How Much Higher Can I Jump If I Lose 25lbs?

[Kombaiyashii]

Hi,

This is not homework, I'm 29 years old. I'm just curious...

Lets say I weigh 180lbs and I can jump 20 inches off the ground.

If I lose 25lbs making me 155lbs. How much higher should I be able to jump? (assuming that I'm not losing any muscle mass).

Is the answer around 23 inches?

 

[DennisN]

I get the same estimate when I did a quick calculation. But it would be more physically illuminating if you specified how you calculated your estimate . And perhaps which assumptions you used.

 

[Kombaiyashii]

I basically just turned the new weight into a percentage of the old weight i.e.

155/180 = 86%. 100 - 86 = 14%

Then I added 14% onto the height. 20/100*14=2.8+20 = 22.8ish

However, I heard that NASA scientists use a different method for stuff like this. Is there one?

 

[HallsofIvy]

More precisely, F= ma. I interpret your "assuming that I'm not losing any muscle mass" as meaning F stays the same while m is reduced. If your old mass and initial acceleration are m and a', and your new mass and acceleration are m' and a', then we have ma= m'a' so that a'= (m/m')a.

If you lose 25 out of 180 pounds then (m/m')= (180/155)m= (36/31)m and so a'= (36/31) a. Essentially that says that the acceleration is inversely proportional to the mass. Assuming, further, that your feet stay in contact with the ground for the same length of time, with v= at, we have that v'= (36/31)v.

Now, it gets more complicated because "height of jump" is not directly proportional to initial speed. We have that s= -(g/2)t^2+ v_0t which has derivative -gt+ v_0. At the top of the jump, that will be 0: t= v_0/g and the height of the jump is given by -(g/2)(v_0^2/g^2)+ v_0^2/g= -v_0^2/2g+ v_0^2/g= v_0^2/(2g). That is, the height of the jump is proportional to the square of the initial speed. With v' now equal to (36/31) times v, h', the new height, will be (36/31)^2= 1.35 times the previous height.

That is, all other things being equal, if, at 180 pounds, you were able to jump 20 inches, at 155 pounds you should be able to jump 1.35(20)= 26.9 or about 27 inches, not the "23 inches" you get by assuming a direct, linear, proportion.

 

[jbriggs444]

If you lose 25 out of 180 pounds then (m/m')= (180/155)m= (36/31)m and so a'= (36/31) a. Essentially that says that the acceleration is inversely proportional to the mass. Assuming, further, that your feet stay in contact with the ground for the same length of time, with v= at, we have that v'= (36/31)v.

Which assumptions are made in the model figure crucially into the conclusion, of course.

The explicit assumption above is that of fixed force over fixed time. The implicit assumption above is that the fixed force is large enough and the fixed time is short enough that we can ignore the impulse due to the force of gravity over the duration of the take-off.

A different assumption might be to consider a fixed force over a fixed distance instead, again simplifying the problem by assuming a distance short enough so that one can ignore the work done by the force of gravity over the distance of the take-off.

A more realistic model might reach different conclusions.

 

[DennisN]

I assumed same energy/work for the two cases, and used potential energy W = mgh.
This yields W₁ = m₁gh₁ and W₂ = m₂gh₂. W₁ = W₂ yields h₂ = m₁*h₁/m₂ = 180*20/155 = 23,2 (ca)
Excluding drag and assuming constant g. And conversions to kg/m/s cancel. Considering what a jumper really does, I think HallsofIvy's assumption/calculation is more realistic than this. A nice example that different assumptions yield different results .

 

[Khashishi]

I suppose it depends on if your muscles are limited by force or power or speed. One problem with trying to keep a constant force is that your muscles are limited in terms of speed. So you can't throw a light ball much faster than you can throw a heavy ball because your muscles just don't move that fast. I think the result will be somewhat higher than 23in, but less than 27in. It has something to do with how much fast twitch and slow twitch muscle you have in your legs. I know I can kick much faster than I can jump, so I'm limited more by force than by speed, so I think the number will be closer to 27in.

 

[sophiecentaur]

I assumed same energy/work for the two cases, and used potential energy W = mgh.
This yields W₁ = m₁gh₁ and W₂ = m₂gh₂. W₁ = W₂ yields h₂ = m₁*h₁/m₂ = 180*20/155 = 23,2 (ca)
Excluding drag and assuming constant g. And conversions to kg/m/s cancel. Considering what a jumper really does, I think HallsofIvy's assumption/calculation is more realistic than this. A nice example that different assumptions yield different results .

I like this approach best. As with any other approach have to assume that your body makes the best use of the Energy it has available so mgh should be the same in both cases. No need for quadratic equations of motion. Back of the proverbial fag packet.

