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

1. Jun 11, 2007

### dongfe

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. Jun 11, 2007

### Danger

Well, let's see... you'll have to factor in the aerodynamics of your sveldt new shape...
Sorry, dude... I can't help on this.

3. Jun 11, 2007

### dongfe

haha, you are right, I might take off like a rocket. I better do a follow up on escape velocity :S haha

4. Jun 11, 2007

### rbj

well, if the 50 lbs came solely out of your thighs, i would guess less high than you can jump now.

5. Jun 11, 2007

### dongfe

"and lets assume I dont lose any strength in my legs.."

6. Jun 11, 2007

### Danger

Rbj is a genius, but he doesn't read very well. :tongue:

7. Jun 11, 2007

### smallphi

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 ....

8. Jun 11, 2007

### ice109

there is absolutely no way to predetermine this. just lose the weight, i guarantee you jump a lot higher.

9. Jun 11, 2007

### dongfe

LOL, there has to be a way to figure this out.

10. Jun 11, 2007

### ice109

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

$$PE=PE$$

$$m_igh_i=m_fgh_f$$

$$\frac{m_ih_i}{m_f}=h_f$$

edit

which i just noticed is exactly what you said.

11. Jun 11, 2007

### nrqed

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

$m_b g h_b = m_a g h_a$ where a=after and b = before so that

$h_a = h_b \frac{ m_b}{m_a} = h_b (1 + \frac{\Delta(m)}{m_a})$
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!

12. Jun 11, 2007

### ice109

:rofl: ......

13. Jun 11, 2007

### dongfe

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???

14. Jun 11, 2007

### ice109

you yourself have already figured out the answer and 2 more people suggested the same method.

15. Jun 11, 2007

### dongfe

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)

16. Jun 11, 2007

### gabee

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

$\frac{m_1 v_1}{t_c} = \frac{m_2 v_2}{t_c} \\$
$\frac{m_1 \sqrt{2gh_1}}{t_c} = \frac{m_2 \sqrt{2gh_2}}{t_c} \\$
$\frac{m_1}{m_2} = \sqrt{\frac{2gh_2}{2gh_1}} \\$
$(\frac{m_1}{m_2})^2 = \frac{h_2}{h_1} \\$

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,

$\frac{m_1}{m_2} = \frac{h_2}{h_1}$

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
17. Jun 12, 2007

### ice109

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

18. Jun 12, 2007

### KingNothing

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".

19. Jun 12, 2007

### dongfe

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

20. Jun 12, 2007

### ice109

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.

21. Jun 12, 2007

### dongfe

If you did say that, I would still lose the weight. If you said the opposite I would still lose the weight. Isn't this a site where you can post questions about physics? My question was simple, the things that you mentioned are obviously effected by losing 50lbs otherwise you wouldn't have mentioned them correct? HOW are they effected?? Not a very unreasonable question. How are each of those things effected by losing weight and what effect do they have on jumping? I understand that its not possible to calculate how high someone will be able to jump but if all of these things will add to the jump then its safe to say what we calculated is a MIN increase. You offered an opintion and now I'm asking you to explain it. You did, I'm asking you to go further and explain how each of the things you mentioned would effect a jump (+'ive or -'ive). If you don't want to then thats ok, but the line of questions is fine, nothing ridic. about it.

22. Jun 12, 2007

### ice109

my original point still stands, it is very difficult tell how any of those things will be affected. you realize elite athletes pay trainers and doctors and a lot of other people to design exercise routines to maximize gains. this isn't physics, this is a lot of different sciences.

23. Jun 12, 2007

### KingNothing

Ice: Physics questions that neglect things have been asked for as long as the study of physics has existed. Trainers and doctors make assumptions everyday, too.

dongfe: If you have serious motivation to get a better vertical, please private message me.

Last edited: Jun 12, 2007