Question

A ball of moist clay falls 17.0 m to the ground. It is in contact with the ground for 19.0 ms before stopping,

Answer 1

Explanation:

It is given that,

Initial height of the ball, $x_o=17\ m$

As it stops, its position, x = 0

It is in contact with the ground for 19.0 ms before stopping, $t=19\times 10^{-3}\ s$

Initial speed of the ball, u = 0

(a) Let a is the acceleration of the ball during the time it is in contact with the ground. Its can be calculated suing equation of kinematics as :

$v^2-u^2=2a(x-x_o)$

a =-g

$v^2=-2g(x-x_o)$

$v^2=-2g(-17)$

$v^2=-2\times 9.8\times (-17)$    

v = 18.25 m/s

(b) The average acceleration can be calculated as :

$a_{avg}=\dfrac{v-u}{t}$

$a_{avg}=\dfrac{18.25-0}{19\times 10^{-3}}$

$a_{avg}=960.52\ m/s^2$

As the value of acceleration is positive, so the ball is accelerating in upward direction.