 

[jbriggs444]

How far down would a typical jumper crouch when taking off for this 20 inch vertical jump? How much of the jump is assisted by standing on tip-toe?

As a rough approximation, I would suggest that we're talking about a 16 inch crouch plus 2 inches of tip-toe standing for the takeoff. Then we're talking about 18 additional inches of free-fall trajectory.

Nice and symmetric. 18 inches of take-off and 18 inches of free fall.

If we assume constant acceleration during the take-off phase, That means that the legs must supply 2 g's worth of acceleration on the jumper's 180 pound mass for the original jump. That's 360 pounds force.

Now switch to the 155 pound jumper.

360 pounds force divided by 155 pounds mass gives a nominal vertical acceleration of 2.32 g's during take-off. Subtract gravity and that's 1.32 g's of actual upward acceleration.

1.32 g over 18 inches of take-off should be equal and opposite to 1 g over <free-fall-trajectory-height> inches. That gives a free fall height of 18*1.32 = 23.76 inches. But we were assuming 2 inches of tip-toe. That totals to 25.76 inches of vertical jump. Call it 26 inches.

 

[sophiecentaur]

How far down would a typical jumper crouch when taking off for this 20 inch vertical jump? How much of the jump is assisted by standing on tip-toe?

As a rough approximation, I would suggest that we're talking about a 16 inch crouch plus 2 inches of tip-toe standing for the takeoff. Then we're talking about 18 additional inches of free-fall trajectory.

Nice and symmetric. 18 inches of take-off and 18 inches of free fall.

If we assume constant acceleration during the take-off phase, That means that the legs must supply 2 g's worth of acceleration on the jumper's 180 pound mass for the original jump. That's 360 pounds force.

Now switch to the 155 pound jumper.

360 pounds force divided by 155 pounds mass gives a nominal vertical acceleration of 2.32 g's during take-off. Subtract gravity and that's 1.32 g's of actual upward acceleration.

1.32 g over 18 inches of take-off should be equal and opposite to 1 g over <free-fall-trajectory-height> inches. That gives a free fall height of 18*1.32 = 23.76 inches. But we were assuming 2 inches of tip-toe. That totals to 25.76 inches of vertical jump. Call it 26 inches.

This is making loads of detailed assumptions - like constant acceleration, which is very doubtful, and also his foot length etc etc. Bear in mind that the guy was interested in losing all that weight. That means he would just be in better muscular shape by the time he actually shed all those pounds and would probably have improved on his motor skills through all those work-outs. So you are dealing with two situations which would really not be comparable. That's why I say cut out the long winded pseudo calculations and simply work on the principle of assuming he can develop the same amount of energy, which will give him the same GPE at the top of his jump. Without knowing a lot more details then how can you improve on that assumption?

 

[jbriggs444]

You are correct, of course, that the crouch analysis is depending on additional assumptions which are not neccessarily realistic.

The assumption of a 16 inch crouch is the result of experiment. I crouched as if in preparation for a jump and made a measurement.

The assumption of 2 inches up on tiptoes is the result of experiment. I measured four inches but figured that the acceleration would be tailing off and adjusted it to two inches.

The assumption of constant acceleration comes out of thin air.

I like to think that the addition of detail results in a more accurate result. But in the absence of actual experimental data (anybody got a jumper who weighs 155 pounds and has a 35 pound suit he can put on) a simple analysis may indeed be as good as a complicated one. My apologies if I had implied otherwise.

 

[A.T.]

BTW: You can jump higher if you go down and up continuously, rather starting the upwards acceleration from a static crouch. This is due to a delay in muscle force build up.

 

[sophiecentaur]

BTW: You can jump higher if you go down and up continuously, rather starting the upwards acceleration from a static crouch. This is due to a delay in muscle force build up.

And the resilience / energy storage of ligaments and tendons too, no doubt.

 

[d3mm]

And the resilience / energy storage of ligaments and tendons too, no doubt.

Absolutely correct, and the quoted poster too.

If anyone is interested there are many videos on the internet that teach the correct technique for the standing high jump, and it can be found that the correct technique makes a huge difference to the result.

 

[BelowAverage]

Like a previous comment said, the variables of force and speed are just as important as the mass. Strength will allow greater force to be applied into the ground which will increase you vertical. However, if your legs aren't quick enough it will limit your maximal height as your speed will cease to accelerate. Therefore, it is optimal to incorporate both speed training, such as sprints, as well as heavy weighted exercises, like squats. An interesting sight is http://www.dunkcalculator.com which gives a fairly accurate estimation of how high you need to jump as well as how much force you need to generate. If you want details on a programme that will increase your vertical you can consult me further :)

Sources -
Me ((completed a 2 year sport course and I have an extensive interest in physical development) evidence provided by me in a sexually intimidating pose))

My 32 inch vertical