I am studying from the OpenStax university physics textbook and am having trouble reproducing the results after checking the answer to the problem. I have searched for a better explanation, but without a teacher I am hoping I can get the help I need here.

An athlete can jump a distance of 8.0 m in the broad jump. What is the maximum distance the athlete can jump on the Moon, where the gravitational acceleration is one-sixth that of Earth?

It's clear from the problem $a_y = -\frac{9.8}{6} \frac{m}{s^2}$, and additionally $x_0 = y_0 = 0$.

What isn't clear is that there is no given velocity. No long jumper is jumping 8m without an initial velocity! This had me confused enough to check the answer, which was $48m$, or $8m * 6$. Implying that since it "takes 6 times as long to reach the ground" the long jumper can jump 6 times as far (in their total displacement). Though intuitive, I am not aware of this type of proportionality from the textbook.

I can't come up with a toy problem off the top of my head to check this. Would someone be kind enough to explain to me how displacement scales with changes in acceleration mathematically?


1 Answer 1


Let the resp. $g$s be: $$g_m=\frac{g_e}{6}$$ $$v_{0,y}=v_0\sin\theta$$ where $v_0$ is the initial (launch) velocity and $\theta$ the angle with the horizontal.

The vertical velocity component is: $$v_{t,y}=v_0\sin\theta-gt$$ which reaches $0$ when: $$v_0\sin\theta-gt=0\Rightarrow t=\frac{v_0\sin\theta}{g}$$ The total time airborne is thus: $$2t=\frac{2v_0\sin\theta}{g}$$ The horizontal displacement is given by: $$x(\Delta t)=v_0\cos\theta \Delta t$$ So that: $$x=v_0\cos\theta \frac{2v_0\sin\theta}{g}=\frac{2v_0^2\sin\theta \cos\theta}{g}$$ So that, all other things being equal: $$\boxed{\frac{x_m}{x_e}=\frac{g_e}{g_m}}$$ Or: $$x_m=6x_e$$

  • $\begingroup$ Now, the big question is whether all other things can be equal (in a realistic scenario)! Can a runner achieve the same maximum velocity with a different strength of gravity? I would argue not! Then, one could ask what strength of gravity is optimal... I assume more friction with the ground is always better for speed, but maybe one can achieve faster speeds by repeated "bounding" in a weaker gravitational field. $\endgroup$ Commented Apr 28, 2021 at 22:26
  • 1
    $\begingroup$ @QuantumMechanic Completely agreed, of course. But I think 'all being things equal' is in the sprit of the question. $\endgroup$
    – Gert
    Commented Apr 28, 2021 at 22:40
  • $\begingroup$ Fantastic reply! I only have a couple follow ups. The $g$ is the equations is referring to gravity in general, and not a particular gravity (earth/moon) correct? The follow up to that is the leap from your final equation to the proportional equation confused me slightly. I was following the math up completely up to that point. Is the result because we have the x value as a fraction of gravity, and we can do the same for moon's gravity, and derive a proportion from that? If so, I understand and I can green check this! $\endgroup$
    – CL40
    Commented Apr 28, 2021 at 23:25
  • $\begingroup$ I'll answer these 2morrow, ta. $\endgroup$
    – Gert
    Commented Apr 29, 2021 at 3:46
  • $\begingroup$ @Gert will you have some time soon? $\endgroup$
    – CL40
    Commented May 1, 2021 at 2:49

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