Problem 64 - Algebra and Trigonometry

Question

Path of a Ball 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

Answer

(a) Max height: 8.125 ft; (b) Horizontal distance: 16.31 ft.

Step 1: Identify the Parabola

The equation given is in the form of a quadratic function: \( y = -\frac{32}{400} x^2 + x + 5 \). This can be simplified to \( y = -0.08x^2 + x + 5 \). We recognize this equation as a downward-opening parabola (since its leading coefficient is negative).

Step 2: Find the Vertex Formula Components

The vertex of a parabola given by \( ax^2 + bx + c \) occurs at \( x = \frac{-b}{2a} \). In our case, \( a = -0.08 \) and \( b = 1 \).

Step 3: Calculate the x-coordinate of the Vertex

Using the formula \( x = \frac{-b}{2a} \), substitute in \( b = 1 \) and \( a = -0.08 \):\[ x = \frac{-1}{2(-0.08)} = \frac{-1}{-0.16} = 6.25 \].

Step 4: Calculate the Maximum Height

Substitute \( x = 6.25 \) back into the original equation to find the maximum height:\[ y = -0.08(6.25)^2 + 6.25 + 5 \]\[ y = -0.08(39.0625) + 6.25 + 5 \]\[ y = -3.125 + 6.25 + 5 \]\[ y = 8.125 \].The maximum height attained by the ball is 8.125 feet.

Step 5: Set y to 0 for Ground Impact

To find when the ball hits the ground, set \( y = 0 \) in the equation:\[ 0 = -0.08x^2 + x + 5 \].

Step 6: Use the Quadratic Formula

Solve \( -0.08x^2 + x + 5 = 0 \) using the quadratic formula: \( x = \frac{-b\pm\sqrt{b^2-4ac}}{2a} \) where \( a = -0.08 \), \( b = 1 \), and \( c = 5 \).

Step 7: Calculate the Roots

Substitute the values into the quadratic formula:\[ x = \frac{-1 \pm \sqrt{1^2 - 4(-0.08)(5)}}{2(-0.08)} \]\[ x = \frac{-1 \pm \sqrt{1 + 1.6}}{-0.16} \]\[ x = \frac{-1 \pm \sqrt{2.6}}{-0.16} \].Let's focus on the positive root:\[ x = \frac{-1 + \sqrt{2.6}}{-0.16} \].\( \sqrt{2.6} \approx 1.61 \), so:\[ x = \frac{-1 + 1.61}{-0.16} \]\[ x = \frac{0.61}{-0.16} \approx -3.81 \].Since the distance must be a positive value, consider the other scenario:\[ x = \frac{-1 - 1.61}{-0.16} \]\[ x = \frac{-2.61}{-0.16} \approx 16.31 \].The horizontal distance traveled when it hits the ground is approximately 16.31 feet.

Key Concepts

Vertex of a Parabola

The vertex of a parabola is a significant point that represents either the highest point or the lowest point on the graph, depending on the orientation of the parabola. In the case of a downwards-facing parabola, like the path of the ball described by the quadratic function, the vertex corresponds to the maximum point.

To find the vertex, we employ the formula for the x-coordinate: \( x = \frac{-b}{2a} \), where \( a \), \( b \), and \( c \) are the coefficients from the quadratic equation \( ax^2 + bx + c \). For our specific equation \( y = -0.08x^2 + x + 5 \), we identify \( a = -0.08 \) and \( b = 1 \). Substituting these values into the vertex formula gives us \( x = 6.25 \).

Once we find this x-coordinate, we can compute the maximum height (y-coordinate) by substituting \( x = 6.25 \) back into the original quadratic equation.

Maximum Height Calculation

The maximum height of the ball's path is determined by calculating the y-value of the quadratic function at the x-coordinate of the vertex, which we've already found to be \( x = 6.25 \).

Substitute \( x = 6.25 \) into the equation \( y = -0.08x^2 + x + 5 \):
\[ y = -0.08(6.25)^2 + 6.25 + 5 \]
Calculate \( 6.25^2 \) to find 39.0625, and proceed:
\[ y = -0.08(39.0625) + 6.25 + 5 \]
\[ y = -3.125 + 6.25 + 5 \]
\[ y = 8.125 \].

Thus, the maximum height of the ball reached during its trajectory is 8.125 feet. This value gives us an insight into how high the ball travels at its peak before descending back towards the ground.

Quadratic Formula

The quadratic formula is a powerful tool used to find the roots of a quadratic equation. It is especially useful when the quadratic does not easily factor. The formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].

For the quadratic equation \( -0.08x^2 + x + 5 = 0 \), we have \( a = -0.08 \), \( b = 1 \), and \( c = 5 \). Insert these into the formula to find the solutions for \( x \):
\[ x = \frac{-1 \pm \sqrt{1^2 - 4(-0.08)(5)}}{2(-0.08)} \]
This simplifies to
\[ x = \frac{-1 \pm \sqrt{1 + 1.6}}{-0.16} \]
\[ x = \frac{-1 \pm \sqrt{2.6}}{-0.16} \].

Since the context is physical (the path of a ball), we need the positive root for distance. Solving this yields \( x = 16.31 \), which is the horizontal distance the ball travels before hitting the ground.