Question

A ball of moist clay falls 17.5 m to the ground. It is in contact with the ground for 19.0 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

Answer on Question #51371-Physics-Mechanics-Kinematics-Dynamics

A ball of moist clay falls $h = 17.5 \, \text{m}$ to the ground. It is in contact with the ground for $t = 19.0 \, \text{ms}$ 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?

Solution

(a) The speed of the ball before the contact is

$$v = \sqrt{2 g h},$$

where $g = 9.81 \frac{\text{m}}{\text{s}^2}$ is the acceleration due to the gravity.

The average acceleration of the ball during the time it is in contact with the ground is

$$\ddot{a} = \frac{v_f - v_i}{t} = \frac{0 - \sqrt{2 g h}}{t} = - \frac{\sqrt{2 \cdot 9.81 \frac{\text{m}}{\text{s}^2} \cdot 17.5 \, \text{m}}}{19.0 \cdot 10^{-3} \, \text{s}} = -975 \frac{\text{m}}{\text{s}^2}.$$

(b) The sign “-” in average acceleration means that it directed up (the velocity of the ball before the contact is down, but the final velocity of the ball is zero).

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