Question

A ball of moist clay falls 19.3 m to the ground. It is in contact with the ground for 21.0 ms before stopping. (a) What is the average acceleration of the ball during the time it is in contact with the ground? (Treat the ball as a particle.) (b) Is the average acceleration up or down?

Answer 1

-926.67 m/s²

The negative sign indicates the direction is upward

Step-by-step explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s² (positive downward and negative upward)

`v^2-u^2=2as\\\Rightarrow v=√(2as+u^2)\\\Rightarrow v=√(2* 9.81* 19.3+0^2)\\\Rightarrow v=19.46\ m/s`

The ball will hit the ground at 19.46 m/s

`v=u+at\\\Rightarrow a=(v-u)/(t)\\\Rightarrow a=(0-19.46)/(0.021)\\\Rightarrow a=-926.67\ m/s^2`

The acceleration of the clay ball is -926.67 m/s²

The negative sign indicates the direction is upward.