A person standing on an open ground hears the sound of a jet aeroplane, coming from north at an angle $$60°$$ with ground level, but he finds the aeroplane right vertically above his position. If $$v$$ is the speed of sound, speed of the plane is:

Let us imagine the instant when the observer actually hears the sound. At that very instant the jet aeroplane is right above him, i.e., vertically over his head. However, the sound seems to come from the north at an elevation of $$60^{\circ}$$ above the horizontal. This means the sound that is being heard now was emitted a little earlier, when the aeroplane was still to the north of the observer.

Denote the following quantities:

$$h=$$ constant height (altitude) of the aircraft above the ground,

$$x=$$ horizontal distance (due south-north direction) that the plane had to cover in that time to reach the observer’s vertical line,

$$u=$$ actual speed of the plane (what we have to find),

$$v=$$ speed of sound in air (given).

At the earlier instant (let us call it time $$t=-\Delta t$$) the aircraft was at point $$A$$, a distance $$x$$ to the north of the observer. Its coordinates relative to the observer’s position $$O$$ are therefore $$(-x,\,0,\,h)$$ if we take the north-south axis as the $$x$$-axis, east-west as $$y$$ and height as $$z$$. The sound ray that has just reached the observer travelled along the straight line $$AO$$ during the interval $$\Delta t$$.

Because the sound ray reaches the observer at an elevation angle of $$60^{\circ}$$, the line $$AO$$ makes an angle $$60^{\circ}$$ with the ground. Hence, in the right-angled triangle formed by the altitude and the ground distance, we have

$$\tan 60^{\circ} \;=\; \frac{\text{opposite side}}{\text{adjacent side}} \;=\; \frac{h}{x}.$$

Using $$\tan 60^{\circ}=\sqrt{3}$$, we get

$$\frac{h}{x}=\sqrt{3},\qquad\text{so}\qquad h=\sqrt{3}\,x.$$

The length of the slant path $$AO$$ travelled by the sound is

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

The sound, moving with speed $$v$$, took a time $$\Delta t$$ given by the elementary relation

$$\text{Time}=\frac{\text{Distance}}{\text{Speed}}.$$

Therefore,

$$\Delta t=\frac{\sqrt{h^{2}+x^{2}}}{v}.$$

During this same interval $$\Delta t$$ the aircraft moved horizontally from $$A$$ to $$O$$, covering the distance $$x$$ with speed $$u$$. Hence

$$u=\frac{\text{distance travelled by the plane}}{\text{corresponding time}} \;=\;\frac{x}{\Delta t}.$$

Substituting the value of $$\Delta t$$ found above, we obtain

$$u=\frac{x}{\dfrac{\sqrt{h^{2}+x^{2}}}{v}} =\frac{x\,v}{\sqrt{h^{2}+x^{2}}}.$$

But we have already related $$h$$ and $$x$$ as $$h=\sqrt{3}\,x$$. Squaring and adding:

$$h^{2}=3x^{2},\qquad h^{2}+x^{2}=3x^{2}+x^{2}=4x^{2}.$$

Thus,

$$\sqrt{h^{2}+x^{2}}=\sqrt{4x^{2}}=2x.$$

Putting this back into the expression for $$u$$:

$$u=\frac{x\,v}{2x}=\frac{v}{2}.$$

So the speed of the jet aeroplane is exactly one half of the speed of sound.

Hence, the correct answer is Option C.