# A ball of moist clay falls 15.0m to the ground. It is in contact with the ground for 20ms before stopping?

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(a) What is the magnitude of the average acceleration of the ball during the time it is in contact with the ground?

(b) Is the average acceleration up or down?

(a) What is the magnitude of the average acceleration of the ball during the time it is in contact with the ground?

(b) Is the average acceleration up or down?

##### 2 Answers

(a)

(b) The acceleration was upwards.

#### Explanation:

To find the velocity when it reached the ground, use

(a) To find the average acceleration over the time of 0.020 s, use

Since

(b) Since

I hope this helps,

Steve

The velocity of the clay ball as it hits the ground is found from the kinematic expression

#v^2 - u^2 = 2gh#

Substituting given values we get

#v^2 - 0^2 = 2xx9.81xx15.0#

#=>v=17.16\ ms^-1#

After it hits ground it is brought to stop. Acceleration can be found from the kinematic expression

#v=u+at#

Substituting various values we get

#0=17.16+a_"ave"xx(20xx10^-3)#

#=>a_"ave"=-17.16/(20xx10^-3)#

#=>a_"ave"=-857.8\ ms^-2#

(a)

(b)