The angle of elevation of an aeroplane from a point on the ground is `60^(@)` .After 15 seconds flight ,the elevation changes to `30^(@)` .If the aeroplane is flying at a height of `1500sqrt(3)` m , find the speed of the plane

Options

A. 300 m / sec
B. 200 m /sec
C. 100 m /sec
D. 150 m/sec

Answer

To solve the problem, we will use trigonometric relationships involving the angles of elevation and the height of the aeroplane.

### Step 1: Understand the Problem
We have an aeroplane flying at a height of \(1500\sqrt{3}\) m. The angle of elevation from a point on the ground to the aeroplane changes from \(60^\circ\) to \(30^\circ\) after 15 seconds.

### Step 2: Set Up the Problem
Let:
- \(h = 1500\sqrt{3}\) m (height of the aeroplane)
- \(d_1\) = horizontal distance from the point on the ground to the aeroplane when the angle of elevation is \(60^\circ\)
- \(d_2\) = horizontal distance from the point on the ground to the aeroplane when the angle of elevation is \(30^\circ\)

### Step 3: Use Trigonometric Ratios
Using the tangent function, we can express the distances \(d_1\) and \(d_2\):
1. For \(60^\circ\):
\[ \tan(60^\circ) = \frac{h}{d_1} \Rightarrow d_1 = \frac{h}{\tan(60^\circ)} = \frac{1500\sqrt{3}}{\sqrt{3}} = 1500 \text{ m} \]

2. For \(30^\circ\):
\[ \tan(30^\circ) = \frac{h}{d_2} \Rightarrow d_2 = \frac{h}{\tan(30^\circ)} = \frac{1500\sqrt{3}}{\frac{1}{\sqrt{3}}} = 1500 \times 3 = 4500 \text{ m} \]

### Step 4: Calculate the Distance Covered
The distance covered by the aeroplane in 15 seconds is:
\[ \text{Distance covered} = d_2 - d_1 = 4500 \text{ m} - 1500 \text{ m} = 3000 \text{ m} \]

### Step 5: Calculate the Speed of the Aeroplane
Speed is defined as distance divided by time. Therefore:
\[ \text{Speed} = \frac{\text{Distance covered}}{\text{Time}} = \frac{3000 \text{ m}}{15 \text{ s}} = 200 \text{ m/s} \]

### Final Answer
The speed of the aeroplane is \(200 \text{ m/s}\).

---