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The angle of elevation of an aeroplane from a point on the ground is $60^{0}$ . After 15 seconds flight, the elevation changes to $30^{0}$ . If the aeroplane is flying at a height of $1500 \sqrt{3} m$ , then find the speed of the plane.

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300 m/sec

200 m/sec

100 m/sec

150 m/sec

answer is B.

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Detailed Solution

It is given that the angle of elevation of an aeroplane from a point on the ground is

60^{0}

. After 15 seconds flight, the elevation changes to

30^{0}

ΔAED

\tan 60^{o} = \frac{1500 \sqrt{3}}{x}

⇒

\sqrt{3} x = 1500 \sqrt{3}

⇒

x = 1500 m

ΔBEC,

\tan 30^{o} = \frac{1500 \sqrt{3}}{y}

⇒

\frac{1}{\sqrt{3}} = \frac{1500 \sqrt{3}}{y}

⇒

y = 1500 \times 3

⇒

y = 4500

m
Now,

CD = y - x

⇒

CD = 4500 - 1500

m
⇒

CD = 3000

m
Therefore, distance =

3000 m

,
Time taken =

15 s

,
⇒ Speed

= \frac{3000}{15}

.
⇒ Speed

= 200 m / s

.
Hence, the correct option is 2.

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