An aeroplane flies in a straight horizontal line at a constant height of \(1000\sqrt{3}\) m, moving directly away from an observer on the ground. The observer first sees it at an angle of elevation of \(60^\circ\), and 10 seconds later at \(30^\circ\). What is the speed of the aeroplane?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 720 km/h
At each moment the height of the aeroplane is the side opposite the angle of elevation, so its horizontal distance from the observer \(=\displaystyle\frac{\text{height}}{\tan(\text{angle})}\). The plane flies away from the observer, so the distance it covers in 10 s is the difference of the two horizontal distances.
At \(60^\circ\): distance \(=\displaystyle\frac{1000\sqrt{3}}{\sqrt{3}}=1000\) m
At \(30^\circ\): distance \(=\displaystyle\frac{1000\sqrt{3}}{1/\sqrt{3}}=1000\times 3=3000\) m
Distance flown in 10 s \(=3000-1000=2000\) m
Speed \(=\displaystyle\frac{2000}{10}=200\) m/s \(=200\times\displaystyle\frac{18}{5}=720\) km/h
