1. Interpret the situation

The observer hears the sound coming from the north at an angle $60^\circ$ with the ground.

But at that same instant, he actually sees the aeroplane vertically above him.

This means:

• the plane is currently just above the observer,
• but the sound reaching him now was emitted earlier, when the plane was somewhere to the north.

So during the time taken by sound to travel from that earlier position to the observer, the plane moved from that earlier position to the point vertically above the observer.

2. Set up geometry

Let the earlier position of the plane (when the sound was emitted) be $P$. Let the observer be at $O$. At the instant of hearing, the plane is vertically above $O$.

Since the sound is heard coming at angle $60^\circ$ with the ground, the line $PO$ makes angle $60^\circ$ with the horizontal ground.

Let the height of the plane be $h$.

Then in triangle formed:

$an 60^\circ = \frac{h}{x}$

where $x$ is the horizontal distance of the earlier position from the observer.

Thus,

$\sqrt{3} = \frac{h}{x} \Rightarrow x = \frac{h}{\sqrt{3}}$

The distance traveled by sound is:

$PO = \frac{h}{\sin 60^\circ} = \frac{h}{\sqrt{3}/2} = \frac{2h}{\sqrt{3}}$

3. Use time relation

Time taken by sound to travel from $P$ to $O$:

$t = \frac{PO}{v} = \frac{2h}{\sqrt{3}v}$

In this same time, the plane moves horizontally from the earlier position to the point vertically above the observer, i.e. distance $x = \frac{h}{\sqrt{3}}$.

If speed of plane is $u$, then

$u t = \frac{h}{\sqrt{3}}$

Substitute $t$:

$u \left(\frac{2h}{\sqrt{3}v}\right) = \frac{h}{\sqrt{3}}$

Cancel $\frac{h}{\sqrt{3}}$:

$u \cdot \frac{2}{v} = 1$

$u = \frac{v}{2}$

4. Check options

• A: $\frac{\sqrt{3}}{2}v$ ❌
• B: $\frac{2v}{\sqrt{3}}$ ❌
• C: $v$ ❌
• D: $\frac{v}{2}$ ✅

Thus the speed of the plane is

$\boxed{\frac{v}{2}}$