Question

A rugby player runs with the ball directly toward his opponent’s goal, along the positive direction of an x axis. He can legally pass the ball to a teammate as long as the ball’s velocity relative to the field does not have a positive x component. Suppose the player runs at speed $\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$relative to the field while he passes the ball with velocitylocalid="1660895439009" ${\overrightarrow{\mathbf{V}}}_{\mathbf{B}\mathbf{P}}$relative to himself. If localid="1660895452337" ${\overrightarrow{\mathbf{V}}}_{\mathbf{BP}}$has magnitude $\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$, what is the smallest angle it can have for the pass to be legal?

Short answer

Answer

The smallest angle the ball can have for the pass to be legal is ${130}^{°}$

Step-by-step solution

Given data

1. The velocity of the player relative to the field is $\overrightarrow{V_{PF}}=4.0m/s$

2. The velocity of the ball relative to the player is $\overrightarrow{V_{BF}}=6.0m/s$

To understand the concept of relative velocity

The relative velocity of an object is defined as its velocity in relation to some other observer. The player is running towards goal. The player runs relative to the field and he passes the ball. According to the concept of relative motion, we can write the required equations. From it draw the vector diagram, it is right angled triangle. Use trigonometry and find minimum angle.

Formula:

$\overrightarrow{V_{BF}}=\overrightarrow{V_{PF}}+\overrightarrow{V_{BP}}$

$\sin \theta =\frac{oppositeside}{hypotenuse}$

Calculate the smallest angle made by the velocity of the ball can have for the pass to be legal

Since the player is running forward, the pass has to be in a backward direction at such an angle that the addition of these vectors gives resultant. According to the relative motion,

$$\begin{gathered} \overrightarrow{V_{BF}}=\overrightarrow{V_{PF}}+\overrightarrow{V_{BP}} \\ \overrightarrow{V_{BP}}=\overrightarrow{V_{BF}}-\overrightarrow{V_{PF}} \end{gathered}$$

From the vector diagram, by using trigonometry,

$$\begin{gathered} \cos \left(180-\theta \right)=\frac{V_{PF}}{V_{BP}} \\ \cos \left(180-\theta \right)=\frac{4m/s}{6m/s} \\ \theta =131.8^{°} \\ \approx {130}^{°} \end{gathered}$$

Therefore, the smallest angle the ball can have for the pass to be legal is ${130}^{°}$