Question

A ball of moist clay falls 18.7 m to the ground. It is in contact with the ground for 21.0 ms before stopping. What is the average acceleration of the clay during the time it is in contact with the ground? (Treat the ball as a particle.)

Answer 1

`a=-911.42\ m/s^2`

Step-by-step explanation:

It is given that,

A ball of moist clay falls 18.7 m to the ground, s = 18.7 m

It is in contact with the ground for 21.0 ms before stopping,
`t=21\ ms=21* 10^(-3)\ s`

Let v is the velocity when it reached the ground. Using third equation of motion as :

`v^2=u^2+2as`

u = 0 and a = g

`v^2=2* 9.8* 18.7`

v = 19.14 m/s

Let a is the acceleration during time t. It can be calculated as :

`v=u+at`

Now, u = 19.14 m/s and v = 0

`v=u+at`

`a=(-u)/(t)`

`a=(-19.14)/(21* 10^(-3))`

`a=-911.42\ m/s^2`

So, the average acceleration of the clay during the time is
`-911.42\ m/s^2`. Hence, this is the required solution.