Ball Falling Velocity Calculation

The question asks us to find the velocity of a ball when it reaches the ground after falling freely from a height of 20 meters. We are given the initial height, the acceleration due to gravity, and the fact that the ball starts falling freely, meaning its initial velocity is zero.

Given Information:

• Initial height (\(h\)): 20 m
• Initial velocity (\(u\)): 0 m/s (since it falls freely)
• Acceleration due to gravity (\(g\)): 10 m/s\(^2\)

We need to find the final velocity (\(v\)) of the ball just before it hits the ground.

Applying Equations of Motion

We can use the equations of motion under constant acceleration. Since the acceleration is due to gravity and is constant (10 m/s\(^2\)), we can use the following kinematic equation:

\(v^2 = u^2 + 2as\)

Here, \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration, and \(s\) is the displacement. In this case:

• \(a\) is the acceleration due to gravity, \(g\). Since the ball is falling downwards, and gravity acts downwards, we take \(a = g = 10 \text{ m/s}^2\).
• \(s\) is the vertical displacement, which is equal to the initial height \(h\) when the ball reaches the ground. So, \(s = h = 20 \text{ m}\).

Substitute the values into the equation:

\(v^2 = (0 \text{ m/s})^2 + 2 \times (10 \text{ m/s}^2) \times (20 \text{ m})\)

\(v^2 = 0 + 2 \times 10 \times 20 \text{ m}^2\text{/s}^2\)

\(v^2 = 400 \text{ m}^2\text{/s}^2\)

To find the final velocity \(v\), we take the square root of both sides:

\(v = \sqrt{400 \text{ m}^2\text{/s}^2}\)

\(v = 20 \text{ m/s}\)

Final Velocity of Ball

The velocity of the ball when it reaches the ground after falling freely from a height of 20 m, assuming the acceleration of gravity is 10 m/s\(^2\), is 20 m/s.