A 4.0 kg mud ball drops from rest at a height of 12 m. If the impact between the ball and the ground lasts 0.30 s, what is the average net force exerted by the ball on the ground?

force*time= mass*changevelocity

where velocity= sqrt(2gh)

solve for force.

To find the average net force exerted by the ball on the ground, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration.

First, let's find the acceleration of the mud ball as it falls to the ground from a height of 12 m. We can use the equation for gravitational potential energy:

Potential Energy (PE) = mass (m) * acceleration due to gravity (g) * height (h)

The potential energy is converted into kinetic energy when the ball reaches the ground. So, we can equate the potential energy to the kinetic energy:

PE = KE

m * g * h = 1/2 * m * v^2

Where g is the acceleration due to gravity (approximately 9.8 m/s^2) and v is the final velocity of the mud ball just before it hits the ground.

Rearranging the equation to solve for v:

v^2 = 2 * g * h

v^2 = 2 * 9.8 m/s^2 * 12 m

v^2 = 235.2 m^2/s^2

v ≈ 15.32 m/s

Now, we can find the acceleration of the ball using the equation:

Acceleration (a) = (final velocity - initial velocity) / time

Since the mud ball starts from rest, the initial velocity is 0 m/s. Therefore:

a = (15.32 m/s - 0 m/s) / 0.30 s

a ≈ 51.07 m/s^2

Finally, we can find the average net force exerted by the ball using Newton's second law:

Average Net Force (F) = mass (m) * acceleration (a)

F = 4.0 kg * 51.07 m/s^2

F ≈ 204.28 N

Therefore, the average net force exerted by the ball on the ground is approximately 204.28 Newtons.