Would it help if you jump inside a free falling elevator?

While everyone agrees that jumping in a falling elevator doesn't help much, I think it is very instructive to do the calculation.

General Remarks

The general nature of the problem is the following: while jumping, the human injects muscle energy into the system. Of course, the human doesn't want to gain even more energy himself, instead he hopes to transfer most of it onto the elevator. Thanks to momentum conservation, his own velocity will be reduced.

I should clarify what is meant by momentum conservation. Denoting the momenta of the human and the elevator with $p_1=m_1 v_1$ and $p_2=m_2 v_2$ respectively, the equations of motion are

$$ \dot p_1 = -m_1 g + f_{12} $$ $$ \dot p_2 = -m_2 g + f_{21} $$

Here, $f_{21}$ is the force that the human exerts on the elevator. By Newton's third law, we have $f_{21} = -f_{12}$, so the total momentum $p=p_1+p_2$ obeys

$$ \frac{d}{dt} (p_1 + p_2) = -(m_1+m_2) g $$

Clearly, this is not a conserved quantity, but the point is that it only depends on the external gravity field, not on the interaction between human and elevator.

Change of Momentum

As a first approximation, we treat the jump as instantaneous. In other words, from one moment to the other, the momenta change by

$$ p_1 \to p_1 + \Delta p_1, \qquad p_2 \to p_2 + \Delta p_2 .$$

Thanks to momentum "conservation", we can write

$$ \Delta p := -\Delta p_1 = \Delta p_2 .$$

(Note that trying to find a force $f_{12}$ that models this instantaneous change will probably give you a headache.)

How much energy did this change of momentum inject into the system?

$$ \Delta E = \frac{(p_1-\Delta p)^2}{2m_1} + \frac{(p_2+\Delta p)^2}{2m_2} - \frac{p_1^2}{2m_1} - \frac{p_2^2}{2m_2} .$$ $$ = \Delta p(\frac{p_2}{m_2} - \frac{p_1}{m_1}) + (\Delta p)^2(\frac1{2m_1}+\frac1{2m_2}) .$$

Now we make use of the fact that before jumping, the velocity of the elevator and the human are equal, $p_1/m_1 = p_2/m_2$. Hence, only the quadratice term remains and we have

$$ (\Delta p)^2 = \frac2{\frac1{m_1}+\frac1{m_2}} \Delta E .$$

Note that the mass of the elevator is important, but since elevators are usually very heavy, $m_1 \ll m_2$, we can approximate this with

$$ (\Delta p)^2 = 2m_1 \Delta E .$$

Energy reduction

How much did we manage to reduce the kinetic energy of the human? After the jump, his/her kinetic energy is

$$ E' = \frac{(p_1-\Delta p)^2}{2m_1} = \frac{p_1^2}{2m_1} - 2\frac{\Delta p\cdot p_1}{2m_1} + \frac{(\Delta p)^2}{2m_1}.$$

Writing $E$ for the previous kinetic energy, we have

$$ E' = E - 2\sqrt{E \Delta E} + \Delta E = (\sqrt E - \sqrt{\Delta E})^2 $$

or

$$ \frac{E'}{E} = (1 - \sqrt{\Delta E / E})^2 .$$

It is very useful to estimate the energy $\Delta E$ generated by the human in terms of the maximum height that he can jump. For a human, that's roughly $h_1 = 1m$. Denoting the total height of the fall with $h$, we obtain

$$ \frac{E'}{E} = (1 - \sqrt{h_1/h})^2 .$$

Thus, if a human is athletic enough to jump $1m$ in normal circumstances, then he might hope to reduce the impact energy of a fall from $16m$ to a fraction of

$$ \frac{E'}{E} = (1 - \sqrt{1/16})^2 \approx 56 \% .$$

Not bad.

Then again, jumping while being weightless in a falling elevator is likely very difficult...

As an addition to already posted answers and while realising that experiments on Mythbusters don't really have the required rigour of physics experiments, the Mythbusters have tested this theory and concluded that:

The jumping power of a human being cannot cancel out the falling velocity of the elevator. The best speculative advice from an elevator expert would be to lie on the elevator floor instead of jumping. Adam and Jamie speculated the attendant survived because the tight elevator shaft created an air cushion. This together with spring action from slack elevator cable could have slowed the car to survivable speeds.

(This myth is fueled by the story of an elevator attendant found alive but badly injured in an elevator car that had fallen down a shaft in the Empire State Building after a B-25 Medium Bomber crashed into it in 1945.)

The reason that jumping can make a relatively large difference is that the kinetic energy is proportional to the square of the velocity. Thus relatively small changes to the velocity can result in relatively large changes to the kinetic energy. In addition, the velocity which a human can achieve in jumping is a substantial percentage of the velocity of fatal falls.

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Let the human weigh $m$, let him jump with upward velocity $v$ and let the elevator fall from a height $H$. Then the human's initial potential energy will be $10mH$. What fraction of this potential energy can he avoid having turned into kinetic energy?

At any given point before jumping, the human's kinetic energy and potential energy add up to $10mH$. If he jumps at a height of $h$, his potential energy will be $mgh$ and his kinetic energy will be $mg(H-h) = 0.5mV^2$ where $V$ is the elevator (and human before jumping) velocity, taken as a positive number so that $V=\sqrt{2g(H-h)}$.

At the moment of jumping, he will not reduce potential energy, but instead will decrease his velocity. So his kinetic energy decreases from $0.5mV^2$ to $0.5m(V-v)^2$. Therefore his total energy will become:
$$mgh + 0.5m(V-v)^2$$ $$= mgh + 0.5m(\sqrt{2g(H-h)}-v)^2$$ $$= mgH +0.5mv^2 - mv\sqrt{2g(H-h)}.$$ The terms have a simple interpretation. $mgH$ is the energy in the absence of any jumping. $0.5mv^2$ is the energy of the jump (in the frame of reference of the human). And the remaining term is the reduction in energy due to the reference frame conversion.

We wish to make the third term as negative as possible. This occurs when h is small so we put $h=0$ (as our intuition suggests, indeed, the best time to jump is just as the elevator impacts). Then the remaining kinetic energy is: $$mgH +0.5mv^2 -mv\sqrt{2gH}.$$

An example of a height $H$ which is generally fatal for a human is $H=10m$. A maximum velocity for a very athletic human jump is on the order of $v=3.64$ m/s. Such a jump would give a maximum height of 0.66 meters. See: Vertical Jump Test calculator for data on human jumping capabilities by sex, age, and athletic ability. Using $g=10$ and $m=50$, the kinetic energy before and after jumping are: $$mgH = 5000J$$ $$mgH+0.5mv^2-mv\sqrt{2gH} = 2757J$$

Thus, in fact, jumping could reduce the kinetic energy suffered by a factor of two. The final collision with the floor would be reduced from a height of 10m = 32.8 feet, to a height of 5.5m = 18 feet.