An aeroplane flying at a constant speed, parallel to the horizontal ground, 3   km above it is observed…

Let from point $C$ the angle of elevation of plane at $B$ is ${60}^{°}$.

and after $5$ seconds it reach at $B^{\text{'}}$

In $\mathit{∆}ABC, AC=\sqrt{3}\cot {60}^{°}=1$

In $\Delta C{A}^{\text{'}}{B}^{\text{'}}, A^{\text{'}}C=\sqrt{3} \cot {30}^{°}=3$

Hence, distance $A{A}^{\text{'}}=A^{\text{'}}C-AC=3-1=2$

Now, speed equals total distance travelled divided by total time taken.

Thus, speed $=\frac{2}{\frac{5}{60\times 60}}=\frac{2\times 60\times 60}{5}=\frac{7200}{5}=1440 \mathrm{km}/\mathrm{hr}$