A ball of moist clay falls to the ground from a height of 12 m. It is in contact with the ground for 20.5 ms before coming to rest. What is the average acceleration of the clay during the time it is in contaact with the ground?

To find the average acceleration of the clay, we can use the equation:

Acceleration = (Final Velocity - Initial Velocity) / Time

In this case, the clay starts from rest and comes to rest after 20.5 milliseconds. Since it starts from rest, the initial velocity is 0 m/s.

We need to find the final velocity of the clay when it hits the ground. To do that, we can use the equation for free fall:

Final Velocity = √(2 * g * h)

Where g is the acceleration due to gravity (approximately 9.8 m/s²) and h is the height (12 m).

Final Velocity = √(2 * 9.8 m/s² * 12 m)
Final Velocity = √(235.2 m²/s²)
Final Velocity ≈ 15.33 m/s

Now we can calculate the acceleration:

Acceleration = (Final Velocity - Initial Velocity) / Time
Acceleration = (15.33 m/s - 0 m/s) / (20.5 ms)
Acceleration = 15.33 m/s / 0.0205 s
Acceleration ≈ 749.76 m/s²

Therefore, the average acceleration of the clay during the time it is in contact with the ground is approximately 749.76 m/s².