Question

A tennis ball is dropped from a height of 7 m and rebounds to a height of 3 m. If it is in contact with the floor for 0.029 s, what is the average acceleration during this period?

Answer 1

Answer on Question 54936, Physics, Mechanics | Kinematics | Dynamics

Question:

A tennis ball is dropped from a height of $7m$ and rebounds to a height of $3m$. If it is in contact with the floor for $0.029s$, what is the average acceleration during this period?

Solution:

Let's first find the final velocity of the tennis ball when it hit the ground from the kinematic equation:

$$v^2 = v_0^2 + 2gy,$$

where, $v_0$ is the initial velocity of the tennis ball which is zero, $v$ is the final velocity of the tennis ball when it hit the ground, $g$ is the acceleration of gravity and $y$ is the displacement.

Thus, the final velocity of the tennis ball when it hit the ground will be (since, we take the direction of the $y$-axis downward, the displacement of the tennis ball will be $y = 7m$, and $g = 9.8\frac{m}{s^2}$):

$$v = \sqrt{2gs} = \sqrt{2 \cdot 9.8 \frac{m}{s^2} \cdot (7m)} = \sqrt{137.2 \frac{m^2}{s^2}} = 11.71 \frac{m}{s}.$$

Let's secondly find the initial velocity of the ball when it rebounds to a height of $3m$. We use the same kinematic equation as in the previous case. When the tennis ball reaches the height of $3m$ its final velocity will be zero, so we can rewrite our equation (in this case the displacement of the tennis ball will be $y = -3m$ and $g = -9.8\frac{m}{s^2}$):

$$0 = v_0^2 + 2gy,$$ $$0 = v_0^2 + 2 \cdot \left(-9.8 \frac{m}{s^2}\right) \cdot (-3m),$$ $$v_0 = \sqrt{2 \cdot \left(-9.8 \frac{m}{s^2}\right) \cdot (-3m)} = \sqrt{58.8 \frac{m^2}{s^2}} = -7.67 \frac{m}{s}.$$

The initial velocity of the ball have sign minus because we take the direction of the $y$-axis downward.

Finally, we can find the acceleration from the definition of the impulse:

$$F \Delta t = m \Delta v = m v - m v _ {0},$$ $$m a \Delta t = m v - m v _ {0},$$ $$a = \frac {v - v _ {0}}{\Delta t} = \frac {1 1 . 7 1 \frac {m}{s} - (- 7 . 6 7 \frac {m}{s})}{0 . 0 2 9 s} = \frac {1 9 . 3 8 \frac {m}{s}}{0 . 0 2 9 s} = 6 6 8. 3 \frac {m}{s ^ {2}}.$$

Answer:

$$a = 6 6 8. 3 \frac {m}{s ^ {2}}.$$

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