# What is the longest distance that can be jumped after swinging from a rope?

In the movie Mission Impossible 3, the main character Ethan Hunt tries to enter a building in Shanghai by swing through the sky, as shown below:

The jump consists of 2 sections, the red part, which is an arc, and the orange part, which is a parabola.

A question naturally arises: For a target building with a given height, what is the furthest horizontal distance it can be placed such that we can jump to its roof using this method?

Before answering this question, we must consider the following fact:

Observing the path when we release at different heights, we see that if we release very late, we can reach great heights, but with reduced horizontal distance. If we release early, we increase horizontal distance, but with reduced height. Because of this trade off, the area that can be reached is restricted within the yellow line。

The question is then:

(1) What is the equation x(z) that describes the orange line above?

(2) For a given point (x,z) that is inside the region bounded by the orange line, what is the proper release height h(x,z) such that the parabola will go through (x,z). The answer should be a set of h. This is the same as asking what is the proper release height if we want to reach a building with height z and is located at x distance away from the center building.

For the ease of discussion, I propose that we use the follow coordinate system and notation :

Here are some calculations that are already done: At the release point, $$\text{Upward shooting angle}=\psi$$ $$v=\sqrt{2gh}$$ $$h(\psi)=r\cos\psi-H$$ $$\psi(h)=\arccos\frac{h+H}{r}$$

Upward and horizontal speeds are: $$v_{u}=v\sin\psi$$ $$v_{h}=v\cos\psi$$ Time for Ethan to fly through the orange part is: $$t=\frac{2v_{u}}{g}=\frac{2v\sin\psi}{g}$$ Thus $$d_{h}=v_{h}\cdot t$$

$$=v\cos\psi\frac{2v\sin\psi}{g}$$

$$=\frac{v^{2}}{g}\sin2\psi$$ If we were to use $h$ as the variable, then $$d(h)=\frac{v^{2}}{g}\sin\left(2\arccos\frac{h+H}{r}\right)$$

$$d(h)=\frac{\left(\sqrt{2gh}\right)^{2}}{2g}\left(\frac{h+H}{r}\right)\sqrt{1-\left(\frac{h+H}{r}\right)^{2}}$$

$$d(h)=h\left(\frac{h+H}{r}\right)\sqrt{1-\left(\frac{h+H}{r}\right)^{2}}$$

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1
– David Zaslavsky♦ Sep 24 '12 at 23:38
Thank you very much David! – Xiaowen Li Sep 25 '12 at 1:48

## 1 Answer

Amazingly, not only has someone recently done an analysis of the Tarzan swing, but they've posted it on the Arxiv. See http://arxiv.org/abs/1208.4355 for the article or the Arxiv Blog for a summary.

To quote the Arxiv Blog article:

In fact there is no simple rule for maximising the horizontal flight distance. It turns out this depends on a number of factors, such as rope swing's distance off the ground, the length of the rope and the angle of the rope when Tarzan begins his swing as well as the angle of the rope at the point of release.

However the optimal angle of release is always less than 45 degrees.

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thank you very much for the answer. The paper provided some very useful analysis. Now question(2) is kind of trivial(purely mathematical problem). But I believe question 1 is unanswered, and remained interesting. – Xiaowen Li Sep 24 '12 at 10:48
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