An aeroplane flying at a constant speed, parallel to the horizontal ground, $$\sqrt{3}$$ km above it, is observed at an elevation of 60$$^\circ$$ from a point on the ground. If, after five seconds, its elevation from the same point is 30$$^\circ$$, then the speed (in km/hr) of the aeroplane is:

Let us denote the fixed horizontal ground point of observation by $$P$$ and the aeroplane by $$A$$. The aeroplane is flying in a straight line parallel to the ground at a constant height of $$\sqrt{3}\text{ km}$$.

At the first instant, the angle of elevation of $$A$$ from $$P$$ is $$60^\circ$$. Using the right-angled triangle formed by the vertical height, the horizontal distance and the line of sight, we have

$$\tan 60^\circ \;=\;\frac{\text{vertical height}}{\text{horizontal distance }(x_1)}.$$

We know the height is $$\sqrt{3}\text{ km}$$, and $$\tan 60^\circ = \sqrt{3}$$. Substituting,

$$\sqrt{3} \;=\; \frac{\sqrt{3}}{x_1} \;\;\Longrightarrow\;\; x_1 \;=\; \frac{\sqrt{3}}{\sqrt{3}} = 1\text{ km}.$$

Five seconds later, the angle of elevation falls to $$30^\circ$$. Again applying the tangent ratio,

$$\tan 30^\circ \;=\;\frac{\text{vertical height}}{\text{horizontal distance }(x_2)},$$

with $$\tan 30^\circ = \dfrac{1}{\sqrt{3}}$$. Hence,

$$\frac{1}{\sqrt{3}} \;=\; \frac{\sqrt{3}}{x_2} \;\;\Longrightarrow\;\; x_2 \;=\; \sqrt{3}\times\sqrt{3} = 3\text{ km}.$$

The aeroplane therefore travels horizontally from $$x_1 = 1\text{ km}$$ to $$x_2 = 3\text{ km}$$ in the given five-second interval. The horizontal distance covered is

$$\Delta x \;=\; x_2 - x_1 = 3 - 1 = 2\text{ km}.$$

Speed is distance divided by time. First we find the speed in km per second:

$$v \;=\;\frac{\Delta x}{\Delta t} = \frac{2\text{ km}}{5\text{ s}} = 0.4\text{ km/s}.$$

To convert kilometres per second to kilometres per hour, we multiply by the number of seconds in an hour, $$3600$$:

$$v_{\text{(km/hr)}} \;=\; 0.4 \times 3600 = 1440\text{ km/hr}.$$

Hence, the correct answer is Option D.