This article is translated from a Chinese article on my Zhihu account. The original article was posted at 2020-02-17 18:49 +0800.

(Continue to review middle school knowledge!)

Suppose a supersonic airplane has Mach number MM (Mach number is the speed divided by sound speed). It flies horizontally. At some time, it flies past over your head at height hh. What is the distance between you and it when you have just heard it?

Suppose the time when the airplane flies past over your head is t=0t=0; the sound speed is vv. Then, the motion of the airplane is x=Mvt.x=Mvt. The distance between you and the airplane is l=x2+h2.l=\sqrt{x^2+h^2}.

Let τ\tau be the time at which you hear the sound emitted by the airplane at tt, then τ=t+lv=t+τ02+M2t2,\tau=t+\frac lv=t+\sqrt{\tau_0^2+M^2t^2}, where τ0≔hv\tau_0\coloneqq\frac hv.

Then, we can find the minimum value of τ\tau by letting dτdt=0\frac{\mathrm d\tau}{\mathrm dt}=0. However, here we use a middle school technique to reduce calculation.

Write the relationship between τ\tau and tt as a quadratic equation w.r.t. tt (M2−1)t2+2τt+τ02−τ2=0.\left(M^2-1\right)t^2+2\tau t+\tau_0^2-\tau^2=0. It must have real solutions, so the determinant is non-negative, i.e. τ2−(M2−1)(τ02−τ2)≥0.\tau^2-\left(M^2-1\right)\left(\tau_0^2-\tau^2\right)\ge0.

We can solve the inequality to get τ2≥(1−M−2)τ02.\tau^2\ge\left(1-M^{-2}\right)\tau_0^2. Here we can see why do we require the airplane to be supersonic (otherwise the right-hand side of the inequality is negative, and τ\tau does not have a minimum value).

By this, we have τmin=τ01−M−2.\tau_{\mathrm{min}}=\tau_0\sqrt{1-M^{-2}}. Therefore, the answer to our question is l∣t=τmin=Mh.\left.l\right|_{t=\tau_{\mathrm{min}}}=Mh.

Elegant!