(Continue to review middle school knowledge!)

Suppose a supersonic airplane has Mach number $M$ (Mach number is the speed divided by sound speed). It flies horizontally. At some time, it flies past over your head at height $h$. 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 $t=0$; the sound speed is $v$. Then, the motion of the airplane is

$x=Mvt.$

The distance between you and the airplane is

$l=\sqrt{x^2+h^2}.$

Let $\tau$ be the time at which you hear the sound emitted by the airplane at $t$, then

$\tau=t+\frac lv=t+\sqrt{\tau_0^2+M^2t^2},$

where $\tau_0:=\frac hv$.

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

Write the relationship between $\tau$ and $t$ as a quadratic equation w.r.t. $t$

$\left(M^2-1\right)t^2+2\tau t+\tau_0^2-\tau^2=0.$

It must have real solutions, so the determinant is non-negative, i.e.

$\tau^2-\left(M^2-1\right)\left(\tau_0^2-\tau^2\right)\ge0.$

We can solve the inequality to get

$\tau^2\ge\left(1-M^{-2}\right)\tau_0^2.$

Here we can see why do we require the airplane to be supersonic (otherwise the right-hand side of the inequality is negative, and $\tau$ does not have a minimum value).

By this, we have

$\tau_{\mathrm{min}}=\tau_0\sqrt{1-M^{-2}}.$

Therefore, the answer to our question is

$\left.l\right|_{t=\tau_{\mathrm{min}}}=Mh.$

Elegant! 