If you stand on the top of a falling ladder you will hit the ground at a higher speed (and therefore presumedly sustain more injury) if you hold on to the ladder than if you jump off it. This was solved here.

Where is the "break even" height on the ladder, from where you will hit the ground with the same speed if you jump off it or if you follow it down? This question just makes an assumption that you would hit the ground more softly if you stay on to the ladder (compared to jumping off it), if you are located at the lower part of it.

I don't think the midpoint is the break even point. I quickly calculated and I think you should stay on the ladder if you are on its midpoint. (The following is just a quick computation, there could be errors in it.)

$$ v_{midpoint}^2 = \frac{1}{2}gl \frac{m + m_L}{\frac{1}{4}m + \frac{1}{3}m_L} $$

$m$ and $m_L$ are the respective masses of man and ladder, $l$ is the length of the ladder.

  • 6
    $\begingroup$ Incidentally, even better is to ride the ladder partway by standing on it (but not holding on) until gravity pulls it away from you. $\endgroup$ – Rex Kerr Aug 19 '16 at 16:00
  • 8
    $\begingroup$ Tho' both the answer here and at your link ignore the real-world issue of What's Down There. You want to jump off when you're more likely to land on soft ground than rocks or a poison ivy patch. :-) $\endgroup$ – Carl Witthoft Aug 19 '16 at 16:40
  • 1
    $\begingroup$ @RexKerr and what do you do if it's falling over backwards? I'm pretty sure it'd be impossible to "stand on it" without holding on. $\endgroup$ – Doktor J Aug 20 '16 at 5:26
  • 3
    $\begingroup$ When you say "jump" do you mean "let go" or do you mean "jump" (like, propel yourself away from the ladder with force)? $\endgroup$ – Jason C Aug 20 '16 at 8:50
  • 1
    $\begingroup$ From the title I expected this question to be "How long should I stay on a falling ladder before jumping off?" $\endgroup$ – trichoplax Aug 20 '16 at 17:38

You should stand at 2/3 of the height of the ladder.

If you land with the same kinetic energy as without a ladder, then the ladder should land with the same kinetic energy as without you. Equating the kinetic energy of the ladder with its potential energy at the beginning:

$$\frac{1}{2} mgL = \frac{1}{2} I_L \omega^2 = \frac{1}{2} \left(\frac{1}{3} mL^2\right) \omega^2$$ gives: $$\omega = \sqrt{\frac{3g}{L}}$$

where $L$ is the length, $m$ the mass, $I_L$ the moment of inertia and $\omega$ is the angular velocity of the ladder.

For you the same equation holds, but now $\omega$ is known:

$$MgH = \frac{1}{2} I_M \omega^2 = \frac{1}{2} (MH^2) \left(\frac{3g}{L}\right)$$ with $M$ your mass, $I_M$ your moment of inertia and $H$ your height. Solving for $H$ gives:


or of course $H=0\;.$


We can do this with a minor modification to the calculation described in the earlier question. As before we'll take the ladder length to be $\ell$, but now we'll take your height to be $\alpha\ell$, where $\alpha$ ranges from zero to one. Our reference point is if you let go, in which case your speed when you hit the ground will be:

$$ v^2 = 2g\alpha\ell \tag{1} $$

Now suppose you hold onto the ladder. As before we calculate the total potential energy change of both you and the ladder, which is:

$$ V = mg\alpha\ell + \frac{1}{2}m_Lg\ell \tag{2} $$

And this must be equal to the increase in angular kinetic energy $\tfrac{1}{2}I\omega^2$. The combined moment of inertia of you and ladder is:

$$ I = m(\alpha\ell)^2 + \frac{1}{3}m_L\ell^2 $$

And setting the kinetic energy equal to the potential energy gives:

$$ mg\alpha\ell + \frac{1}{2}m_Lg\ell = \tfrac{1}{2}\left(m\alpha^2 + \frac{m_L}{3}\right)\ell^2\omega^2 $$

And since $v=r\omega$ your velocity is $v=\alpha\ell\omega$ giving:

$$ mg\alpha\ell + \frac{1}{2}m_Lg\ell = \tfrac{1}{2}\left(m\alpha^2 + \frac{m_L}{3}\right)\ell^2\frac{v^2}{\alpha^2\ell^2} $$

Which rearranges to:

$$ v^2 = g\ell\alpha^2 \frac{2m\alpha + m_L}{m\alpha^2 + \frac{m_L}{3}} $$

And finally substitute for $v$ from equation (1) to get:

$$ 2g\alpha\ell = g\ell\alpha^2 \frac{2m\alpha + m_L}{m\alpha^2 + \frac{m_L}{3}} $$

And this rearranges to:

$$ \alpha = \frac{2}{3} $$

So if you are more than $\tfrac{2}{3}$ of the way up the ladder you should let go, while if you are lower than $\tfrac{2}{3}$ of the way up the ladder you should hang on.

  • $\begingroup$ Thanks for both replies. I hadn't really "expected" the masses of ladder and man to be left out of the solution. $\endgroup$ – ycc_swe Aug 19 '16 at 15:04
  • 1
    $\begingroup$ @ycc_swe: Actually I was a bit surprised to find that everything cancelled out in the last equation. It looked as though it was going to be complicated then all the terms cancelled! $\endgroup$ – John Rennie Aug 19 '16 at 15:06
  • 1
    $\begingroup$ "You should let go" - not necessarily, if you are at the top of the ladder you have some angular momentum as the ladder tips, making it easier to transfer the impact energy into an acrobatic roll. Granted, you'd better have your wits about you. $\endgroup$ – David Wilkins Aug 19 '16 at 16:21
  • $\begingroup$ @DavidWilkins - If your size is negligible compared to the size of the ladder (and, let's face it, who doesn't go around climbing ladders that allow you to be approximated as a dimensionless point, when one finds oneself in a physics problem), then your velocity when the ladder hits will be exactly normal to the ground. No acrobatic rolling for you! If you let go partway down, then yes, you may be able to reduce the forces in the lateral direction by rolling (thus extending your deceleration time). $\endgroup$ – Rex Kerr Aug 20 '16 at 2:04
  • 3
    $\begingroup$ I can only say that having jumped off a roof with forward momentum, which translated to angular momentum as gravity influenced my trajectory, and having rolled out of the fall instead of falling flat, that it does make a difference. Yes I scared a few bystanders, but physically it made sense. And worked at the time $\endgroup$ – David Wilkins Aug 20 '16 at 2:11

If one has the presence of mind, I vote for sliding down the ladder at any cost - gloves and shoes acting as a control brake. Shortening the diameter (height of your feet from ground) will surely reduce any impact. The risk is loss of control in the vertical decent. Gloves might be a recommended safety feature of any ladder climbing to keep confidence while executing a rapid slide. I would think you could get 6 feet lower before getting into the lateral arc. I admit that my own experience is limited to sliding off a wet cedar shingle roof and kicking the ladder away as I slid into it. It was only 10 feet onto damp lawn so I was saved by that and walked away. Yeah, I've had the training but I landed on my butt. No roll was possible.

  • $\begingroup$ Your theory seems wrong $\endgroup$ – QuIcKmAtHs May 5 '18 at 2:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.