A Tennis Serve. In the fastest measured tennis serve, the ball left the racquet at $\mathbf{73}.\mathbf{14}m/s$. A served tennis ball is typically in contact with the racquet for$\mathbf{30}.\mathbf{0}$and starts from rest. Assume constant acceleration. (a) What was the ball’s acceleration during this serve? (b) How far did the ball travel during the serve?

• The ball left the racquet at $73.14m/s$.
• The tennis ball was in contact with the racquet at about .$30.0m/s=30.0\times {10}^{-3}s\$=0.03s$

The law states that an object will be at rest or in motion unless it is acted by an external force.

The acceleration of the ball can be identifiedby dividing the subtraction of the final and the initial velocity divided by time. Moreover, the product of the average velocity with time provides the displacement of the ball.

a) From Newton’s first law, the velocity of the ball can be expressed as:

$v=v_0+at$

Here,$v_0$is the initial velocity, $v$is the final velocity of the ball, anda and t are the acceleration and the time taken by the ball.

Substituting the above value, we get-

$$\begin{gathered} 73.14m/s=\text{0m/s+}\left(0.03\text{s}\right)\times a \\ \text{a}={\text{2438m/s}}^2 \end{gathered}$$

Thus, the acceleration of the ball during the serve is$\text{a}=\text{2438}m/s^2$ .

b) From Newton’s first law, the displacement of the ball can be expressed as,

$s=v_{0}t+\frac{1}{2}a{t}^2$

Here, s is the displacement of the ball,$v_0$ is the initial velocity and t is the time taken by the ball.

Substituting the values in the above expression, we get-

$\begin{array}{rcl}s& =& \left(0m/s\right)\times 0.03s+\frac{1}{2}\left(2438m/s^2\right)\times {\left(0.03s\right)}^2\\ s& =& \left(1219m/s^2\right)\times {\left(0.03\right)}^2\\ s& =& 1.10m\end{array}$

Thus, the ball has traveled$1.10m$during the serve.