A supersonic jet traveling at Mach 2.0 at an altitude of 9500 m passes directly over an observer on the ground. Where will the plane be relative to the observer when the latter hears the sonic boom? (See Fig. 12–39.)

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. By the time the plane reaches this position then the person on the ground will finally hear it, that's the time when the shock wave reaches the person and the plane will have traveled some distance, x, that we have to figure out. And we know that the angle of this shock wave cone is gonna, the tangent of that angle is going to be the height of the planes traveling at divided by the distance that has traveled horizontally. So, tan θ is h over x in other words. And we can solve for x by multiplying both sides by x over tan θ. And we get x on the left and equals h over tan θ on the right. Now, we can't do anything with this because we don't know what θ is, but we can figure it out because we know what the speed of sound and the speed of the plane are. We know the sine of this shock wave cone angle is the speed of sound divided by the speed of plane and the speed of the plane is the Mach number times the speed of sound. So, the speeds of sounds cancel on top and bottom there. And so sine θ is 1 over the Mach number. And so that means θ is the inverse sine of the reciprocal of the Mach number which we can then substitute into our formula for x. So, θ is in resign of 1 over M and that's what we've replaced θ with here. So, x is h over tangent of the inverse sine of 1 over M. So, that's 9,500 meters the altitude of the plane divided by tangent of the inverse sine of 1 over 2. And that's about 16 kilometers the plane will have traveled by the time the person hears it.