Mathematics: To test the quality of a tennis ball, you drop it onto the floor from a height of 4.23 m. it rebounds to a height of 2.54 m. if the ball is in contact with the floor for 12.2 ms, what is its average acceleration during that contact?

To test the quality of a tennis ball, you drop

mikeydog5291

02.01.2022

$a_{avg}=1324.69*\frac{m}{s^{2} }$

Step-by-step explanation:

The average acceleration is given by:

$a_{avg}=\frac{\Delta v}{\Delta t} =\frac{v_{f}-v_{0}}{\Delta t}$

where $v_{0}$ is the velocity of the ball an instant before hitting the ground and $v_{f}$ an instant after.

we can obtain $v_{0}$ by applying the following formula:

$v_{0}^{2} =v'_{0}^{2} -2*g*\Delta y$

where $v'_{0}=0$ since the ball falls from rest, g is the acceleration of gravity and $\Delta y=(0-4.23)*m=-4.23*m$

then:

$v_{0}=\sqrt{v'_{0}^{2} -2*g*\Delta y}$

$\pm v_{0}=\sqrt{0^{2} -2*9.8*(-4.23)}$

$v_{0}=-9.1054*\frac{m}{s}$ (note that here we select the negative result because of the downwards direction of the velocity)

Similarly, we calculate $v_{f}$ with the formula:

$v_{f}^{2} =v'_{f}^{2} -2*g*\Delta y$

where $v'_{f}=0$ is the velocity at the maximum height, and $\Delta y=(2.54-0)*m=2.54*m$

then:

$v_{f}=\sqrt{2*9.8*2.54}$

$v_{f}=7.0558*\frac{m}{s}$

Finally the average acceleration is:

$a_{avg}=\frac{7.0558+9.1054}{12.2*10^{-3} }*\frac{m}{s^{2}}=1324.69* \frac{m}{s^{2}}$

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