1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Jumping force, and weight lifting equivalence

  1. Aug 25, 2012 #1
    Hi I like to play basketball and am curious as to how much force is required to jump a certain height I have tried to calculate it but I get stuck at energy and am not sure how to find the force as it is difficult to measure without equipment and being very accurate and precise (unless there is something obvious I could be doing and am missing).

    I have done the following calculations already:
    my average vertical jump is 27 inches and for calculations and 27 inches = .6858 meters and my desired vertical is 30 inches = .762 meters

    so using the change in gravitational potential energy Eg=mgΔh (my mass is 67.1317kg)
    Eg=(67.1317)(9.8)(.6858) and for my desired 30 inch vertical use (.762)m
    Eg=451.18 J ^ which Eg=501.31 J
    and with that if I wanted I could find out my take off speed by using the kinetic energy equation. But my main problem is what amount of force did I provide against the ground to get that speed.

    finally once I have found the force could I use the equation Fg=mg and using the force calculated to jump that high could I find the mass if I were to squat or lift I would then be able to jump that high?

    Thanks in advance for insight/comments
  2. jcsd
  3. Aug 25, 2012 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    At it's simplest the Kinetic Energy you have when you leave the ground converts to Potential Energy on the way up..

    KE = 0.5mv2
    PE = mgh


    0.5mv2 = mgh

    mass cancels giving

    V2= 2gh

    However when you jump you start by bending your knees so your trunk starts off lower. Then as you start to jump you initially accelerate different parts of your body at different rates. For example your feet/foot stays on the ground until the legs are straight again. Your arms may also move downwards initially? So the peak force on the ground might be higher than calculated?

    I think I would film a jump and work out how much lower your belt/waist goes as you bend to prepare for the jump and what height it reaches at the top of the jump. Use that data to calculate both the height of the jump "h" and the distance over which the body is accelerated "s" before it leaves the ground. Then use the equation of motion..

    V2 = U2 +2as

    where the initial velocity U = 0

    So using the earlier equation for V2..

    2gh = 0 + 2as

    a = gh/s

    F=ma so

    F = mgh/s
  4. Mar 24, 2014 #3
    use two equations imo. e=f.d then e=m.g.h, if you can find the distance your upswing takes, then you find e through m.g.h, you will be able to find force, and see how it correlates with your weight lifting. Sorry, I originally was going to reply to a similar thread where these equations would have been more appropriate....
  5. Mar 24, 2014 #4


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    If you can estimate the distance over which your muscles are acting then you can equate this work done to the gain in gravitational PE.
    Muscle force times 'stroke' length = mass times maximum height gained (from crouched position, of course)

    You could measure your mass and the stroke length and also the height attained. The only unknown in the above equation is Force - so you can calculate it.

    You would have to get a lot more complicated than that if you wanted to get a significantly better estimate of the actual Force from your muscles. The geometry of your skeleton is such that the effective lifting force is not constant over the whole of the stroke and that introduces just one extra problem.
  6. Mar 24, 2014 #5
    This looks reasonable to me except I think it assumes that acceleration is constant over the interval s. Maybe a better approximation might be to assume that the acceleration and therefore the force is proportional to s:

    a = ks
    F = mks

    Work done by F:

    W = mks2/2
    = mgh

    ∴ k = 2gh/s2
    → F = 2mgh/s

    or twice the original estimate.

    I would guess the reality is that the actual measured force would lie somewhere in between the two.
    Last edited: Mar 24, 2014
  7. Mar 25, 2014 #6
    Sounds like an interesting way of designing a workout program.

    I'm going to suggest an experimental way to do this, since the calculations written before me are similar to what I would have done.
    How about modifying a digital scale so that it records weight vs time (I'm assuming a commercial product is not available or maybe it's too expensive, but I haven't googled it). Record several jumps from the scale to get the average maximum weight recorded from which you can get maximum force applied. Actually with such a graph and if you also record your movements you could get a bunch of different quantities which might help tracking the results from your workouts, such as impulse, work, power, etc.
  8. Mar 25, 2014 #7


    User Avatar

    Staff: Mentor

    Google for "accelerometer" if only to see how these things are constructed.

    However, I'd be surprised to find that the linear approximation is not good enough.
  9. Mar 25, 2014 #8
    Hi Nugatory, what do you mean by the linear approximation?
  10. Mar 25, 2014 #9


    User Avatar

    Staff: Mentor

    from #5 in this thread: "I think it assumes that acceleration is constant over the interval s"
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook