Plane Flying Overhead

You are standing outside in the sunshine. An airplane at 3000 feet is going to fly directly over your head in a few minutes. It is traveling 200 mph. You don’t notice it until you hear it, and the instant you hear it you spot it at x degrees above the horizon. However, the sound you hear came from the plane somewhat earlier, when it was 20° above the horizon. Take the speed of sound at 1100 ft/sec, find x, the angle at which you looked up and saw the plane.

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So here’s the scene: When you first heard the plane, the sound arrived from point A. But at that point in time, you looked up and saw the plane at point B. The plane flew from A to B (distance a) while the sound was traveling from A to Y (distance b). Once we find a and b, we can concentrate on △ABY, find w*, and thus find x.

So:  3000  = sin 20 → b 3000  ≈ 8771 feet.
b sin 20
At 1100 ft/sec, the sound took 7.97 sec to get to Y. In 7.97 sec, the plane flew from A to B.
200 mi/hr =  200  mi/sec, so AB =  200  · 7.97 = 0.443 mi, or 2339 feet. That’s a.
60 · 60 3600
We need ∠w a  =  c . We can get c using the law of cosines.
sin w sin 20
c2 = (2339)2 + (8771)2 – 2(2339)(8771) cos 20= 43845080 → c ≈ 6622 ft.
Then sin w a sin 20  =  2339 sin 20  = 0.1208 → w = 6.94° → x = 26.94°.
c 6622

*Note: This problem is not made up using the right information. You could measure angle x, because you could see the plane. You’d never know that 20°. Could the problem be recast, starting with ∠x as known and solving for ∠AYC?

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