Chapter 3, Problem 64GP

To determine

To calculate: Minimum speed required by the ball to clear the net, distance at which the ball lands and the time for which the ball will be in the air. Also state if the ball lands at a good distance.

Answer to Problem 64GP

Minimum speed required by the ball to clear the net is 2.63 m/s.

Distance at which the ball lands is 18.7 m. Since, this is within 15 m to 22 m, the ball is in good region.

Time for which the ball is in the air is 0.71 s.

Explanation of Solution

Given:

Height of the net = 0.90 m

Distance between net and the server = 15 m

Height at which the ball is launched = 2.50 m

Formula used:

The following kinematic equations:

  $y=y_0+v_{y0}t+\frac{1}{2}{a}_y{t}^2$

Where, $y$ is displacement, $y_0$ is the initial position, $v_{y0}$ is the initial velocity in the vertical direction, t is time taken and $a_y$ ( $9.8{\text{ m/s}}^{\text{2}}$ ) is acceleration.

  $\Delta x=V_{x}t$

Where, $\Delta x$ is the horizontal distance, $V_x$ is velocity in the horizontal direction and t is time taken.

Calculation:

Initial position of the ball is taken as 2.50 m, initial velocity is taken as 0, acceleration is taken as negative (against gravity) and y is taken as height of the net.

Time taken for the ball to reach the net is calculated as:

  $\begin{array}{l}y=y_0+v_y{}_{0}t+\frac{1}{2}{a}_y{t}^2\\ 0.90=2.50+0t+\frac{1}{2}\left(-9.8\right)t^2\\ 0.9=2.50-4.9t^2\\ 4.9t^2=1.6\\ t^2=\frac{1.6}{4.9}\\ t=\sqrt{\frac{1.6}{4.9}}\\ t=0.57\text{ s}\end{array}$

Therefore, time for which the ball to reach the net 0.57 seconds.

Distance between net and server is taken as horizontal distance.

Velocity is calculated as:

  $\begin{array}{l}\Delta x=V_{x}t\\ V_x=\frac{\Delta x}{t}\\ V_x=\frac{15.0}{0.57}\\ V_x=26.3\text{ m/s}\end{array}$

Therefore, minimum required to clear the net is 2.63 m/s.

To find time for which the ball is in the air, the final displacement, y should be 0. Therefore,

  $\begin{array}{l}y=y_0+v_{y0}t+\frac{1}{2}{a}_y{t}^2\\ 0=2.50+0t+\frac{1}{2}\left(-9.8\right)t^2\\ 4.9t^2=2.50\\ t^2=\frac{2.50}{4.9}\\ t=\sqrt{\frac{2.50}{4.9}}\\ t=0.71\text{ s}\end{array}$

Therefore, time for which the ball is in the air is 0.71 seconds.

Distance at which the ball lands is calculated as:

  $\begin{array}{l}\Delta x=V_{x}t\\ \Delta x=\left(26.3\right)\left(0.71\right)\\ \Delta x=18.7\text{ m}\end{array}$

Therefore, distance at which the ball lands is 18.7 m.

Since, this is within 15 m to 22 m, the ball is in good region.

Conclusion:

Minimum speed required by the ball to clear the net is 2.63 m/s.

Distance at which the ball lands is 18.7 m. Since, this is within 15 m to 22 m, the ball is in good region.

Time for which the ball is in the air is 0.71 s.