An aeroplane flying horizontally $ 1 \mathrm{~km} $ above the ground is observed at an elevation of $ 60^{\circ} $. After 10 seconds, its elevation is observed to be $ 30^{\circ} $. Find the speed of the aeroplane in $ \mathrm{km} / \mathrm{hr} $.

Given:

An aeroplane flying horizontally \( 1 \mathrm{~km} \) above the ground is observed at an elevation of \( 60^{\circ} \). After 10 seconds, its elevation is observed to be \( 30^{\circ} \). 

To do:

We have to find the speed of the aeroplane in \( \mathrm{km} / \mathrm{hr} \).

Solution:

Let $C$ be the plane flying in the sky at a height of $1\ km$. After a flight of 10 seconds plane is at point $E$.

From the figure,

$BC=DE=1\ km$

Let the distance between $A$ and $B$ be $y\ m$ and the distance between the points $C$ and $E$ be $x\ m$.

This implies,

$BD=CE=x\ m$

In right $\Delta \mathrm{CAB}$,

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

$\sqrt3=\frac{1}{y}$

$y=\frac{1}{\sqrt3}\ km$..........(i)

Similarly,

In right $\Delta \mathrm{EAD}$,

$\tan 30^{\circ}=\frac{ED}{AD}$

$\frac{1}{\sqrt3}=\frac{1}{x+y}$

$x+y=\sqrt3\ km$

$\Rightarrow x=\sqrt3-\frac{1}{\sqrt3} \mathrm{~m}$                 [From(i)]

$\Rightarrow x=\frac{\sqrt3(\sqrt3)-1}{\sqrt3}$

$=\frac{3-1}{\sqrt3}$

$=\frac{2}{1.732}$

$=1.15 \mathrm{~km}$

Therefore,

The distance \( 1.15 \mathrm{~km} \) is covered in 10 seconds.

We know that,

$Speed=\frac{Distance}{Time}$

Speed of the plane $=\frac{1.15}{\frac{10}{60\times60}} \mathrm{~km} / \mathrm{hr}$

$=\frac{1.154(3600)}{10} \mathrm{~km} / \mathrm{hr}$

$=\frac{4156.8}{10} \mathrm{~km} / \mathrm{hr}$

$=415.68 \mathrm{~km} / \mathrm{hr}$

Therefore, the speed of the aeroplane is $415.68 \mathrm{~km} / \mathrm{hr}$. 

Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

37 Views

Kickstart Your Career

Get certified by completing the course

Get Started