Problem 99 - University Physics with Modern Physics

Question

Completed Pass. At Enormous State University (ESU), the football team records its plays using vector displacements, with the origin taken to be the position of the ball before the play starts. In a certain pass play, the receiver starts at \(+1.0 \hat{\imath}-5.0 \hat{\jmath}\) , where the units are yards, \(\hat{\mathbf{i}}\) is to the right, and \(\hat{\boldsymbol{j}}\) is downfield. Subsequent displacements of the receiver are \(+9.0 \hat{\imath}\) (in motion before the snap), \(+11.0 \hat{y}\) (breaks downfield), \(-6.0 \hat{\imath}+4.0 \hat{\jmath}\) (zigs), and \(+12.0 \hat{i}+18.0 \hat{j}\) (zags). Meanwhile, the quarterback has dropped straight back to a position \(-7.0 \hat{\jmath}\) . How far and in which direction must the quarterback throw the ball? (Like the coach, you will be well advised to diagram the situation before solving it numerically.)

Step-by-Step Solution

Answer

The quarterback must throw 38.48 yards at an angle of 65.37 degrees downfield towards the right.

Step 1: Diagram the Situation

To visualize the problem, draw the initial position of the receiver and the quarterback. The receiver starts at the position given by the vector \(+1.0 \hat{\imath} - 5.0 \hat{\jmath}\) in the coordinate system where \(\hat{\i}\) is to the right and \( \hat{\j}\) is downfield. The quarterback begins at the origin (0,0).

Step 2: Calculate Total Displacement of the Receiver

Determine the total displacement of the receiver by adding up all subsequent displacement vectors. Starting at \(+1.0 \hat{\imath} - 5.0 \hat{\jmath}\), add \(+9.0 \hat{\imath}\), \(+11.0 \hat{\jmath}\), \(-6.0 \hat{\imath} + 4.0 \hat{\jmath}\), and \(+12.0 \hat{\imath} + 18.0 \hat{\jmath}\). This results in the total displacement: \(+1.0 + 9.0 - 6.0 + 12.0 = 16.0 \hat{\imath}\) and \(-5.0 + 11.0 + 4.0 + 18.0 = 28.0 \hat{\jmath}\). Thus, the receiver's final position is \(16.0 \hat{\imath} + 28.0 \hat{\jmath}\).

Step 3: Determine Quarterback's Position

The problem states the quarterback moves to \(-7.0 \hat{\jmath}\). Thus, the quarterback's position in vector form is \(0 \hat{\imath} - 7.0 \hat{\jmath}\).

Step 4: Calculate Throw Vector

To find the direction and distance the quarterback must throw, calculate the vector from the quarterback's position \(0 \hat{\imath} - 7.0 \hat{\jmath}\) to the receiver's final position \(16.0 \hat{\imath} + 28.0 \hat{\jmath}\). This is done by subtracting the quarterback's position vector from the receiver's position vector: \((16.0 \hat{\imath} + 28.0 \hat{\jmath}) - (0 \hat{\imath} - 7.0 \hat{\jmath}) = 16.0 \hat{\imath} + 35.0 \hat{\jmath}\).

Step 5: Calculate Magnitude of the Throw Vector

The magnitude of the vector \(16.0 \hat{\imath} + 35.0 \hat{\jmath}\) can be calculated using the Pythagorean theorem. \[\sqrt{(16.0)^2 + (35.0)^2} = \sqrt{256 + 1225} = \sqrt{1481} = 38.48 \text{ yards}\].

Step 6: Determine the Direction of the Throw

To determine the direction from the horizontal for the throw, calculate the angle \(\theta \) such that \(\tan \theta = \frac{35.0}{16.0}\). This gives \(\theta = \tan^{-1}\left(\frac{35.0}{16.0}\right) \approx 65.37^{\circ}\). The throw is 65.37 degrees downfield to the right.

Key Concepts

Displacement Vectors

In physics, a displacement vector helps us understand how an object moves from one point to another. It includes both magnitude (how far) and direction. In this exercise, the player's path consists of several vector displacements.
Each displacement vector represents a segment of the player's journey on the football field, denoted using coordinates such as \(+1.0 \, \hat{\imath} - 5.0 \, \hat{\jmath}\). This notation tells us the receiver starts 1 yard to the right and 5 yards downfield. The journey continues with several more vectors:

• \(+9.0 \, \hat{\imath}\) signifies a 9-yard movement to the right.
• \(+11.0 \, \hat{\jmath}\) indicates an 11-yard move downfield.
• \(-6.0 \, \hat{\imath} + 4.0 \, \hat{\jmath}\) shows a zig: 6 yards left, 4 yards downfield.
• \(+12.0 \, \hat{\imath} + 18.0 \, \hat{\jmath}\) reflects a zag: 12 yards to the right, 18 yards downfield.

The total displacement is found by combining all vectors to determine the receiver's final position on the field. Thus, understanding displacement vectors provides clarity on the cumulative movement direction and distance.

Coordinate System in Physics

A coordinate system in physics provides a reference framework to describe the position and movement of objects. In this exercise, the football field serves as our coordinate grid.
The system uses \( \hat{\imath} \) and \( \hat{\jmath} \) to define directions. \(\hat{\imath}\) points to the right, while \(\hat{\jmath}\) indicates the direction downfield. Each position or movement on the field is described using these directional units.
For example, a position like \(+1.0 \, \hat{\imath} - 5.0 \, \hat{\jmath}\) tells us that the receiver starts one yard right and five yards downfield from the origin, the original position of the ball at (0,0) coordinate. This structure eases calculations and visualizing the football play.
In a simpler sense, think of it like reading directions on a map, where every step is precisely measured and defined using coordinates.

Pythagorean Theorem in Vector Calculation

The Pythagorean theorem is fundamental in calculating the magnitude of displacement vectors. It helps us determine the straight-line distance (or actual distance) between two points irrespective of the path taken.
When given the vector \(16.0 \, \hat{\imath} + 35.0 \, \hat{\jmath}\), we consider it as forming a right triangle with the x-axis and y-axis on our coordinate grid. The magnitude is then calculated as \( \sqrt{(16.0)^2 + (35.0)^2} \).
This calculation is derived from the classic Pythagorean formula \( a^2 + b^2 = c^2 \), where \( a \) and \( b \) are the legs of the triangle and \( c \) is the hypotenuse, representing the vector's length. For this problem, those dimensions simplify to \( \sqrt{256 + 1225} = \sqrt{1481} \approx 38.48 \text{ yards} \). The Pythagorean theorem elegantly converts component vectors into meaningful, measurable distance on the football field.