**Solution:**

Let's break down the given information:

- The angle of elevation of the bird at some point on the ground is 60 degrees.
- After 50 seconds of flight, the angle of elevation changes to 30 degrees.
- The bird is flying at a distance of 500√3 meters.

To find the speed of the bird, we need to determine the distance it covers in 50 seconds. We can use trigonometry to solve this problem.

**Finding the Distance:**

Let's consider the initial angle of elevation of 60 degrees. We can create a right triangle with the bird's flight path as the hypotenuse and the vertical height as the opposite side. The distance the bird covers in 50 seconds is the adjacent side of this triangle.

Using the trigonometric ratio for the angle of elevation of 60 degrees (tan 60°), we can write:

tan 60° = opposite/adjacent

Since the opposite side is the vertical height and the adjacent side is the distance covered in 50 seconds, we have:

tan 60° = height/distance

Using the value of the height, which is 500√3 meters, we can solve for the distance covered in 50 seconds.

tan 60° = 500√3/distance

Simplifying the equation:

√3 = 500√3/distance

Cross-multiplying:

distance = 500 meters

Therefore, the bird covers a distance of 500 meters in 50 seconds.

**Finding the Speed:**

To find the speed of the bird, we can divide the distance covered by the time taken:

Speed = Distance/Time

In this case, the distance covered is 500 meters and the time taken is 50 seconds.

Speed = 500 meters/50 seconds = 10 meters/second

Therefore, the speed of the bird is 10 meters per second.

**Conclusion:**

The bird is flying at a speed of 10 meters per second.

Let a triangle ABC such that
angle of elevation=ACB=60
by extending the line BC upto D
angle of elevation =ADB=30
in right triangle ABC
tan60=500√3/x=√3
x=500m
similarly in right triangle ABD
tan30=500√3/1/√3

=1500m
distance covered=1500m-500m=1000m
speed=d/t
=1000/50=200m/s