Question

An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10.0 s apart is ${30}^o$, what is the speed of the aircraft ?

Answer 1

In right angled triangle OAB, $\tan {30}^o=\frac{h}{x}$

We get $x=\frac{h}{\tan {30}^o}$

$\therefore$ $x=3400\sqrt{3}=5889$ m

Time taken $t=10.0$ s

Thus speed of the aircraft $v=\frac{x}{t}$

$\therefore$ $v=\frac{5889}{10.0}=588.9$ $m/s$