Question

The angle of elevation of a jet plane from a point $A$ on the ground is ${60}^{\circ }$. After a flight of $15$ second, the angle of elevation changes to ${30}^{\circ }$. If the jet plane is flying at a constant height of $1500\sqrt{2}$m.Find the speed of the jet plane.

Answer 1

Let $D$ and $E$ be the initial and final positions of the plane respectively

Consider the $△ABD$

$\tan {60}^{\circ }=\frac{BD}{AB}$

$\Rightarrow \sqrt{3}=\frac{1500\sqrt{3}}{AB}$

$\Rightarrow AB=1500$

Consider the $△ACE$

$\tan {30}^{\circ }=\frac{CE}{AC}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{1500\sqrt{3}}{AC}$

$\Rightarrow AC=4500$

$BC=AC-AB=4500-1500=3000$m

$DE=BC=3000$m

i.e., the plane travels a distance of $3000$ m in $15$ seconds.

$\therefore$, speed of the plane $=\frac{3000}{15}=200$m\s.