To test the quality of a tennis ball, you drop it into the floor from a height

of 4.00 m. It rebounds to a height of 2m. If the ball is in contact with the floor

for 12.0 ms, (a) What is the magnitude of its average acceleration during the

contact. (b) Is the average acceleration up or down?

The height from which the ball is dropped, ${\mathrm{y}}_1=4\mathrm{m}$.

The height reached by the ball after hitting the floor, ${\mathrm{y}}^2=2\mathrm{m}$.

Time for which the ball is in contact with the floor, $\mathrm{t}=12.0\mathrm{ms}$.

Average acceleration can be determined by using the kinematic equation of kinematic.

The average acceleration of the ball during the contact is,

$a_{avg}=\frac{v_2-v_1}{∆t}$

(i)

$v_f^2=v_i^2+2a∆y$

(ii)

Average acceleration$=\frac{v_2-v_1}{∆t}$ (iii)

$v_1$is the velocity when the ball is about to hit the ground.

$v_2$is the velocity immediately after the ball has rebounded.

Using the third kinematic equation $v_1$can be determined

Substituting the given values,

$$\begin{gathered} V_f^2=v_i^2+2a∆y \\ {V}_i^2=0+2-g-y_i-0 \\ =2\times 9.8\times 4 \\ =78.4 \\ {v}_1=\pm 8.854m/s^2 \end{gathered}$$

Therefore, the velocity of the ball is $-8.75m/s^2$. Value is taken negative due to

downward direction.

In this journey the final velocity is taken as zero. Substituting the values,

$$\begin{gathered} a_{avg}=\frac{v_2-v_1}{∆t} \\ =\frac{6.26-8.854}{12ms} \\ =\frac{15.114}{12\times {10}^{-3}} \\ =1.2559\times {10}^{3}m/s^2 \end{gathered}$$

Therefore, the average acceleration ${\mathrm{a}}_{\mathrm{a}\mathrm{vg}}=1.26\times {10}^3\mathrm{m}/{\mathrm{s}}^2$.

The value of the acceleration is positive. The downward gravitational acceleration is taken as negative. It implies that the average acceleration is positive.