From a point on the ground, the angle of elevation of an aeroplane is 60 degrees. After flying for 30 seconds in a straight horizontal line at a constant height of 4500 metres, the angle of elevation becomes 30 degrees. What is the speed of the aeroplane in metres per second?

Introduction / Context:
This problem involves an aeroplane flying in a straight horizontal path at a constant height above the ground. Two different angles of elevation of the aeroplane are observed from the same point on the ground at two different times. Using trigonometry and the fact that the height is constant, we can calculate the horizontal distances at both instants and thus determine the speed of the aeroplane in metres per second.

Given Data / Assumptions:

• Initial angle of elevation of the aeroplane from a fixed ground point is 60 degrees.
• After 30 seconds, the angle of elevation from the same point becomes 30 degrees.
• The aeroplane flies at a constant height of 4500 metres above the ground.
• The aeroplane moves in a straight horizontal line.
• We neglect curvature of the earth or wind effects, and treat the path as a right angled triangle situation at each instant.

Concept / Approach:
In height and distance questions, for a right angled triangle, tan of the angle of elevation is equal to the vertical height divided by the horizontal distance from the observer. Let the horizontal distances at the two instants be x1 and x2. Using tan 60 degrees and tan 30 degrees, we express x1 and x2 in terms of the fixed height. The aeroplane covers a horizontal distance of (x2 − x1) in 30 seconds, so the speed is (x2 − x1) divided by 30, expressed in metres per second.

Step-by-Step Solution:
Step 1: At the first instant, tan 60° = 4500 / x1, so √3 = 4500 / x1, giving x1 = 4500 / √3 = 1500√3 metres. Step 2: After 30 seconds, tan 30° = 4500 / x2, so 1 / √3 = 4500 / x2, giving x2 = 4500√3 metres. Step 3: Horizontal distance covered in 30 seconds is x2 − x1 = 4500√3 − 1500√3 = 3000√3 metres. Step 4: Speed of the aeroplane = distance / time = (3000√3) / 30 = 100√3 metres per second.

Verification / Alternative check:
An approximate numerical check helps. Take √3 approximately equal to 1.732. Then 100√3 is about 173.2 metres per second. In 30 seconds the aeroplane would travel about 5196 metres horizontally. The initial distance is around 2598 metres and the final around 7794 metres, whose difference is very close to 5196 metres, matching our calculation. The angles of 60 degrees and 30 degrees are also consistent with the computed distances and constant height of 4500 metres.

Why Other Options Are Wrong:
Option 50√3 corresponds to half the correct horizontal speed and would imply a much smaller change in horizontal distance, inconsistent with the change in angle. Options 200√3 and 300√3 imply much larger speeds, which would overshoot the required change in distance for the given height. Option 150√3 is also larger than the derived value and does not fit the trigonometric relations for both angles.

Common Pitfalls:
Many learners incorrectly assume that the aeroplane is approaching the observer when the angle decreases, but here the angle of elevation reduces from 60 degrees to 30 degrees so the plane is moving away. Another pitfall is forgetting to convert time or speed units, but in this problem everything is already in seconds and metres. Errors in handling √3 or in subtracting x1 from x2 also lead to wrong answers.

Final Answer:
The speed of the aeroplane is 100√3 metres per second.