Introduction:
When an object is dropped vertically on Earth, it falls under the influence of gravity. The speed of the object changes as it falls towards the ground. In this response, we will discuss the change in speed of the object after falling through a distance d from its highest point.

Acceleration due to gravity:
The acceleration due to gravity on Earth is approximately 9.8 m/s^2. This means that every second the object falls, its speed increases by 9.8 m/s.

Velocity at highest point:
When the object is dropped from its highest point, its initial velocity is zero. This means that the object is not moving at the highest point.

Velocity after falling through distance d:
After the object falls through a distance d, its speed increases. Using the formula for velocity under constant acceleration:

v^2 = u^2 + 2as

where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance fallen.

Since the initial velocity is zero, the formula reduces to:

v^2 = 2as

Solving for v, we get:

v = sqrt(2as)

This means that the speed of the object after falling through a distance d is proportional to the square root of d.

Conclusion:
In conclusion, the change in speed of an object dropped vertically on Earth after falling through a distance d from its highest point is proportional to the square root of d. This is because the acceleration due to gravity is constant, and the velocity of the object under constant acceleration is proportional to the square root of the distance fallen.

Change in velocity = root 2gdAs the initial velocity is zero final velocity is root 2gd from third kinematics equation.