Problem:
Serving at a speed of 170 km/h, a tennis player hits the ball at a height of 2.5 m and an angle θ below the horizontal. The baseline is 11.9 m from the net, which is 0.91 m high. What is the angle θ such that the ball just crosses the net? Will the ball land in the service box, whose service line is 6.40 m from the net?
Solution:
The initial speed is 170 km/h. We convert this to m/s.
170\ \frac{\text{km}}{\text{hr}}\times \frac{1\ \text{hr}}{3600\ \text{s}}\times \frac{1000\ \text{m}}{1\ \text{km}} =47.2222\ \text{m}/\text{s}
Along the horizontal direction, the equation of motion from the baseline to the net is
\begin{align*}
x & = v_{0x}t \\
11.9\ \text{m} & = \left( 47.2222\ \text{m/s} \right) \cos \theta \ t
\end{align*}Solve for t in terms of theta.
t = \frac{11.9\ \text{m}}{\left( 47.2222\ \text{m/s} \right) \cos \theta} \ \qquad \ \color{Blue} \left( \text{Equation 1} \right)Along the vertical direction, the equation of motion considering that downward motion is negative
\begin{align*}
\Delta y & = v_{oy}t+\frac{1}{2}at^{2} \\
0.91\ \text{m}-2.50\ \text{m} & = -\left( 47.2222\ \text{m/s} \right) \left( \sin \theta \right) \ t + \frac{1}{2}\left( -9.80\ \text{m/s}^2 \right)t^{2} \\
-1.59 & = -47.2222 \sin\theta \ t-4.90 t^2 \\
1.59 & =47.2222 \sin \theta \ t +4.9t^2
\end{align*}Substitute t from equation 1 to this equation
\begin{align*}
1.59 & =47.2222 \sin \theta +4.9t^2 \\
1.59 & = 47.2222 \sin \theta \left( \frac{11.9\ }{\left( 47.2222\ \right) \cos \theta} \right) +4.9 \left( \frac{11.9\ }{\left( 47.2222\ \right) \cos \theta} \right)^2 \\
1.59 & = 11.9 \frac{\sin \theta}{\cos \theta} + \frac{0.3112 }{\cos^2 \theta} \\
1.59 & = 11.9 \tan \theta +0.3112 \sec^2 \theta \\
1.59 & = 11.9 \tan \theta +0.3112(1+\tan^2 \theta) \\
1.59 & = 11.9 \tan \theta +0.3112 +0.3112 \tan^2 \theta \\
\end{align*}Rewrite this is quadratic form of tangent theta
1.59 = 11.9 \tan \theta +0.3112 +0.3112 \tan^2 \theta \\ 0.3112 \tan ^2 \theta +11.9 \tan \theta -1.2788 =0
Solve for \tan \theta
\tan \theta = 0.1072 \\ \tan \theta = -38.3462
Solve for \theta
\theta = 6.1187 \degree \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)The time of flight to the ground can be calculated
\begin{align*}
2.50\ \text{m} & =\left( 47.2222 \ \text{m/s} \right)\left( \sin 6.1187 ^\circ \right)t+\frac{1}{2}\left( 9.80\ \text{m/s}^2 \right)t^2 \\
2.50 & = 5.0333t+4.90t^2 \\
\end{align*}
4.90t^2+5.0333t-2.50 = 0 \\
t = 0.3662 \ \text{s}Therefore, the range of the ball is
\begin{align*}
R & = v_x t \\
R & = \left( 47.2222 \ \text{m/s} \right)\left( 0.3662\ \text{s} \right) \\
R & = 17.2928 \ \text{m}
\end{align*}The ball will land 17.2928 meters – 11.9 meters = 5.3928 meters from the net. This is less than 6.40 meters.
Therefore, the ball will land in the service box.