can't help on this.
     
  4. Jun 11, 2007 #3
    haha, you are right, I might take off like a rocket. I better do a follow up on escape velocity :S haha
     
  5. Jun 11, 2007 #4

    rbj

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    well, if the 50 lbs came solely out of your thighs, i would guess less high than you can jump now.
     
  6. Jun 11, 2007 #5
    "and lets assume I dont lose any strength in my legs.."
     
  7. Jun 11, 2007 #6

    Danger

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    Rbj is a genius, but he doesn't read very well. :tongue:
     
  8. Jun 11, 2007 #7
    Assumptions:

    1. your muscles will appy the same average force F during the jump
    2. your body will move the same distance L while pushing on the ground with your legs, that distance is determined by the length cleared by legs from folded to unfolded position

    The work = F L remains the same, independent of your weight and will lead to the same kinetic energy of your body at the moment your feet leave the ground. The rest is conservation of your kinetic+potential energy while your in the air ....
     
  9. Jun 11, 2007 #8
    there is absolutely no way to predetermine this. just lose the weight, i guarantee you jump a lot higher.
     
  10. Jun 11, 2007 #9
    LOL, there has to be a way to figure this out.
     
  11. Jun 11, 2007 #10
    thats like saying theres a way to predict the whether if you knew all of these initial conditions of the earth. how high you jump is dependent on a lot of extremely complicated things. you can only approximate. if you assume your power output doesn't increase this relationship is as good as anything else

    [tex] PE=PE[/tex]

    [tex] m_igh_i=m_fgh_f[/tex]

    [tex] \frac{m_ih_i}{m_f}=h_f[/tex]

    edit

    which i just noticed is exactly what you said.
     
  12. Jun 11, 2007 #11

    nrqed

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    My two cents...
    If we assume that the power(force per unit time) delivered by the legs remains the same and that it is applied over the same distance, then the initial energy is the same as before. And we have

    mgh = initial energy

    Therefore

    [itex] m_b g h_b = m_a g h_a [/itex] where a=after and b = before so that

    [itex] h_a = h_b \frac{ m_b}{m_a} = h_b (1 + \frac{\Delta(m)}{m_a})[/itex]
    so yes, the percent increase is the same as the percent decrease of the mass IF we define the percent decrease by the difference of mass divided by the mass after!
     
  13. Jun 11, 2007 #12
    :rofl: ......
     
  14. Jun 11, 2007 #13
    all I meant is only one thing is being changed, the mass of the person jumping, this should be fairly easy to figure out shouldn't it???
     
  15. Jun 11, 2007 #14
    :confused::confused: you yourself have already figured out the answer and 2 more people suggested the same method.
     
  16. Jun 11, 2007 #15
    oh, i wrote that in repsonse to you, I hadn't checked for msgs after yours when I read that :)

    just curious as to what other variables there are other than the weight that have changed?

    thanks for your help everyone, i guess I was right but it still feels wrong lol. I guess I will lose the weight and work on the time it takes to fully staighten the legs (fast which muscles)
     
  17. Jun 11, 2007 #16
    If you assume the average applied force is the same (as you originally asked), and that the time of contact with the ground remains the same, then

    [itex]
    \frac{m_1 v_1}{t_c} = \frac{m_2 v_2}{t_c} \\
    [/itex]
    [itex]
    \frac{m_1 \sqrt{2gh_1}}{t_c} = \frac{m_2 \sqrt{2gh_2}}{t_c} \\
    [/itex]
    [itex]
    \frac{m_1}{m_2} = \sqrt{\frac{2gh_2}{2gh_1}} \\
    [/itex]
    [itex]
    (\frac{m_1}{m_2})^2 = \frac{h_2}{h_1} \\
    [/itex]

    However if you assume, as in the other posts, that your legs can do the same amount of work in each jump, you get a linear relationship,

    [itex]
    \frac{m_1}{m_2} = \frac{h_2}{h_1}
    [/itex]

    Intuition seems to suggest that the second of these is the "more correct" approach...it takes longer contact time with the ground to jump up if you are holding heavy weights than it does without.
     
    Last edited: Jun 11, 2007
  18. Jun 12, 2007 #17
    muscle recruitment, efficiency, vO2 max, maximum power output, lots and lots and lots of things
     
  19. Jun 12, 2007 #18
    This forum impresses me sometimes. You ask what time it is, and ten people explain why you can't possibly know for sure, before one person says "four o'clock".
     
  20. Jun 12, 2007 #19
    Ice, of the following things

    "muscle recruitment, efficiency, vO2 max, maximum power output, lots and lots and lots of things"

    because I'm not sure what they are and how they work.. would losing 50lbs have a negative effect in anyway or do all this variables also add to the vertical if 50lbs was lost. Atleast that way I I would know what I have calculated is a MIN
     
  21. Jun 12, 2007 #20
    no offense dude but this is a ridiculous line of questioning? if i say no quite possibly all these things will suffer and your maximum jump will decrease are you going to not lose the 50lbs? whats the point of figuring out exactly how beneficial losing 50lbs is because you know that it will be.
     
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