Question

The angle of elevation of a jet plane from a point A on the ground is ${60}^{°}$. After a flight of $20sec$ at the speed of $432km/hr$ the angle of elevation changes to ${30}^{°}$. If the jet plane is flying at a constant height, then its height is:

• A. $1200\sqrt{3}m$
• B. $1800\sqrt{3}m$
• C. $3600\sqrt{3}m$
• D. $2400\sqrt{3}m$

Answer 1

The correct option is A

$1200\sqrt{3}m$

Step 1: Given data

The initial angle of elevation of the Jet =${60}^{°}$

After the flight, the angle of elevation of the jet=${30}^{°}$

Speed of jet=$432km/hr$

Time $\left(t\right)=20sec$

Step 2: Determine the distance travelled by flight

Speed of jet=$432\times \frac{5}{18}=120m/sec$

It is understood that, $\text{distance}\left(d\right)\text{=speed}\left(s\right)\text{×time}\left(t\right)$. So,

$$\begin{gathered} d=120\times 20 \\ d=2400m \end{gathered}$$

Step 3: Determine the angle of elevation

Consider the following figure,

We know that,$Angleofelevation=\frac{height}{horizontaldis\tan ce}$. So,

$$\begin{gathered} \tan 60=\frac{h}{AC} \\ AC=\frac{h}{\sqrt{3}}----\left(1\right) \end{gathered}$$

Similarly in $△AQD$,

$\begin{array}{l}\tan 30=\frac{h}{AD}\\ AD=\sqrt{3}h----\left(2\right)\end{array}$

According to the figure, $AD-AC=\text{distance travelled}$. So,

$$\begin{gathered} \sqrt{3}h-\frac{h}{\sqrt{3}}=2400 \\ 3h-h=2400\sqrt{3} \\ h=1200\sqrt{3} \end{gathered}$$

So, when the jet plane is flying at a constant height, then its height would be $1200\sqrt{3}$.

Hence, option A is the correct answer.