# Weight loss and vertical leap

March 17, 2005 12:16 PM Subscribe

BasicPhysicsFilter: Everything else being equal, how much would my vertical leap improve if I lost 10% of my weight without losing any muscle?

Best answer: I figured something like this, but I'm no physics expert:

F = M * A

100 = 100 *A = 1

100 = 90 * A = 1.1111111

so 11.1% would be right.

posted by Sonic_Molson at 12:44 PM on March 17, 2005

F = M * A

100 = 100 *A = 1

100 = 90 * A = 1.1111111

so 11.1% would be right.

posted by Sonic_Molson at 12:44 PM on March 17, 2005

Best answer: If the energy is the same, then that means at the height of your leap the new and old potential energy match. This is just MGH, where M-mass, G-grav constant, H-height.

So, if M1*G*H1=M2*G*H2 just do some arithmetic and you get H2/H1=M1/M2.

In the Case of 10% weight loss, M2=0.9*M1, so H2/H1=1.11...

Or, what drpynchon said.

posted by vacapinta at 12:46 PM on March 17, 2005

So, if M1*G*H1=M2*G*H2 just do some arithmetic and you get H2/H1=M1/M2.

In the Case of 10% weight loss, M2=0.9*M1, so H2/H1=1.11...

Or, what drpynchon said.

posted by vacapinta at 12:46 PM on March 17, 2005

Damn, now that I thought it over; that's all wrong. Your acceleration will improve by 11.1%, but it will definetly not improve your jump height by 11.1%. At least I don't think so.

I didn't do so well in Physics :-(

posted by Sonic_Molson at 12:46 PM on March 17, 2005

I didn't do so well in Physics :-(

posted by Sonic_Molson at 12:46 PM on March 17, 2005

Best answer: Newton's 2d law: Force = mass * acceleration

F = ma

trivially,

a = F/m

Distance = .5 * acceleration * time^2, when starting at rest (details)

d = .5 * a * t^2

So, original distance:

d1 = .5 * (F/m) * t^2

Distance with 90% of the mass and the same force:

d2 = .5 * (F/.9m) * t^2

d2/d1 = (.5 * (F/m) * t^2)/(.5 * (F/.9m) * t^2)

= 1/.9

= 10/9

= 1.11

= 11.1% gain

You were right the first time, Sonic Moison.

posted by Zed_Lopez at 12:50 PM on March 17, 2005

F = ma

trivially,

a = F/m

Distance = .5 * acceleration * time^2, when starting at rest (details)

d = .5 * a * t^2

So, original distance:

d1 = .5 * (F/m) * t^2

Distance with 90% of the mass and the same force:

d2 = .5 * (F/.9m) * t^2

d2/d1 = (.5 * (F/m) * t^2)/(.5 * (F/.9m) * t^2)

= 1/.9

= 10/9

= 1.11

= 11.1% gain

You were right the first time, Sonic Moison.

posted by Zed_Lopez at 12:50 PM on March 17, 2005

You were right the first time Sonic, and you were semi-right the second time. If we assume that his strength is unchanged, neither is the force or energy he exerts. All of these things are conserved. Once you have conservation of energy, the problem is solved as shown above.

But also, potential energy can be thought of as the path integral of the force over a distance, which is what gets you the simplification of Ep = mgh. The force in question is constant, and while your acceleration will go up 11.1% your mass will fall 10% so F = m1a1 = m2a2 = (0.9m1)(1.11a1) = m1a1.

You might gain a few inches, which makes sense if you think about it.

posted by drpynchon at 12:57 PM on March 17, 2005

But also, potential energy can be thought of as the path integral of the force over a distance, which is what gets you the simplification of Ep = mgh. The force in question is constant, and while your acceleration will go up 11.1% your mass will fall 10% so F = m1a1 = m2a2 = (0.9m1)(1.11a1) = m1a1.

You might gain a few inches, which makes sense if you think about it.

posted by drpynchon at 12:57 PM on March 17, 2005

Yep, I got 11.1% as well (before peaking into the thread). So apparently I can jump 1/0.8, or 1.25 times (or 25%) higher than at the beginning of the year (since I've lost about 20% of my weight).

posted by Doohickie at 1:40 PM on March 17, 2005

posted by Doohickie at 1:40 PM on March 17, 2005

assuming your muscles are equally efficient at pushing two different distributions of weight over your body up in the air, yes.

posted by plexiwatt at 2:03 PM on March 17, 2005

posted by plexiwatt at 2:03 PM on March 17, 2005

There is that whole thing about ancient greek standing long jumpers using weights in their hands to improve their jump.

Then there is the Lance Armstrong vs. Jan Ulrich thing... If your muscles can fire really fast then you want to be light, but if you can't extend as quickly you probably get more energy invested in the jump with a little more weight. A little more energy may or may not mean more height of course. You have to do the math, or preferably the experiment, to know for sure.

Humph... This biomechanics article says that my Armstrong analogy is the same effect as the weights... well sort of anyway...

posted by Chuckles at 8:28 PM on March 17, 2005

Then there is the Lance Armstrong vs. Jan Ulrich thing... If your muscles can fire really fast then you want to be light, but if you can't extend as quickly you probably get more energy invested in the jump with a little more weight. A little more energy may or may not mean more height of course. You have to do the math, or preferably the experiment, to know for sure.

Humph... This biomechanics article says that my Armstrong analogy is the same effect as the weights... well sort of anyway...

posted by Chuckles at 8:28 PM on March 17, 2005

This thread is closed to new comments.

posted by drpynchon at 12:26 PM on March 17, 2005