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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: enter image description here

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: enter image description here

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 : enter image description here


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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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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    $\begingroup$ 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. $\endgroup$ – Xiaowen Li Sep 24 '12 at 10:48
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The above isn't drawn correctly. And it is far too complicated in my opinion. I think I was able to solve this using high school physics:

This bugged me a long time. I am surprised no one ever posted an answer on the internet. The length of his rope was 60.8 meters, assuming a drop above the target building of about 12 meters. The difference in height between the buildings is 64 meters. Using that fact and the distance between the buildings of 47.55, the angle of the pendulum from the vertical is 36.6 degrees. Double checking with the period of pendulum, which is the entire arc, we get 15.7 seconds of actual arc time. In the movie, he free falls for a bit, but the clock doesn't start until he reaches the end of the rope. The time in the movie is about 12.5 seconds, but these calculations are close. I think the only reason he was able to have a rope longer than the distance between the buildings and still make it is because the target roof was angled down. The calculation used the top of the roof where the slope started. This was enough, but you could probably also calculate the force on the rope when his free fall stopped. You could calculate how fast he was going as his free fall time was 14 seconds before he got to the end of the rope.

The bottom line: The stunt was feasible, not impossible. I of course eliminated a bunch of variables: friction, weight of equipment. wind resistance, materials strength of the fulcrum given the force of the stopping rope, etc.

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