91Ó°ÊÓ

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,

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

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,

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

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

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