A tennis ball launcher is placed on a grass lawn and a target is placed flat on the ground beside it. This target is attached to a car by a rope (see figure) and at the instant the launcher fires a ball at an angle $\theta $ and a speed $v_0$, the target travels to the right at constant speed $W$. When the tennis ball lands, it hits the center of the target.

1. At the instant this ball lands, a second ball is launched at the same angle but at a speed $v$. What must $v/v_0$ be so that the second tennis ball also hits the center of the target? (You should get a pure number).

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We will determine the velocities $v_0$ and $v$ of each toss by considering the tosses separately and putting into equations the fact that the ball hits the center of the target in both cases.

Ball toss #1
The first ball is launched with velocity $v_0$ at an angle $\theta $ from the horizontal. Choosing the origin to be at the location of the ball launched and $y$ to be positive upward, we conclude that the kinematic equations describing the motion of the first tennis ball launched are

\left\{ \begin{array}{c} x_1\left(t\right)=v_0{\mathrm{cos} \left(\theta \right)\ }t \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \\ \displaystyle{y_1\left(t\right)=-\frac{1}{2}gt^2+v_0{\mathrm{sin} \left(\theta \right)\ }t} \end{array} \right.

The target is initially located at the ball launcher and moves with constant speed $W$. Thus, the kinematic equation describing the motion of the target is

x_T\left(t\right)=Wt

The ball hits the target (at ground level) at time $t_1$ which satisfies $y_1\left(t_1\right)=0$ i.e.

\begin{aligned} y_1\left(t_1\right)=0\ \ \ \ &\Rightarrow \ \ \ -\frac{1}{2}gt^2_1+v_0{\mathrm{sin} \left(\theta \right)\ }t_1=0 \\ \\ &\Rightarrow \ \ \ \ t_1\left(-\frac{1}{2}gt_1+v_0{\mathrm{sin} \left(\theta \right)\ }\right)=0 \\ \\ &\Rightarrow \ \ \ \ \ t_1=0\ \ \ \ \ or\ \ \ \ \ t_1=\frac{2v_0{\mathrm{sin} \left(\theta \right)\ }}{g} \end{aligned}

Ruling out $t_1=0$ which corresponds to the instant the first ball is launched, we keep the second root for $t_1$. In addition, we then argue that at time $t_1$, the $x$ positions of the ball and the target are the same since the ball hits the center of the target. Thus, we write

\begin{aligned} x_1\left(t_1\right)=x_T\left(t_1\right)\ \ \ \ &\Rightarrow \ \ \ \ -\frac{1}{2}gt^2_1+v_0{\mathrm{sin} \left(\theta \right)\ }t_1=Wt_1 \\ \\ &\Rightarrow \ \ \ \ \ 2v^2_0{\mathrm{sin} \left(\theta \right)\ }{\mathrm{cos} \left(\theta \right)\ }=\frac{2Wv_0{\mathrm{sin} \left(\theta \right)\ }}{g} \\ \\ &\Rightarrow \ \ \ \ \ v_0{\mathrm{cos} \left(\theta \right)\ }=W \end{aligned}

This actually makes perfect sense as it describes the fact that both the ball and the target, which have constant horizontal velocity, must share the same horizontal velocity if the ball is going to hit the center of the target when it reaches ground level.

Ball toss #2
The second ball is launched with velocity $v$ at an angle $\theta $ from the horizontal. Choosing the origin to be at the location of the ball launched and $y$ to be positive upward, we conclude that the kinematic equations describing the motion of the second tennis ball launched are

\left\{ \begin{array}{c} \displaystyle{x_2\left(t\right)=v{\mathrm{cos} \left(\theta \right)\ }t} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \\ \displaystyle{y_2\left(t\right)=-\frac{1}{2}gt^2+v{\mathrm{sin} \left(\theta \right)\ }t} \end{array} \right.

When the second ball is launched, the target is located at $x_T\left(t_1\right)=Wt_1$ and moves with constant speed $W$. Thus, the kinematic equation describing the motion of the target is

x_T\left(t\right)=Wt+Wt_1

The second ball must be launched so that at time $t_2$ when it reaches ground level, it hits the center of the target i.e. so that $x_2\left(t_2\right)=x_T\left(t_2\right)$.

With the same reasoning as followed previously, we conclude that the time $t_2$ is given by

t_2=\frac{2v \mathrm{sin}\left(\theta \right)}{g}

and we then write $x_2\left(t_2\right)=x_T\left(t_2\right)$.

\begin{aligned} x_2\left(t_2\right)=x_T\left(t_2\right)\ \ \ &\Rightarrow \ \ \ \ -\frac{1}{2}gt^2_2+v \mathrm{sin}\left(\theta \right)t_2=Wt_2+Wt_1 \\ \\ &\Rightarrow \ \ \ \ \frac{2v^2{\mathrm{sin} \left(\theta \right)\ }{\mathrm{cos} \left(\theta \right)\ }}{g}=Wt_2+\frac{2v^2_0{\mathrm{sin} \left(\theta \right)\ }{\mathrm{cos} \left(\theta \right)\ }}{g} \\ \\ &\Rightarrow \ \ \ \ \ \frac{2v^2{\mathrm{sin} \left(\theta \right)\ }{\mathrm{cos} \left(\theta \right)\ }}{g}=\frac{2vv_0{\mathrm{sin} \left(\theta \right)\ }{\mathrm{cos} \left(\theta \right)\ }}{g}+\frac{2v^2_0{\mathrm{sin} \left(\theta \right)\ }{\mathrm{cos} \left(\theta \right)\ }}{g} \end{aligned}

Thus, we solve

v^2=vv_0+v^2_0\ \ \ \ \Rightarrow \ \ \ \ \frac{v^2}{v^2_0}-\frac{v}{v_0}-1=0

We let $X=v/v_0$ and we solve the following quadratic

X^2-X-1=0\ \ \ \ \Rightarrow \ \ \ \ \ \left\{ \begin{array}{c} \displaystyle{X_1=\frac{1-\sqrt{5}}{2} < 0} \\ \\ \displaystyle{X_2=\frac{1+\sqrt{5}}{2} > 0} \end{array} \right.

Given that a ratio of two speeds cannot be negative, we conclude that

\boxed{X_2=\frac{v}{v_0}=\frac{1+\sqrt{5}}{2}}

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