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 3.3 m/s relative to the field while he passes the ball with velocity BP relative to himself. If BP has magnitude 5 m/s, what is the smallest angle it can have for the pass to be legal?
To determine the smallest angle at which the pass is legal, we need to analyze the velocities involved and apply the conditions given.
Let's break down the problem:
1. The rugby player is running with a velocity of 3.3 m/s relative to the field along the positive x-axis.
2. The ball's velocity relative to the rugby player (BP) has a magnitude of 5 m/s.
To find the smallest angle, we can find the x-component of the ball's velocity relative to the field, as that should not be positive for the pass to be legal.
Let's consider the velocities involved:
1. The player's velocity relative to the field (Vp) = 3.3 m/s along the positive x-axis.
2. The ball's velocity relative to the player (BP) has magnitude 5 m/s.
We can break down the ball's velocity relative to the field (BV) using the player's velocity relative to the field (Vp) and the ball's velocity relative to the player (BP).
BV = Vp + BP
To find the x-component of BV, we need to find the x-components of Vp and BP. Since Vp is along the positive x-axis, its x-component is 3.3 m/s. The x-component of the ball's velocity relative to the player (BPx) can be found using the magnitude of BP and the angle it makes with the x-axis.
Given that BP has a magnitude of 5 m/s, we need to find the smallest angle θ at which the x-component of the ball's velocity relative to the field (BVx) is not positive.
BVx = Vpx + BPx
Since the ball's velocity relative to the field (BV) is given by the sum of the player's velocity relative to the field (Vp) and the ball's velocity relative to the player (BP), we can substitute the values:
BVx = Vpx + BP*cos(θ)
BVx should not be positive for the pass to be legal. Therefore:
BVx = 3.3 + 5*cos(θ) ≤ 0
Solving this inequality, we can find the smallest angle θ for which the pass is legal.
To determine the smallest angle at which the pass can be legal, we need to consider the relative velocity of the ball relative to the field.
Given:
Player's speed relative to the field (v) = 3.3 m/s
Ball's velocity relative to the player (BP) = 5 m/s
Let's assume the angle between the direction of player's velocity (x-axis) and the direction of the pass (BP) is θ.
The x-component of the ball's velocity relative to the field is given by:
VBx = v + BP cos(θ)
Since the pass is legal only when the x-component of the ball's velocity is not positive, we can set VBx to zero and solve for θ.
0 = v + BP cos(θ)
Re-arranging the equation:
BP cos(θ) = -v
Taking the magnitude of both sides:
|BP cos(θ)| = |v|
Since both BP and v are positive, we can write:
BP cos(θ) = v
Substituting the given values:
5 cos(θ) = 3.3
Simplifying the equation further:
cos(θ) = 3.3/5
cos(θ) = 0.66
To find the smallest angle, we need to take the inverse cosine (arc cosine) of both sides:
θ = arccos(0.66)
Using a calculator or mathematical software, we find:
θ ≈ 47.2 degrees
Therefore, the smallest angle at which the pass can be legal is approximately 47.2 degrees.