Problem 64

Question

A ball is thrown across a playing field from a height of $5 \mathrm{ft}$ above the ground at an angle of $45^{\circ}$ to the horizontal at a speed of $20 \mathrm{ft} / \mathrm{s}$. It can be deduced from physical principles that the path of the ball is modeled by the function $$y=-\frac{32}{(20)^{2}} x^{2}+x+5$$ where $x$ is the distance in feet that the ball has traveled horizontally. (a) Find the maximum height attained by the ball. (b) Find the horizontal distance the ball has traveled when it hits the ground.

Step-by-Step Solution

Verified

Answer

Max height: 13.06 ft; Distance traveled: 16.33 ft.

1Step 1: Identify the given function

The path of the ball is modeled by the quadratic function $y = -\frac{32}{400} x^2 + x + 5$, which simplifies to $y = -\frac{x^2}{12.5} + x + 5$. This function represents a parabola opening downwards since the coefficient of $x^2$ is negative.

2Step 2: Calculate the maximum height (vertex)

The maximum height of the parabola occurs at its vertex. For a quadratic function in the form $ax^2 + bx + c$, the x-coordinate of the vertex is $x = -\frac{b}{2a}$. Here, $a = -\frac{1}{12.5}$ and $b = 1$. Plug these values into the formula:\[x = -\frac{1}{2 \times -\frac{1}{12.5}} = 6.25.\]Substitute $x = 6.25$ back into the function to find the maximum height:\[y = -\frac{(6.25)^2}{12.5} + 6.25 + 5.\]Calculate the value to find the maximum height:\[y = -\frac{39.0625}{12.5} + 6.25 + 5 = 7.8125 + 6.25 + 5 = 13.0625.\]So, the maximum height is approximately $13.0625$ ft.

3Step 3: Find horizontal distance when it hits the ground

The ball hits the ground when $y = 0$. Set the function equal to zero and solve for $x$:\[ -\frac{x^2}{12.5} + x + 5 = 0.\]Multiply the equation by 12.5 to clear the fraction:\[-x^2 + 12.5x + 62.5 = 0.\]Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = -1$, $b = 12.5$, and $c = 62.5$:\[x = \frac{-12.5 \pm \sqrt{(12.5)^2 - 4(-1)(62.5)}}{-2} = \frac{-12.5 \pm \sqrt{156.25 + 250}}{-2}.\]Calculate the discriminant:\[x = \frac{-12.5 \pm \sqrt{406.25}}{-2}.\]\[x = \frac{-12.5 \pm 20.1563}{-2}.\]The roots are:\[x_1 = \frac{-12.5 + 20.1563}{-2} \approx -3.828\](reject, as distance cannot be negative)\[x_2 = \frac{-12.5 - 20.1563}{-2} \approx 16.328.\]So, the horizontal distance is approximately $16.33$ ft when the ball hits the ground.

Key Concepts

Quadratic FunctionVertex of a ParabolaQuadratic Formula

Quadratic Function

A quadratic function is a type of polynomial function that plays a crucial role in projectile motion. It is generally expressed in the form $ f(x) = ax^2 + bx + c $, where $a$, $b$, and $c$ are constants with $a eq 0$.
This type of function describes a curve known as a parabola.
In the given exercise, the ball follows a path described by the quadratic function:

$ y = -\frac{32}{400}x^2 + x + 5 $.

The negative coefficient of $x^2$ indicates that the parabola opens downwards, modeling the trajectory of the ball. As the ball moves up and down along its path, its height over distance forms this downward-opening curve. Quadratic functions are widely used to describe various real-life scenarios involving parabolic motion, making them vital in physics and engineering applications.

Vertex of a Parabola

The vertex of a parabola is a key point that represents either the maximum or minimum value of a quadratic function. For a parabola that opens downwards, like in our problem, the vertex represents the maximum point.
In the standard form of a quadratic equation $ ax^2 + bx + c $, the x-coordinate of the vertex can be found using the formula:

$ x = -\frac{b}{2a} $.

For the function $ y = -\frac{x^2}{12.5} + x + 5 $, we substitute $ a = -\frac{1}{12.5} $ and $ b = 1 $ into the vertex formula to find:

$ x = -\frac{1}{2 \times -\frac{1}{12.5}} = 6.25 $.

Substituting back into the function gives the height at this point:

$ y = -\frac{(6.25)^2}{12.5} + 6.25 + 5 = 13.0625 $.

Therefore, at $ x = 6.25 $ feet, the ball reaches a maximum height of approximately 13.06 feet.

Quadratic Formula

The quadratic formula is a valuable tool used to find the roots of quadratic equations. It provides the solutions for $ x $ in any quadratic equation $ ax^2 + bx + c = 0 $. The formula is:

$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $.

In the context of the problem, we use the quadratic formula to determine when the ball hits the ground, i.e., when $ y = 0 $. This leads to the equation: