a ball of moist clay falls t the gournd from a height of 12m . 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 contact with the ground?

Compute the speed of the clay when it hits the ground. It is

V = sqrt (2 g H)= 15.3 m/s
Divided that by the 0.0205 s that it takes to stop, and you will have the average deceleration rate as it deforms after hitting the ground.

I get 746 m/s^2 which is 76 g's

I love you drwls!

To find the average acceleration of the clay during the time it is in contact with the ground, we can use the equation:

acceleration = change in velocity / time

Since the clay falls from a height of 12m and comes to rest, its initial velocity is zero and its final velocity is also zero. We can find the change in velocity by subtracting the initial velocity from the final velocity.

change in velocity = final velocity - initial velocity
= 0 - 0
= 0

Given that the clay is in contact with the ground for 20.5 ms, we can convert this to seconds by dividing it by 1000.

time = 20.5 ms / 1000
= 0.0205 s

Now we can substitute the values into the equation:

acceleration = change in velocity / time
= 0 / 0.0205
= 0 m/s^2

Therefore, the average acceleration of the clay during the time it is in contact with the ground is 0 m/s^2.