Question

A ball of moist clay falls 15.0m to the ground. It is in contact with the ground for 20.0m sbefore stopping. (a) What is the magnitude of 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?

Short answer

a) Average acceleration of the ball during contact$857\mathrm{m}/{\mathrm{s}}^2$.

b) Direction of average acceleration is upward.

Step-by-step solution

Given Data

Distance covered by the ball, y = -15.0 m

Time for which ball is in contact with the ground, (t),

$$\begin{gathered} \mathrm{t}=20\mathrm{ms} \\ =20\times {10}^{-3}\mathrm{s} \end{gathered}$$

Understanding the concept

Using Kinematics equation, velocity of the ball at the time of hitting the ground can be determined, and from the calculation average acceleration of the ball can be calculated.

The required kinematic equation can be written as,

$v^2=v_0^2+2ay$ (i)

The average acceleration is calculated using the equation,

$a=\frac{changeinvelocity}{Time}$

(ii)

(a) Determination of average acceleration of the ball

Substitute the given data in equation (i) to calculate the velocity of the ball when it

hits the ground.

$$\begin{gathered} {\mathrm{v}}^2=0+2\times -\frac{9.8}{{\mathrm{s}}^2}\times -15\mathrm{m} \\ \mathrm{v}=\sqrt{294} \\ =17.146\mathrm{m}/\mathrm{s} \end{gathered}$$

Now, use the equation (ii) to calculate the average acceleration from the velocity and

time.

$$\begin{gathered} a=\frac{17.416m/s-0}{20\times {10}^{-3}s} \\ =857.3m/s^2 \\ =857m/s^2 \end{gathered}$$

Therefore, the average acceleration of the ball is$857m/s^2$

(b) Determination of direction of average acceleration

The acceleration due to gravity is considered negative because the downward direction is assumed to be negative for this problem. The average acceleration calculated in part (a) has a positive sign. It implies that the acceleration is positive and acting in the upward direction.