There's a particular myth regarding an elevator and a person in it that goes like this:

If a person is in a free-falling dive in an elevator (say the steel wires have broken and the emergency brakes do not work for some reason) for the person to save him/her self from the ground hit, it would be sufficient to jump, energetically, at the exact last instant before the elevator hits the ground.

Usually, in the way this thought experiment is presented, the hypothesis are the following (although typically not stated):

  • No friction between the elevator frame and the walls.
  • Air friction is neglected (there is no theoretical terminal speed of the elevator to be reached).
  • The person would be able to perfectly time the jump, if the impact occurs at $t = 0$ then the jump would occur at $t=0^-$.
  • The initial height of the free-fall is such as to allow the elevator/person system to reach a significant speed (say 45 $m/s$ for example) before reaching the ground.

Why does the conclusion of the myth is wrong? I thought I could find at least 2 ways of showing it is plainly wrong:

The first one uses relative motion composition by considering a fixed and a uniformely accelerated reference frame:

  • The fixed reference frame (reference frame 1) is the one of a person from the ground watching the elevator move.
  • The uniformely accelerated one is the one of the person in the elevator which is accelerating at $g=9.81$ meters per second squared towards the ground.

The instantaneous absolute speed in the $y$ direction is given by: $$v_{abs_y}(t) = v_{refframe1}(t) + v_{rel_y}(t) = \frac{1}{2}gt^2$$ where $t = 0$ is the initial time of the free-fall. If the person jumps $\delta t$ before the ground hit, then $v_{rel_y}(t+\delta t)=-|k|$ but $|k| << v_o(t+ \delta t) = 45 m/s$ due to the fact that it simply is not possible for a person to reach 45 $m/s$ speed with a jump. The absolute speed before touchdown would then be roughly the absolute speed of 45 $m/s$.

The second one is far simpler. Consider the fixed reference frame: one instant prior to the hit, at time $t=0^-$ the person's body has kinetic energy equal to $\frac{1}{2}mv^2$. If at time $t=0$ the person has zero speed, that means that the kinetic energy has been dissipated in an infinitesimal amount of time leading to the development of infinite power from the jump which is absurd/impossible. Even if you allow for some small (but not infinitesimal) amount of time the power required would still be far too much for a person's leg muscles to develop.

Even if you take into account terminal speed by considering air friction and other frictions, both the resonings still hold. What I would like you to answer in this question is if you find any flaws with my two solutions to this problem or inconsistent arguments from the physics point of view. While you are at it, feel free to add your explanation of why it would/would not work.

I know this has already been covered in SE questions such as here and here but in this question I'd like to debate the possible flaws of my two particular arguments.

  • 3
    $\begingroup$ The conclusion isn't wrong. Certainly if you could jump off of the elevator with enough force to make your downward velocity $0$ then you're good to go. The problem is that you probably can't supply enough force to do this with just your legs $\endgroup$ Commented Sep 1, 2019 at 12:54
  • 1
    $\begingroup$ I think the myth busters tried this once & were unable to time it properly, but I recall the wreckage was sufficient to "kill" their ballistics dummy they shoved in there. $\endgroup$
    – Kyle Kanos
    Commented Sep 1, 2019 at 13:49
  • 1
    $\begingroup$ @AaronStevens All in all I believe that jumping off of the elevator with enough force to make your downward velocity 0 is just exposing the poor guy to same acceleration as smashing together with the elevator itself. In short, to survive he has to reduce the acceleration to safe values, so he just needs to start slowing down earlier, but he can't becouse he's not at ground level yet, $\endgroup$
    – carloc
    Commented Sep 1, 2019 at 17:45
  • $\begingroup$ Mythbusters did this exact experiment some time back. I'm sure you can find the video on YouTube. $\endgroup$ Commented Sep 1, 2019 at 18:02
  • $\begingroup$ @carloc True. Another reason why in practice this is a no go $\endgroup$ Commented Sep 1, 2019 at 21:01

3 Answers 3


Both of your approaches are fine in principle, but your logic is rather loose and inconsistent.

The points to bear in mind are as follows:

  1. The maximum vertical speed a very fit person might attain by jumping from standstill will be at most of the order of a few metres per second. If you could jump at 5m/s you would reach a height of c1.2 metres, which would be a pretty spectacular achievement from a standing start. I suspect the average person would struggle to achieve 0.5m, which would be a speed of around 3m/s. Either way, it would make almost no difference to the outcome if the elevator was to hit the ground at 45m/s- an impact at 40m/s or 42m/s would kill you just as well.

  2. The point above ignores the fact that the effect of jumping will be to increase the downward speed of the elevator somewhat, thus decreasing the upward speed of the jump, but for an elevator that is many times the mass of a person, you can probably ignore the effect.

  3. Jumping at the very last moment is a crazy refinement- you want to time your jump so that you maximise your upward speed at the moment of impact. Your body will take time to accelerate to its maximum speed in a jump, so you should take-off slightly earlier.

  4. The condition you impose in your second argument, of losing all one's KE almost instantaneously by jumping is clearly nonsense. The acceleration required would be more or less the same as the acceleration due to the impact itself.

  5. The urban myth is patently false. It is like suggesting that a pedestrian could survive a head-on collision with a truck at 45m/s by walking backwards from it just before impact.


Whether or not the person is "saved" would depend on the duration of the free fall which would determine the persons velocity upon impact. For example, If a person jumps from some height above the ground the person can lesson the impact force by bending the knees immediately on contact. This reduces the average force on the person by the work energy principle. On the other hand the action wouldn't save the person if the jump was from a tall building.

The same applies to the elevator if the elevator cables failed shortly before reaching the ground. But I'm not sure jumping alone would reduce the force if the person still lands straight legged, i.e., without bending the knees on impact, because that would not reduce the average impact force. Suppose the person was suspended from the ceiling of the elevator by a rope so that the person's feet were several feet from the ground (the height of a jump) and the rope was cut just prior to impact. It's the same situation as the person jumping from a height. The person still needs to bend the knees to lessen the average impact force.

Of course if the cable failed several floors from the bottom, no action would be sufficient to save the person.

Hope this helps.


Conservation of Energy (as discussed):

Person does work on elevator, because direction of force = direction of motion. Therefore, Energy increases for elevator and decreases for person. Whether “significant” depends.

Conservation of Momentum:

System momentum is $(M+m)v$ before and after jump (before impact). Person’s speed change $- \Delta v$. For elevator: $\Delta V = \tfrac{m}{M} ~\Delta v$.

If elevator mass is high, and if one could normally jump a meter, then $\Delta v = 4.5 \tfrac{m}{s}$. (From: $d=1= vt = (0.5gt)t \implies t= 0.45, \text{ and}~v_f=0=v_i-gt$)

I’d rather hit at 40.5 than 45! And impact energy $E=\tfrac{1}{2} mv^2$ has reduced by almost 20%!

Note: the Timing doesn’t matter except if one would jump so early as to hit the ceiling.

There is nothing saying the jump takes zero time or is infinitely intense. If strong enough, could get to $v=0$.


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