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A few weeks ago I was on an airplane and to pass the time started thinking about this problem.

Using the following information, I wanted to know how far away a cloud I could see was.

Under some simplifications here were my hypotheses:

  • The plane is moving at 800 km/h (ground speed).
  • The width of the window is 10 inches.
  • My head is being kept still 5 inches from the centre of the window.
  • Fix a point on the cloud, say one of the very edges, enters the window frame and then leaves the window frame 20 seconds later.
  • Assume that the wind speed on the plane is the same as the wind speed on the cloud. That is, the cloud is stationary with respect to an observer on the ground.

How far is the cloud?

My answer is 2.22km.

I'm not sure if my method is the best or if there are other interesting ways to do it. I thought I would post this problem here as I found it interesting!

JonHerman
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  • I have a question: how the last assumption affects the result? – Jaehyeon Seo Aug 05 '14 at 16:47
  • @TheGreatSeo I think he means to say that the cloud is stationary w.r.t ground, i.e, he knows the relative velocity of cloud and plane. – Guy Aug 05 '14 at 16:54
  • Yes, thank you. That is what I mean – JonHerman Aug 05 '14 at 16:59
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    It's pretty straight-forward, and your solution seems along the optimal lines. The relative distance travelled by the cloud is $vt$, and similar triangles give you $vt/x=w/d$ if $x$ is the unknown distance, $w$ width of the window and $d$ distance to the window. Fortunately, you get $x=vt \frac{d}{w}$ so even ridiculous units cancel out. – orion Aug 05 '14 at 18:54

1 Answers1

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  1. If your head is 5 inches from the centre of 10 inches wide window, then your view is $90^{\circ}$ wide and the limits of your view creates with the window (and the trajectory of the cloud relative to the position of your head) the isosceles right triangle. The position of the cloud when it was in the nearest position is then equal to the height of this triangle, which is equal to the half of the hypotenuse.

  2. The plane travels $20[s]*800[km/h] = \frac{40}{9}[km]$. It means, that the cloud traveled $\frac{40}{9}[km]$ relative to the position of your head.

  3. The distance to the cloud in its' nearest position is then equal $$\frac{1}{2}\frac{40}{9}[km] = \frac{20}{9}[km]\approx2.22[km]$$

Jaroslaw Matlak
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