# Why can't we jump very high?

If somebody is able to output $$3750$$ $$\text{N}$$ lifting weight on the back doing squats, why can't they jump at $$150$$ $$\text{km}/\text{h}$$ without weight?

With weight:

$$(300 \text{ kg of load} + 75\% \text{ of }100 \text{ kg of the bodymass}) \times (9.8\text{ m}/\text{s}^2 + 20 \text{ cm}/\text{s}^2 = 10\text{m}/\text{s}^2) = 3750 \text{ N}$$

Without weight: $$3750 \text{ N } / ~75\text{ kg} = 50 \text{ m}/\text{s}^2-g = 40.2\text{ m}\times 3600 \text{ h}^{-1}\text{s}^{-1} = 144720 \text{ m } \text{h}^{-1} \text{s}^{-1} \approx 145~\text{km} \text{ h}^{-1} \text{s}^{-1}$$

I don't measure work because the force is applied over the same distance.

1. What are the restraining factors? Is it more a matter of frequency of micro-impulses? A failure in coordination? Do the cells fail to keep up with their own speed?
2. Where does the energy released to produce the force lifting the weight go if it's not converted into kinetic energy as much as it should without weight? Into heat?

Because I assume the same neurologic signal can be sent to fire the cell with or without the weight.

I've been told that physics in physiology was tricky and misleading however, also I thought something like ATP might have a limited potential to travel inside the cell in terms of speed and reach what it has to reach. I'm not very familiar with the concept of power.

• I've removed a number of comments that were attempting to answer the question and/or responses to them. Please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. Commented Jun 3, 2020 at 20:09

Force x Distance = Work. Mr. Muscles can press 3750 N, but over what distance can he apply this force? At most 0.4 m, starting from a squat. That makes 1500 J. If he were to jump with that much kinetic energy at takeoff, at 6.3 m/s, gravity would exert 735 N on him, and he could jump 2.05 m high. Given the realities of human legs and muscles, this is a wild exaggeration. (Good figure skaters doing triple jumps rise only 0.4 m.)

Why is it hard to impart so much kinetic energy? Not enough time. Given your boy’s takeoff speed, he would have had only 0.127 seconds to impart the energy.

As for 150 km/h = 41.6 m/s, your boy would need 65 kJ of kinetic energy. Vroom. (Psst. Please learn to use MKS consistently: seconds, not hours.)

A muscle moving slowly can perform a lot of lifting work against a load like a massive barbell. But to jump high, you have to set yourself in fast motion, and for quick movement, the load experienced by the muscle becomes the muscle's own mass. This problem can't be fixed by adding more muscle, since that just adds more mass. This sets a natural limit on how quickly a muscle can propel the skeleton to which it is attached, even if the skeleton is hinged so as to act as a lever. This in turn means there is a natural limit to how high you can jump straight up, because your maximum upward velocity is limited.

• The muscle can push much more than its own mass, so the extra mass by adding fibers can't be the restraining factor in gathering speed; my point was that usually when you cut the mass by three, you expect the velocity in the expression of impulse to be multiplied by three for impacts as well as the acceleration in the expression of force if some work is done on an object. Commented Jun 3, 2020 at 9:09
• Have you read a treatment of jumping from a biomechanics standpoint? There you will learn the reasons why a grasshopper can jump hundreds of times its body length while an elephant cannot jump at all. Commented Jun 3, 2020 at 16:37

## How our body "works"

Let's figure out what factors does our ability to do something depends on. Let's say you had to lift a $$10 \:\rm kg$$ weight and a $$100 \:\rm kg$$ weight upto the same height (and starting from the same height) in about ten seconds). Without getting into the specific numbers, we can conclude that lifting the heavier weight is definitely going to cost us more energy and thus it's expected to be tiresome. Thus, we get our first factor as load.

But now imagine if you only had to lift the $$100 \:\rm kg$$ block (to the same height), however, you have to do it now by lifting it $$1\: \rm cm$$ every hour until the block reaches that height$$^{\dagger}$$. This would definitely be less tiring than the first case. But according to physics, the work done in both the scenarios is exactly the same (ignoring any other activities you did during that boring hour ;)). Which indicates that load isn't sufficient enough to decide the feasibility of doing something. And thus we get another factor, time.

