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Q.

The angle of elevation of an airplane from a point on the ground is 60º. After a flight of 30 seconds, the angle of elevation becomes 30º. If the airplane is flying at a constant height of 3000√3 m, find the speed of the airplane.

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a

The speed of the airplane is 200 m/s.

b

The speed of the airplane is 300 m/s.

c

The speed of the airplane is 400 m/s.

d

The speed of the airplane is 500 m/s.

answer is A.

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Detailed Solution

Given that, height of plane from ground is constant and it is 3000√3 m.
We know that, when an object is above the horizontal level, the angle of elevation is the angle between the line of sight and the horizontal.
Let, the initial position of the plane is at C and after 30 seconds it moves to D.
Therefore, the distance traveled by plane in 30 secs = CD = y m.
Let the point where the line of sight from C and D meets the horizontal be B.
Consider the perpendiculars from points C and D meet the horizontal at points A and E respectively.
From the given information and above assumptions, we get the following figure.
Question Image

Question Image

Let, AB = x m
Consider right-angled ΔABC and let us use trigonometric ratios to simplify.

tan 60 ° = Perpendicular Base tan 60 ° = AC AB tan 60 ° = 3 3 = 30003 x 3 × x = 30003 x = 300033 x = 3000 m

Now, consider the right-angled ΔBDE and let us use trigonometric ratios to simplify.

tan 30 ° = Perpendicular Base

tan 30 ° = DE BE

13 = 30003 x + y

tan 30 ° = 13 and BE = x + y

13 = 30003 3000 + y

But,

x = 3000 m

.
So,

x = 3000 m 3000 + y = (30003) × 3 3000 + y = 3000 × 3 3000 + y = 9000 y = 9000 − 3000 y = 6000 m

Thus, the distance traveled by plane in 30 secs is 6000 m.
Now let us find the speed of the plane.

Speed of the plane = Distance travelled by plane Time taken Speed of the plane = 6000 m 30 sec Speed of the plane = 200 m / sec

Hence, the speed of the airplane is 200 m/s.
Therefore, the correct option is (1).

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