Problem 49 - Introductory Algebra for College Students

Question

A ball is thrown upward and outward from a height of 6 feet. The height of the ball, \(y,\) in feet, can be modeled by $$y=-0.8 x^{2}+3.2 x+6$$ where \(x\) is the ball's horizontal distance, in feet, from where it was thrown. a. What is the maximum height of the ball and how far from where it was thrown does this occur? b. How far does the ball travel horizontally before hitting the ground? Round to the nearest tenth of a foot. c. Graph the equation that models the ball's parabolic path.

Step-by-Step Solution

Answer

a. The maximum height of the ball is when x = 2, i.e., 2 feet from where it was thrown. b. The ball will hit the ground at the x-values given by solving the equation \(0=-0.8x^2+3.2x+6\). c. The result of graphing the equation \(y=-0.8 x^{2}+3.2 x+6\) will show the parabolic path of the movement of the ball.

Step 1: Finding the maximum height

The maximum or minimum of a quadratic function is found at the vertex. For a quadratic function in the form \(f(x)=ax^2+bx+c\), the x-coordinate of the vertex is given by the formula \(-b/2a\). Substituting the coefficients of the equation, we find the x-coordinate of the vertex to be \(-3.2/-1.6 = 2\). Substituting \(x=2\) into the equation will then give us the maximum height - which is the y-coordinate of the vertex.

Step 2: Calculate the x-intercept (when the ball hits the ground)

The x-intercept of a parabola (or any function) can be found by setting y = 0 and solve for x. In this case, setting y = 0 we have the equation \(0=-0.8x^2 + 3.2x + 6\). We can solve this quadratic equation using the quadratic formula, which gives us the values of \(x\) when the ball hits the ground.

Step 3: Drawing the graph

Here we plot the equation that models the parabolic path of the ball using the values we obtained in step 1 and step 2. The y-intercept is given by the equation - which is 6. We plot points for these three crucial places: the y-intercept, the vertex and the x-intercept, and draw a smooth curve connecting them.

Key Concepts

Parabolic Motion

Parabolic motion refers to the path that a projectile follows when it is subject to gravity. It often forms a symmetrical arc known as a parabola. When a ball is thrown, like in the exercise, it will rise, slow down, and then descend back to the ground.
The equation given in the problem, \( y = -0.8x^2 + 3.2x + 6 \), describes such a path. Here, \(-0.8x^2\) demonstrates the influence of gravity pulling the ball down. Meanwhile, \(+3.2x\) simulates the initial forward thrust when the ball is thrown. Plus, the constant \(+6\) helps set the starting point of the throw, marking the initial height above ground.
Understanding parabolic motion is useful in many areas like sports, engineering, and even video game design as it helps predict future positions of moving objects. By analyzing its arc, we can find the maximum height, distance traveled, and landing point.

Vertex Calculation

The vertex of a parabola is a crucial point because it tells us either the highest or lowest point on the curve. For parabolic motion, this is usually the maximum height the projectile reaches in its path.
With a quadratic function given as \(y = ax^2 + bx + c\), you can calculate the x-coordinate of the vertex with the formula \(-b/2a\). This formula helps determine when the maximum height occurs.

• For our equation, \(a = -0.8\), and \(b = 3.2\).
• Using the formula, we get \(-3.2/(2 * -0.8) = 2\).
• This means the ball will reach its highest point when it has traveled 2 feet horizontally.

Once we know the x-coordinate, we simply substitute \(x = 2\) into the equation to find the y-coordinate of the vertex. This y-coordinate is the maximum height reached by the ball.

Solving Quadratic Equations

To explore the complete journey of the ball in the exercise, we solve for where it lands by determining when it hits the ground. This involves finding the x-intercepts of the quadratic equation.
The x-intercepts are where the graph crosses the x-axis, meaning \(y = 0\). Let's set the equation \(y = -0.8x^2 + 3.2x + 6\) equal to zero:

• Rewriting, this gives: \(0 = -0.8x^2 + 3.2x + 6\).
• We can apply the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), here \(a = -0.8\), \(b = 3.2\), and \(c = 6\).

Solving this will give us the two x-values where the ball first reaches the ground and again on its descent.
Using these x-intercepts, we can find how far the ball traveled horizontally. Calculating these points allows us to better understand how the quadratic nature charges the ball’s destiny in its trajectory.