# Vertical jump vs. weight

1. Oct 1, 2011

### jon_harris7

Imagine an athelete that weighs 180 pounds (81.6 kg) is able to jump 20" (0.51 meters) vertically. How high could he jump if he were able to decrease his weight to 170 pounds (77.1 kg)?

I think the basic idea here is to find out, given the mass, what force is required to reach the specified height. Then, you would assume the force is constant but the mass is decreased. That yields an increased acceleration for the same force, which leads to a higher max jump. But I haven't figured out the specifics. Any ideas?

It would also be nice to know the answer in a more generalized sense, i.e. increase in jump vs. decrease in weight.

This is motivated by a real-world scenario of an athelete trying to increase his vertical jump and wanting to know how much improvement can be made simply by losing weight.

Thanks for any help!

2. Oct 1, 2011

### Rayquesto

weight doesn't matter. As long as the force applied to go up is the same which will depend upon the energy the body requires for the force to take place, then the initial velocity when jumping will be the same. Therefor, they will be able to jump the same distance. In practice, a person will likely jump at the same force. So, in practice, the acceleration will decrease upon increased mass. So, the distance, in pratctice, will decrease.

3. Oct 1, 2011

### Rayquesto

the real world problem comes down to a combo of biology, psychology and physics.

4. Oct 1, 2011

### Jamma

Incorrect, the weight will matter. Saying that it won't is like saying that a man can jump the same height as a man carrying a boulder.

You could work out how much energy the jumper can put into a jump. Then work out the initial velocity given the mass (which will decrease as you increase the mass) and from this you can work out the height he attains.

5. Oct 1, 2011

### 1mmorta1

What would make this question even more interesting is to ask how high the man could jump if he lost the wait AND considered the fact that he would lose strength when he loses wait.

6. Oct 1, 2011

### 1mmorta1

Weight. Sorry

7. Oct 1, 2011

### davenn

not a logical assumption. there's plenty of people that are leaner than me and are much stronger and could jump higher or longer !!!

Dave

8. Oct 1, 2011

### Rayquesto

I don't understand some people. Your weight does not affect how high you can go. I finally realized that during my studies of motion in space relative to Earth (in the absence of air drag). Once jumped, a person's velocity at which they initiate a jump will affect the maximum height and other factors involved in the motion in air since only gravity and velocity affect the quantities of distance while the person is in air.

the thing I was suggesting was that it is likely a person will push with a more initial velocity if they decreased their mass and so they'd go a higher max height assuming that force is constant.

9. Oct 2, 2011

### Jamma

You are right that two bodies have the same motion in gravity if they start off with the same initial velocity.

However, in this case, the heavier person won't be able to achieve the same initial velocity as the lighter person. Here I am assuming of course that both can put the same amount of energy into the jump, so it's kind of like (but not precisely the same as) looking at the difference between a man jumping and a man jumping with a heavy backpack on (try and do this yourself, you won't be able to jump as high!).

More concretely, the kinetic energy, E=(0.5)mv^2, so if their energies of jump are the same then v=sqrt(2E/m). So the initial velocity goes down and hence (of course) the height of the jump, because from that point, as you noticed, the mass won't matter.

Hope that's clear.

[EDIT]:Oh, I see that you noticed the change in velocity. Your post was a bit misleading: "weight doesn't matter. As long as the force applied to go up is the same ..." It seems a very odd assumption to think that the energy of the jump should scale for a person as E=(0.5)mv^2 where v is constant. Isn't it far more natural to assume that the energy of the jump is constant? (a fat person losing weight doesn't mean they can now put only E=(0.5)dv^2 into the jump where d is the change in their weight).

Last edited: Oct 2, 2011
10. Oct 2, 2011

### jon_harris7

It is certainly possible to lose weight without losing a significant amount of strength. So the loss of strength can be safely neglected for my purposes. But you're right, it would make for a more interesting/more difficult problem. Training methods would be a large factor, e.g. is strength training done along with weight loss vs. weight loss is achieved by diet alone, etc.

11. Oct 2, 2011

### Jamma

I suppose you can keep reducing your weight until you run out of fat and start going into protein and you should still be able to put the same strength into a jump (roughly).

Why not find how high you can jump, your percentage weight in fat, and then find out how high you could jump if all the fat was gone.

12. Oct 2, 2011

### jon_harris7

Jamma, your comment about the energy of the jump being the same put me on the right track. It turns out to be a very simple relationship: increase in vertical jump is directly proportional to decrease in mass/weight. Weigh half as much, jump twice as high!
So to answer my original question, for an individual weighing 180 pounds and jumping 20 inches, decreasing weight to 170 pounds will yield a jump of 21.2 inches.

For the derivation, the key equations are
1) h = 0.5*v02/g, h is the maximum height reached for a given initial velocity v0 and
2) E=0.5*m*v02, energy required to achieve a given initial velocity v0 for a body with mass m
Note here that both h and E are proportional to v02. That is key.

