A ball of moist clay falls to the ground from a height of 19.8 m. It is in contact with the ground for 19.0 ms before coming to rest. What is the average acceleration of the clay during the time it is in contact with the ground? (Treat the ball as a particle.)
PLEASE HELP! :)
find the velocity just before impact.
vf^2=2g*h
solve for vf.
then, acceleration=(0-Vf)/timecontact
To find the average acceleration of the clay while in contact with the ground, we can use the following formula:
average acceleration = change in velocity / change in time
First, let's find the change in velocity of the clay as it falls from a height of 19.8 m. We can do this using the formula for free fall:
change in velocity = sqrt(2 * acceleration * distance)
Since the ball is falling from rest and coming to a stop, the final velocity will be 0. The distance traveled is the height from which it fell, which is 19.8 m. Therefore, the equation becomes:
0 = sqrt(2 * acceleration * 19.8)
Squaring both sides of the equation, we get:
0 = 2 * acceleration * 19.8
Simplifying further, we have:
0 = 39.6 * acceleration
Dividing both sides by 39.6, we find:
acceleration = 0 m/s²
Since the acceleration is 0, it means that the ball does not change its velocity while in contact with the ground. Therefore, the average acceleration of the clay during the time it is in contact with the ground is 0 m/s².