Question

The angle of elevation of a jet plane from a point $A$ on the ground is ${60}^{\circ }$. After a flight of $15$ seconds the angle of elevation changes to ${30}^{\circ }$. The plane is flying at a constant height of $1500\sqrt{3}m$. Calculate the speed of the plane.

• A. 200 m/s
• B. 250 m/s
• C. 215 m/s
• D. 234 m/s

Answer 1

The correct option is A: $200m/s$

Let $BC$ be the distance covered in $15$ seconds. Also, $BE=CD=1500\sqrt{3}m$, be the height of the plane. Also, $\angle EAB={60}^{\circ }$ and $\angle DAC={30}^{\circ }$

In right $\Delta AEB$, we have

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

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

$\Rightarrow AB=1500m$

In right $\Delta ADC$, we have

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

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{1500\sqrt{3}}{AB+BC}$

$\Rightarrow BC=1500\times 3-1500$

$\Rightarrow BC=3000m$

Thus, Speed of the jet plane is $=\frac{3000}{15}=200m/s$