91Ó°ÊÓ

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

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

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:

$\stackrel{→}{V_{BF}}=\stackrel{→}{V_{PF}}+\stackrel{→}{V_{BP}}$

$\sin \mathrm{θ}=\frac{oppositeside}{hypotenuse}$

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,

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

From the vector diagram, by using trigonometry,

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

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