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What is the physics behind a normal human jump ?

a human jump

when a normal human wants to jump.

  1. First, some energy is stored in their thighs and The elastic tendons Just like a spring. In mechanics and physics, Hooke's law of elasticity is an approximation that states that the extension of a spring is in direct proportion with the load applied to it. Many materials obey this law as long as the load does not exceed the material's elastic limit

$$F=-k.x$$ where x is the displacement of the spring's end from its equilibrium position (a distance, in SI units: metres); F is the restoring force exerted by the spring on that end (in SI units: N or kg·m/s^2); and k is a constant called the rate or spring constant (in SI units: N/ m

what is human rate constant k?

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Can you explain what you mean? What kind of jump? If one exactly like in your image, what part of it is mimportant? There are a lot of aspects to take into account, so if you can give context that would help. –  Rory Alsop Aug 18 '12 at 10:10
    
@Rory Alsop ok i'll do it –  jump Aug 18 '12 at 11:32
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Human (and indeed most) tendons are poor springs and it is primarily the active system (i.e. muscles) that power a jump, so the question is a on shaky ground from the get-go. Kangaroos on the other hand have a relatively efficient energy storage and return system in their legs. –  dmckee Aug 18 '12 at 15:56
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1 Answer

Assuming you're asking how to find $h$, you need the equation relating velocity to acceleration and distance:

$$v^2 = u^2 + 2as$$

where $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration and $s$ is the distance travelled.

Your first step is to calculate the velocity you leave the ground. Assuming the muscles in your thighs can generate a force $F$, the acceleration $a$ will be given by Newton's law $a = F/m$ where $m$ is your mass. The distance travelled is just $s_2 - s_1$, and the initial velocity is zero. Putting all this in the equation above gives:

$$v^2 = 2 \frac{F}{m} (s_2 - s_1)$$

Now you know the velocity, the height $h$ is calculated using the first equation and setting the acceleration to be $g$, the acceleration due to gravity ($g = 9.81ms^{-2}$):

$$v^2 = 2gh$$

or substituting for $v^2$:

$$2 \frac{F}{m} (s_2 - s_1) = 2gh$$

to give:

$$h = \frac{F}{mg} (s_2 - s_1)$$

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Or in other words, forget all this spring stuff. The reason you lift off the ground is that you are carried by your own inertia. –  Chris White Dec 16 '12 at 20:38
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