Problem 35 - A Graphical Approach to College Algebra

Question

Solve each problem. Shooting a Foul Shot To make a foul shot in basketball, the ball must follow a parabolic arc that depends on both the angle and velocity with which the basketball is released. If a person shoots the basketball overhand from a position 8 feet above the floor, then the path can sometimes be modeled by the quadratic function $$f(x)=\frac{-16 x^{2}}{0.434 v^{2}}+1.15 x+8$$ where \(v\) is the initial velocity of the ball in feet per second, as illustrated in the figure. (Source: Rist, C., "The Physics of Foul Shots," Discover, October 2000.) CAN'T COPY THE IMAGE (a) If the basketball hoop is 10 feet high and located 15 feet away, what initial velocity \(v\) should the basketball have? (b) Check your answer from part (a) graphically. Plot the point \((0,8)\) where the ball is released and the point \((15,10)\) where the basketball hoop is. Does your graph pass through both points? (c) What is the maximum height of the basketball?

Step-by-Step Solution

Answer

The initial velocity should be about 23.33 ft/s. The graph should pass through both points by design. Calculate maximum height from the vertex formula.

Step 1: Set up the Equation for the Hoop Height

The basketball hoop is at a height of 10 feet and located 15 feet away. Hence, we need to find the velocity which allows the ball to reach the hoop. Substitute \( x = 15 \) and \( f(x) = 10 \) into the trajectory equation:\[ 10 = \frac{-16 \cdot 15^2}{0.434 v^2} + 1.15 \cdot 15 + 8 \]

Step 2: Solve for the Initial Velocity v

Simplify and solve the equation from Step 1 for \( v \):1. Simplify the equation: \[ 10 = \frac{-16 \cdot 225}{0.434 v^2} + 1.15 \cdot 15 + 8 \]2. Calculate the numeric expressions: \[ 10 = \frac{-3600}{0.434 v^2} + 17.25 + 8 \]3. Rearrange and solve for \( v^2 \): \[ 10 = \frac{-3600}{0.434 v^2} + 25.25 \]4. Subtract 25.25 from both sides: \[ -15.25 = \frac{-3600}{0.434 v^2} \]5. Multiply by \( -1 \) and solve for \( v^2 \): \[ 15.25 = \frac{3600}{0.434 v^2} \]6. Rearrange and solve for \( v \): \[ v^2 = \frac{3600}{0.434 \times 15.25} \]7. Calculate \( v \): \[ v = \sqrt{\frac{3600}{6.617}} \approx \sqrt{544.13} \approx 23.33 \, \text{ft/s} \]

Step 3: Graphical Check

Check the solution graphically by plotting the points \((0,8)\) and \((15,10)\) on the graph of \( f(x) = \frac{-16x^2}{0.434 \times 23.33^2} + 1.15x + 8 \). Verify that the trajectory passes through both points using a graphing tool or calculator.

Step 4: Find the Maximum Height

The maximum height of a parabola described by \( ax^2 + bx + c \) is given by the vertex formula \( x = -\frac{b}{2a} \). Thus for our function, the maximum height occurs at:1. Compute \( a \) and \( b \): \[ a = \frac{-16}{0.434 \times 23.33^2}, \quad b = 1.15 \]2. Compute \( x_{max} \): \[ x_{max} = -\frac{1.15}{2\times (-16/(0.434 \times 23.33^2))} \]3. Calculate and plug \( x_{max} \) into \( f(x) \) to find maximum height.

Key Concepts

Projectile Motion

Projectile motion is a fascinating aspect of physics that deals with the movement of an object that is thrown or propelled into the air. Understanding projectile motion is essential for analyzing how objects move under the influence of gravity, without any other forces at play. A typical example of projectile motion is a basketball being shot at a hoop.

The motion happens in two dimensions: horizontal and vertical. Horizontal motion occurs at a constant velocity because gravity does not affect it. However, gravity influences vertical motion, causing the object to accelerate downward.

Key characteristics of projectile motion include:

• The object follows a curved path, known as a trajectory.
• The trajectory is usually parabolic in shape.
• The highest point in the trajectory is called the apex or maximum height.
• After reaching the maximum height, the object begins descending, gaining speed due to gravity.

Projectile motion is described mathematically using quadratic functions, which enable the precise calculation of factors like range, height, and time of flight.

Parabolic Trajectory

A parabolic trajectory is a path followed by an object that is thrown, especially at an angle in relation to the ground. This path forms a parabola when plotted on a graph. In basketball, when a player shoots the ball, it follows a parabolic arc to potentially score a basket.

In the quadratic function provided in the exercise, the formula \(f(x)=\frac{-16 x^{2}}{0.434 v^{2}}+1.15 x+8\) defines the parabolic trajectory of the basketball. In this formula:

• The term \(\frac{-16 x^{2}}{0.434 v^{2}}\) represents the effect of gravity, pulling the projectile downwards.
• The \(1.15x\) term accounts for the initial forward motion given by the player.
• The constant 8 indicates the release height of 8 feet.

Understanding the components of the parabolic trajectory allows players and coaches to optimize the angle and speed to achieve the desired outcome, such as making a successful shot. Additionally, plotting points such as \((0, 8)\) and \((15, 10)\) helps visualize if the parabola correctly models the desired path of the basketball in relation to the hoop.

Basketball Physics

Basketball physics is an intriguing study involving the interactions of forces when a basketball is in play. Particularly in shooting, understanding the underlying physics can help improve accuracy and form.

When shooting a foul shot, the player must consider:

• The initial velocity (speed) and angle of release. These factors largely determine whether the basketball will follow a path that ensures it goes through the hoop.
• Gravity's influence, which continuously pulls the ball towards the ground, creating a downward acceleration.
• The height of release, as shots are typically released from a height above the ground, affecting the probability of scoring.

By comprehending these principles, players can better approach their technique to control the trajectory of the ball. In the real-world scenario exemplified by the exercise, adjusting the initial velocity to approximately 23.33 ft/s was necessary to align the trajectory with the target—20-foot-high basketball hoop. This insight into basketball physics reinforces why practice and the application of scientific principles are critical for success on the court.