Combining these, you get:
3) h = E/(m*g), the maximum height reached for a given energy and mass.

So to find the change in h vs. change in m you get with E constant:
4) h2/h1 = m1/m2, where m1 and m2 are the old/new masses and h1 and h2 are the old/new heights

Or, to find the new vertical jump h2 that could be achieved from decreasing weight from m1 to m2:
5) h2 = h1*m1/m2

Neat!

13. Oct 2, 2011

### Ken G

That is certainly the easiest way to do it, but I doubt it's actually true in practice. You do have to assume something stays the same when you lose weight-- but what is that something? Saying the stored chemical energy is the same is the only simple assumption that comes to mind, but it is probably not true, because converting stored energy in the muscles into kinetic energy is a very lossy process. To see this, imagine squatting like you are going to jump as high as you can. Then squat even farther. You can hardly jump at all! Are you storing less energy in your muscles? No, you are inhibiting the perfect physiological balance you need to get that energy out. If you lose weight, you might find you can squat a little deeper and still maintain that perfect physiological balance, so you might be able to store even more potential energy. I suspect the real effect is therefore even more than what you got.

This is easy to test-- it's hard to lose weight, but easy to add it. Best would be to do it in the "linear" limit of small added weight, but then you'd need more accurate measurements. In the opposite limit of a large weight, it is easy to see that the proportional effect you derive won't hold up-- if you wear a 100 pound backpack, I wager you could hardly get off the ground at all.

14. Oct 2, 2011

### Jamma

This is true Ken.

For some reason it reminds me of the following thing:

How high does the squirt of a water gun go compared to the size of the hole? Many people will tell you that making the hole smaller will make it go higher, because logically if you have a very large hole it won't "squirt". But this isn't really the case at all, but there is a problem in that if you make the hole too large you can't really apply the force to the liquid inside in the same way.

So the problem is that with the change, there is a change in the way the energy is applied into converting the energy into kinetic energy, much like this example. (You can test the above by getting two plastic bottles and pierce them with a hole near the bottom, different sizes for each, at the same height up the bottle. Put them on a ledge full of liquid and see how far the water goes- it will be the same).

It could be that a large person could put more energy into the jump in the same way that I can put more energy into throwing a cricket ball than I can a ball of paper. Perhaps not, it's all a bit too complicated and probably doesn't add too much of an effect if you assume its the same person who just happens to be a bit lighter/heavier.

15. Oct 2, 2011

### Rayquesto

"You are right that two bodies have the same motion in gravity if they start off with the same initial velocity.

However, in this case, the heavier person won't be able to achieve the same initial velocity as the lighter person. Here I am assuming of course that both can put the same amount of energy into the jump, so it's kind of like (but not precisely the same as) looking at the difference between a man jumping and a man jumping with a heavy backpack on (try and do this yourself, you won't be able to jump as high!).

More concretely, the kinetic energy, E=(0.5)mv^2, so if their energies of jump are the same then v=sqrt(2E/m). So the initial velocity goes down and hence (of course) the height of the jump, because from that point, as you noticed, the mass won't matter.

Hope that's clear."

Yea that makes sense! :) It's common sense. haha I knew that as soon as I read this question. How much more obvious is it to know that someone heavier will jump with a less initial velocity than the other with less mass? The original question asks about how high you'd go. So, adding math into the question forced me to think initially about the motion in which the body would be in air. Then I thought about the internal energy that will affect the initial velocity. SO, as a result, I'm sure some biology, psychology and even statistics would have to be taken account for this calculation, without observation unless you know of a different way. Maybe you can use some kinetic energy or potental energy equations maybe even momuntem-impulse equations. the potential energy would be like 0 though and the impulse during the jump would require a change in force over a change in time, like the force to overcome the weight to jump.

16. Oct 2, 2011

### Jamma

As I said, when I reread it was clear that you had noticed that the initial velocity would change. I was just a bit confused when you said "weight doesn't matter" which would mean you'd be implementing some quite unnatural scaling of the energy of the jump to make that so, when the only non-arbitrary thing to do would be to leave the energy the same.

17. Oct 2, 2011

### Ken G

One way to subtract weight without "wasting away" the jumper is to imagine it on the Moon, or on an asteroid with gravity 1/100 that of Earth. It's certainly true that you might find a kind of saturation at very weak gravity, where lessening gravity just makes it harder to "gain purchase" as you try to jump up. The nonlinear effects at low or high weight can't be used to determine what would happen with small weight gains or losses, that's why it would be best to do the experiment with small weights (though you'd need precise measurements). My expectation is that the scaling of height with weight goes from less steep than an inverse proportion for a very light person, and more steep than inverse proportion for a very heavy person, and where the average person is on that scale is hard to say. Most likely it depends on how athletic you are already. But again, given all this complexity, the assumption that it is an inverse proportion is probably the best you can do in general.