An aeroplane flying at a constant speed, parallel to the horizontal...

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

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

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

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

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

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

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