Question Video: Calculation of Aircraft Altitude from a Supersonic Shock Wave

Video Transcript

A plane is flying at Mach 1.2. And an observer on the ground hears the sonic boom 15 seconds after the plane is directly overhead. What is the altitude of the plane? Use a value of 340 meters per second for the speed of sound.

In this scenario, we have a plane flying along at a steady altitude with a speed of 1.2 times the speed of sound, which we’re calling 𝑣 sub 𝑠. As the plane flies along, its engines produce sound. And the sound waves from these engines expand outward as time elapses. Since the plane is moving at a speed greater than the speed of sound, that means that it creates a shock wave, a line at which the sound waves produced by the plane stack on top of one another. This shock wave slowly moves out. And we’re told that an observer on the ground hears the shock wave 15 seconds after the plane is directly overhead.

Based on this and knowing that the speed of sound is 340 meters per second, we want to solve for the altitude of the plane, capital 𝐴. To solve for the attitude 𝐴, we’re first gonna solve for the angle created by the shock wave and a horizontal line through the center of the plane. We’ll call that angle 𝜃. And to solve for it, let’s consider the triangle that 𝜃 is a part of.

If we start at the center of the first sound wave we’ve drawn in and move from that center to the shock wave so that the two lines meet perpendicularly, then we can say that the length of this line we’ve drawn in is equal to the speed of sound, 𝑣 sub 𝑠, times the time elapsed, what we’ll call 𝑡. Likewise, we can draw a line starting at that same center moving in a horizontal direction to the current position of the plane and say that the length of that line segment is equal to 𝑣, the speed of the jet engine, times 𝑡, the time elapsed.

As we look at this right triangle then, we see that 𝜃 is related to 𝑣 sub 𝑠 times 𝑡 and 𝑣 times 𝑡. In particular, we can write that the sin of the angle 𝜃 is equal to 𝑣 sub 𝑠 times 𝑡 over 𝑣 times 𝑡. We see that the time values cancel out from this fraction. And since 𝑣 is equal to 1.2 𝑣 sub 𝑠, that means our fraction reduces to one divided by 1.2. If we take the inverse sine of both sides of this equation, we find that 𝜃 is approximately equal to 56.44 degrees.

Now that we know 𝜃, let’s consider how this can help us solve for the altitude 𝐴. With our diagram as shown, the observer on the ground won’t hear the shock wave until the shock wave has advanced a bit further until it’s reaching the ground. When that takes place, we’ll have a right triangle with an angle 𝜃. And the side length opposite 𝜃 is the altitude 𝐴. In addition, the entire adjacent side of this triangle is equal to the speed of our jet engine, 𝑣, times the time elapsed, 𝑡. This means we can once again use the trigonometry of this right triangle to solve for what we want to find.

This time, we can say that it’s the tangent of 𝜃 which is equal to, in this case, 𝐴 over 𝑣 times 𝑡. Rearranging, we can write that the altitude of the jet is equal to its speed times the time that’s elapsed since it was overhead the observer times the tangent of the angle 𝜃, the angle its shock wave makes with the horizontal. The speed, 𝑣, is equal to the speed of sound, 340 meters per second, times 1.2. The time, 𝑡, is equal to 15 seconds. And 𝜃 as we’ve seen is 56.44 degrees. Plugging all these values in to our calculator, we find that, to two significant figures, 𝐴 is 9.2 kilometers. That’s the elevation of this plane above ground level.