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

Question

Imagine you're trapped inside a free falling elevator. Would you decrease your impact impulse by jumping during the fall? When?

Answer 1

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.)

Answer 2

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...

Answer 3

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.

Answer 4

From the question of simple reduction of velocity, the answer's already been given (yes, but not enough to make a significant difference) ... but there's one other issue at play here -- how the forces are transfered to the body.

If you're standing upright, it'd all be transfered through your legs; as Flaviu mentioned, laying down so the force is spread across a larger area would be a better option to this. But, if you could manage to jump at just the right time, and you knew how to take a fall (bend your knees, roll into it, etc.), it might be possible to spread the force over a greater time and distance, therefore reducing the impulse, and thus the actual damage to your body.

Unfortunately, I don't think the chances of timing it correctly would be very good, so it wouldn't be particularly advisable. You'd have to weigh the risk & benefit of this strategy vs. just lying down.

Answer 5

It's not about whether or not you can jump up fast enough to cancel out a 60mph impact. If you could jump up at 60 mph, you wouldn't need to because passively absorbing the impact (60mph deceleration) would be less stressful than actively accelerating upward to 60 mph (total impact cancellation), because you would be subjecting yourself to the same-if not greater-'g' forces. It seems more practical to jump up at 30 mph (partial impact cancellation) which can effectively cut down the 'g' forces by distributing the braking distance, sort of like the last second "braking" rockets of cosmonaut space craft that fire just before a parachute landing.

Impact severity is largely defined by the shortness of stopping distance. There's little-if any-substitution for distributing stopping distance in alleviation of impact. So, the question is more one of whether or not jumping up is ever the wisest option.

Suppose you're in an elevator that's headed for an uncushioned landing at twice the velocity you can jump up. You're coming down at 10 mph, and can jump up at 5mph. If your feet leave the floor at the precise moment you reach 5 mph, the deceleration would be a 2X5mph impacts, each with 1/4 the kinetic energy of an unmodified 10 mph impact, which equals half the impact. The main problem is, you reach your highest (and lowest) speed of 5 mph at the critical part of your upward jump where you're in the worst position to absorb the balance of the impact by rolling into it. You might be safer ONLY doing that-namely flexing your knees, and rolling into what skydivers refer to as a parachute landing fall, or plf.

Answer 6

If you jumped just before impact, your speed towards the bottom of the elevator shaft would go down a little bit. But consider that the elevator falls tens of meters, while you jump about one meter. Your jumping ability is quite small and probably won't make a noticeable difference

Answer 7

No. You'd reduce your impact velocity, but only with a negligible small fraction.