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

Answer:

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

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, $t=21\ ms=21\times 10^{-3}\ s$

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

$v^2=2\times 9.8\times 18.7$

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

$v=u+at$

$a=\dfrac{-u}{t}$

$a=\dfrac{-19.14}{21\times 10^{-3}}$

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

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