Question

An airplane flying horizontally at a height of $2500\sqrt{3}\text{m}$ above that ground, is observed to be at an angle of elevation ${60}^{\circ }$ from the ground. After a flight of $25\text{sec}$ the angle of elevation is ${30}^{\circ }$. Find the speed of the plane in $\text{m/sec}$.

• A. $100\text{m/sec}$
• B. $200\text{m/sec}$
• C. $300\text{m/sec}$
• D. $400\text{m/sec}$

Answer 1

The correct option is B $200\text{m/sec}$

Let $P$ and $Q$ be the two positions of the airplane.

$QB=2500\sqrt{3}\text{m}$

Let the speed of the aeroplane be $x\text{m/sec}$.

In $△ABQ$,

$\begin{array}{cc}\tan {60}^{\circ }& =\frac{QB}{AB}\\ \sqrt{3}& =\frac{2500\sqrt{3}}{x}\\ x& =2500\text{m}\end{array}$

In $△APC$,

$\begin{array}{cc}\tan {30}^{\circ }& =\frac{2500\sqrt{3}}{x+y}\\ \frac{1}{\sqrt{3}}& =\frac{2500\sqrt{3}}{x+y}\\ x+y& =3\times 2500\\ 2500+y& =7500\\ y& =5000\text{m}\end{array}$

To travel $5000\text{m}$ the aeroplane takes $25 \text{ sec}$.

So, the speed of the plane will be,

$\frac{5000}{25}=200\text{m/sec}$