A helicopter flies in a straight horizontal line at a constant height of \(600\sqrt{3}\) m, moving directly away from an observer on the ground. The observer first sees it at an angle of elevation of \(60^\circ\), and 8 seconds later at \(30^\circ\). What is the speed of the helicopter?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 540 km/h
At each moment the height of the helicopter is the side opposite the angle of elevation, so its horizontal distance from the observer \(=\displaystyle\frac{\text{height}}{\tan(\text{angle})}\). The helicopter flies away from the observer, so the distance it covers in 8 s is the difference of the two horizontal distances.
At \(60^\circ\): distance \(=\displaystyle\frac{600\sqrt{3}}{\sqrt{3}}=600\) m
At \(30^\circ\): distance \(=\displaystyle\frac{600\sqrt{3}}{1/\sqrt{3}}=600\times 3=1800\) m
Distance flown in 8 s \(=1800-600=1200\) m
Speed \(=\displaystyle\frac{1200}{8}=150\) m/s \(=150\times\displaystyle\frac{18}{5}=540\) km/h
