During a tennis match, a player serves the ball at${}^{\mathbf{23}\mathbf{.}\mathbf{6}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}}$, with the center of the ball leaving the racquet horizontally${}^{\mathbf{2}\mathbf{.}\mathbf{37}\mathbf{}\mathbf{m}}$above the court surface. The net is${}^{\mathbf{12}\mathbf{}\mathbf{m}}$away and${}^{\mathbf{0}\mathbf{.}\mathbf{90}\mathbf{}\mathbf{m}}$high. When the ball reaches the net, (a) does the ball clear it and (b) what is the distance between the center of the ball and the top of the net? Suppose that, instead, the ball is served as before but now it leaves the racquet at${}^{\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{°}}$ below the horizontal. When the ball reaches the net, (c) does the ball clear it and (d) what now is the distance between the center of the ball and the top of the net?

Angle${}^{\theta =5°}$

The height of the net is${}^{0.90m}$

This problem delas with the projectile motion. Also it is based on the standard kinematic equation that describes the motion of an object with constant acceleration. Using this equation, the time required for the ball to reach the net will be calculated. With the calculated time, the height of the ball above the ground at that instant can be computed. Once the height of the ball above the ground is found, the distance between the top of the net and ball at that instant can be found.

Here, the angle between the ball and racquet is${}^{5°}$below the horizontal that is the point at which the ball was hit by racquet.

Formulae:

The speed in general is given by,

$\mathrm{v}=\frac{\mathrm{d}}{\mathrm{t}}$ (i)

The vertical distance is given by,

$\mathrm{y}-{\mathrm{y}}_0={\mathrm{v}}_0\mathrm{t}+\frac{1}{2}{\mathrm{at}}^2$ (ii)

Where${}^{{\mathrm{v}}_0}$is the initial velocity

As the ball leaves the racquet at an angle of${}^{0°}$with the horizontal, the velocity is given by

$$\begin{gathered} {\mathrm{v}}_{0\mathrm{x}}={\mathrm{v}}_0 \\ =23.6\mathrm{m}/\mathrm{s} \\ {\mathrm{v}}_{0\mathrm{y}}=0\mathrm{m}/\mathrm{s} \end{gathered}$$

Using equation (i), the time required for ball to reach the net is,

$$\begin{gathered} t=\frac{d}{v_{0x}} \\ =\frac{12}{23.6} \\ t=0.508s \end{gathered}$$

For the vertical distance, Substituting the value of t in equation (ii),

$$\begin{gathered} \mathrm{y}-{\mathrm{y}}_0={\mathrm{v}}_{0\mathrm{y}}\mathrm{t}+\frac{1}{2}{\mathrm{at}}^2 \\ \mathrm{y}-2.37=0+\frac{1}{2}\times -9.8\times 0.{508}^2 \\ \mathrm{y}=2.37-\left(1.27\right) \\ \mathrm{y}=1.10\mathrm{m} \end{gathered}$$

As the height of the ball when it reaches the net is${}^{1.10\mathrm{m}}$which is higher than that of the net’s height, so the ball will clear the net.

$$\begin{gathered} \mathrm{Height}\mathrm{of}\mathrm{the}\mathrm{ball}\mathrm{above}\mathrm{the}\mathrm{net}=\mathrm{height}\mathrm{of}\mathrm{the}\mathrm{ball}\mathrm{above}\mathrm{the}\mathrm{ground}-\mathrm{Height}\mathrm{of}\mathrm{net} \\ =1.1-0.9 \end{gathered}$$

Thus, the height of the ball above the net is${}^{0.2\mathrm{m}}$

As the ball leaves the racquet at an angle of 5.0 below the horizontal,

$$\begin{gathered} {\mathrm{v}}_{0\mathrm{x}}=23.6\times \cos \left(5\right) \\ =23.5\mathrm{m}/\mathrm{s} \\ {\mathrm{v}}_{0\mathrm{y}}=23.6\times \sin \left(5\right) \\ =-2.05\mathrm{m}/\mathrm{s} \end{gathered}$$

Time required for ball to reach the net,

$$\begin{gathered} t=\frac{d}{v_{0x}} \\ =\frac{12}{23.5} \\ t=0.51s \end{gathered}$$

Substituting t in equation (ii).

$$\begin{gathered} \mathrm{y}-{\mathrm{y}}_0={\mathrm{v}}_{0\mathrm{y}}\mathrm{t}+\frac{1}{2}{\mathrm{at}}^2 \\ \mathrm{y}-2.37=-2.06\times 0.51-\frac{1}{2}\times 9.8\times 0.{51}^2 \\ \mathrm{y}=0.04\mathrm{m} \end{gathered}$$

As the ball is only 0.05m above the ground, so it won’t clear the net.

$$\begin{gathered} \mathrm{Height}\mathrm{of}\mathrm{the}\mathrm{ball}\mathrm{above}\mathrm{the}\mathrm{net}=\mathrm{Height}\mathrm{of}\mathrm{the}\mathrm{net}-\mathrm{Height}\mathrm{of}\mathrm{the}\mathrm{ball}\mathrm{above}\mathrm{the}\mathrm{ground} \\ =0.9-0.04 \end{gathered}$$

Thus, the height of the ball above the net is${}^{0.86\mathrm{m}}$