Now both of these factors don't affect our capacity in the same way. Increasing the load makes the task harder (in this instance, to lift the load to a certain height). On the other hand, increasing the time during which the task is to be performed, makes the task easier. So there's some vague sort of relationship such that

$$D\propto F\qquad \text{and} \qquad D\propto \frac 1 t$$

where $$D$$ is the difficulty of doing the task, $$t$$ is the time duration and $$F$$ is the load. But this sort of expression is very similar to the expression of power

$$P=\frac{W}{t}=\frac{Fd}{t}$$

where $$P$$ is the power, $$W$$ is the work done and $$d$$ is the distance across which the force acts. (Do note that the above expression is a crude simplification of the correct power expression). So, we can satisfactorily say that power is quite a nice measure of how infeasible any task is.

## Questions

What are the restraining factors? Is it more a matter of frequency of micro-impulses? A failure in coordination? Do the cells fail to keep up with their own speed?

So, now whenever we jump, we usually take a quarter of a second to accelerate ourselves. And let's say we jump up at the speed of $$10 \rm m/s$$ (it's quite fast, just for context, it's about equal to the average speed with which Usain Bolt ran his record 100 metres race). Also, let's assume the person to person to have a weight of $$80 \rm kg$$. Computing the power required to do this (assuming no losses):

$$P=\frac{W}{t}=\frac{\left(\frac 1 2 mv^2\right)}{t}\approx 16000 \text W$$

Just to get a feel for how big this power is, the average power consumed by a water boiler is $$8000 \:\text W$$. So you can power up two such water boilers with the power you require to jump with a speed of $$10 \rm m/s$$.

Do note that in the above expression, $$P\propto v^2$$ which implies that the power varies quadratically with the jump velocity. This also implies that an increase in a large velocity by a certain amount will require a higher increase in power than the same certain increase in a small velocity, or in other words, it becomes harder and harder to increase the velocity as the velocity gets larger. Mathematically, this is true because $$\displaystyle \frac{\mathrm dP}{\mathrm d v}\propto v$$.

Now, you can see the effect of time factor in the following scenario. Imagine climbing up the stairs to reach a height of $$100 \:\rm m$$ in 5 minutes. Easy enough. Now imagine jumping and reaching the height of $$100 \:\rm m$$. Impossible! But, the interesting thing is that in both the cases the difference between the initial energy (at the ground, before jumping) and the final energy (at the momentary rest on the top) is the same, which implies, we did the same amount of work, or expended the same amount of energy in both the cases. But in the jumping case, we had to do it within a quarter of a second, whereas in the case of climbing the stairs, we did it over a 5 minute period.

Where does the energy released to produce the force lifting the weight go if it's not converted into kinetic energy as much as it should without weight? Into heat?

Well, the most of the energy expended got used up in increasing the gravitational potential energy of that body. There are also other biological losses, but since this is a physics answer, I might not deal with them here :-) So, ideally speaking, all the work that you did was used to lift the body up and, thus, increase it's gravitational potential energy.

$$^{\dagger}$$You don't have to hold the block (after lifting it) for an hour. Lift it a cm, put it at a platform at that height, relax. Lift it another cm after an hour, put it at a platform at that height, and then relax. Repeat this. This process is analogous to climbing the stairs, take step, relax, take another step, relax, and on and on. Whereas the instantaneous lifting is similar to jumping.

• "lifting it 1 cm every hour until the block reaches that height" is definitely much more tiring than simply lifting it up at once. Simply holding a heavy block at a given height is already tiring, and even holding your empty hand in front of yourself for extended periods of time can be tiring. Commented Jun 3, 2020 at 18:44
• @Ruslan Well, I should have mentioned it explicitly but you don't have to hold the block (after lifting it) for an hour. Lift it a cm, put it at a platform at that height, relax. Lift it another cm after an hour, put it at a platform at that height, and then relax. Repeat this. This process is analogous to climbing the stairs, take step, relax, take another step, relax, and on and on. Whereas the instantaneous lifting is similar to jumping.
– user258881
Commented Jun 3, 2020 at 19:57