Question:

A bird is sitting on an 80 m high tree. The angle of elevation of the bird is \( 45^\circ \) from a point on the ground. The bird now flies horizontally such that its height from the ground remains same. After 2 seconds the angle of elevation of the bird from the same point becomes \( 30^\circ \). Find the speed of the flying bird. (Use \( \sqrt{3} = 1.732 \))

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To calculate the speed of an object moving horizontally while maintaining a constant height, use the change in horizontal distance over time.
Updated On: Oct 10, 2025
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Solution and Explanation

Let the initial position of the bird be \( A \), and the position after 2 seconds be \( B \). The height of the tree is \( 80 \, \text{m} \), and the angle of elevation from point \( P \) to the bird at \( A \) is \( 45^\circ \). Step 1: Calculate the horizontal distance at position \( A \) Using the tangent function for \( \angle 45^\circ \), we can write: \[ \tan(45^\circ) = \frac{80}{x_1}, \] where \( x_1 \) is the horizontal distance from point \( P \) to the tree at the initial position. Since \( \tan(45^\circ) = 1 \), we have: \[ 1 = \frac{80}{x_1} \quad \Rightarrow \quad x_1 = 80 \, \text{m}. \] Step 2: Calculate the horizontal distance at position \( B \) After 2 seconds, the angle of elevation becomes \( 30^\circ \). Using the tangent function for \( \angle 30^\circ \), we have: \[ \tan(30^\circ) = \frac{80}{x_2}, \] where \( x_2 \) is the new horizontal distance. Since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), we get: \[ \frac{1}{\sqrt{3}} = \frac{80}{x_2} \quad \Rightarrow \quad x_2 = 80\sqrt{3} \approx 80 \times 1.732 = 138.56 \, \text{m}. \] Step 3: Calculate the distance traveled by the bird The bird travels from \( x_1 = 80 \, \text{m} \) to \( x_2 = 138.56 \, \text{m} \). The distance traveled is: \[ \text{Distance traveled} = x_2 - x_1 = 138.56 - 80 = 58.56 \, \text{m}. \] Step 4: Calculate the speed of the bird The bird travels this distance in 2 seconds, so the speed \( v \) is: \[ v = \frac{\text{Distance}}{\text{Time}} = \frac{58.56}{2} = 29.28 \, \text{m/s}. \]
Conclusion: The speed of the flying bird is \( 29.28 \, \text{m/s} \).
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