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The angle of elevation of a jet plane from a point $^{'} A^{'}$ on the ground is $60^{o}$ . After a flight of $15 seconds$ , the angle of elevation changes to $30^{o}$ . If the jet plane is flying at a constant height of $1500 \sqrt{3} m$ . Find the speed of jet plane $(\sqrt{3} = 1.732)$ .

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200 m / s

600 m / s

300 m / s

400 m / s

answer is A.

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Detailed Solution

It is given that the angle of elevation is

60^{o} .

After a flight of

15 seconds

, the angle of elevation changes to

30^{o} .

The height of the jet plane from ground is

1500 \sqrt{3} m

We have to find the speed of jet plane
We use,

tanθ = \frac{perpendicular}{base} .

Let us assume that D and E are the initial and final positions of the plane respectively.
Question Image

Consider the

△ ABD

,
Applying the above concept,

\tan 60^{o} = \frac{BD}{AB}

............ (i)

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

[since

\tan 60^{o} = \sqrt{3}

]

\Rightarrow AB = 1500

Consider the

△ ACE

\tan 30^{o} = \frac{CE}{AC}

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

[since

\tan 30^{o} = \frac{1}{\sqrt{3}}

]

\Rightarrow AC = 1500 \times 3

\Rightarrow AC = 4500

m
Then,

\Rightarrow BC = AC - AB

\Rightarrow BC = 4500 - 1500

\Rightarrow

BC=3000m

DE = BC = 3000 m

i.e., the jet plane travels a distance of

3000 m

15

sec.
Hence, speed of the jet plane,

\Rightarrow speed = \frac{3000}{15}

\Rightarrow spee d = 200 m / s

Thus, the speed of the jet plane is 200 m/s.
Therefore, the correct option is 1

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