To test the quality of a tennis ball, you drop it onto the floor from a height of $4.00$ m. It rebounds to a height of $2.00$ m. If the ball is in contact with the floor for $12.0$ ms, what is its average acceleration during that contact? Take $g = 98 m / s^{2}$ .

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Solution

Initial height of the ball $h_{1} = 4 x m$
Final height attained $h_{2} = 2 x m$
So, $e^{2} = \frac{h_{2}}{h_{1}} = \frac{4}{2} = 2$
$⟹ x e = \sqrt{2} = 0.7$
Velocity of ball just before collision $v_{1} = \sqrt{2 g h_{1}} = \sqrt{2 \times 9.8 \times 4} = 8.9 x m / s$ (downwards)
Velocity of ball just after collision $V_{2} = e V_{1} = 0.7 \times 8.85 = 6.2 x m / s$ (upwards)
Change in velocity $Δ V = 8.9 - (- 6.2) = 15.1 x m / s$
Average acceleration $a = \frac{Δ V}{t} = \frac{15.1}{12 \times 10^{- 3}} = 1.25 \times 10^{3} x m / s^{2}$

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To test the quality of a tennis ball, you drop it onto the floor from a height of 4.00m. It rebounds to a height of 2.00m. If the ball is in contact with the floor for 12.0ms, what is its average acceleration during that contact? Take g=98m/s2.

To test the quality of a tennis ball, you drop it onto the floor from a height of $4.00$ m. It rebounds to a height of $2.00$ m. If the ball is in contact with the floor for $12.0$ ms, what is its average acceleration during that contact? Take $g = 98 m / s^{2}$ .