Problem 60 - Algebra and Trigonometry

Question

Path of a Ball \(A\) ball is thrown across a playing field. Its path is given by the equation \(y=-0.005 x^{2}+x+5\) where \(x\) is the distance the ball has traveled horizontally, and \(y\) is its height above ground level, both measured in feet. (a) What is the maximum height attained by the ball? (b) How far has it traveled horizontally when it hits the ground?

Step-by-Step Solution

Answer

Max height: 55 feet. Hits ground at ≈ 104.88 feet horizontally.

Step 1: Understand the Equation

The path of the ball is described by the quadratic equation \( y = -0.005x^2 + x + 5 \). Here, \( y \) represents the height of the ball in feet, and \( x \) represents the horizontal distance traveled in feet. This is a downward-opening parabola because the coefficient of \( x^2 \) is negative.

Step 2: Find the Maximum Height (Vertex Form)

The maximum height of a quadratic function \( y = ax^2 + bx + c \) occurs at the vertex. The x-coordinate of the vertex can be found using \( x = -\frac{b}{2a} \). Here \( a = -0.005 \) and \( b = 1 \). Substitute these values to get \( x = -\frac{1}{2(-0.005)} = 100 \).

Step 3: Calculate the Maximum Height

Substitute \( x = 100 \) back into the equation to find \( y \): \[ y = -0.005(100)^2 + 1(100) + 5 \] \[ y = -0.005 \times 10000 + 100 + 5 \] \[ y = -50 + 100 + 5 \] \[ y = 55 \] Thus, the maximum height attained by the ball is 55 feet.

Step 4: Find When the Ball Hits the Ground

To find when the ball hits the ground, set \( y = 0 \) in the equation: \[ 0 = -0.005x^2 + x + 5 \] Solve this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = -0.005 \), \( b = 1 \), and \( c = 5 \).

Step 5: Solve for x Using the Quadratic Formula

Calculate the discriminant: \[ b^2 - 4ac = 1^2 - 4(-0.005)(5) = 1 + 0.1 = 1.1 \] Then find \( x \): \[ x = \frac{-1 \pm \sqrt{1.1}}{2(-0.005)} \] \[ x = \frac{-1 \pm 1.0488}{-0.01} \] This gives two solutions, but we take the positive one: \[ x = \frac{-1 + 1.0488}{-0.01} \] \[ x = rac{0.0488}{-0.01} = 104.88 \] Thus, the ball hits the ground after approximately 104.88 feet of horizontal travel.

Key Concepts

Parabolas

A parabola is a simple curve shaped like a bowl, and it's derived from quadratic functions. In the context of quadratic equations, a parabola can either open upwards or downwards depending on the sign of the coefficient of the squared term. For instance, in our equation, \(y = -0.005x^2 + x + 5\), the parabola opens downwards because the coefficient of \(x^2\) is negative (-0.005).

Parabolas are symmetric shapes, which means if you fold them right across their point of symmetry, both halves will align perfectly. This symmetry center is known as the vertex. Because of this, parabolas are often used to model trajectories of objects, like the path of a ball in our exercise. Here, the vertex gives critical information about the maximum height the ball reaches.

• Downward opening parabola: Negative \(x^2\) coefficient.
• Symmetry of parabolas is key to finding max/min values.
• Used in real-life scenarios to illustrate paths and trajectories.

Vertex Form

The vertex form of a quadratic equation is particularly useful for finding the highest or lowest point of a parabola. In the standard quadratic equation \(y = ax^2 + bx + c\), the vertex gives a lot of valuable information. It shows either the maximum or minimum value of the quadratic function. By converting a quadratic equation into its vertex form, you can easily identify these points without in-depth calculations.

You can find the vertex \((h, k)\) by using the formula for the x-coordinate of the vertex, \(x = -\frac{b}{2a}\). Once you have \(x\), plug it back into the equation to find the corresponding \(y\)-value, giving you point \((h, k)\) which is the vertex. In our exercise case, with \(a = -0.005\) and \(b = 1\), substituting values gives us \(x = 100\). Substituting \(x = 100\) back yields the maximum height of 55 feet for the ball.

• Vertex form is the same as finding maximum/minimum points.
• Use formula: \(x = -\frac{b}{2a}\) to find vertex \(x\)-coordinate.
• Plug \(x\) into the original equation to find corresponding \(y\).
• Vertex \((h, k)\) reveals maximum height or lowest point in the path.

Quadratic Formula

The quadratic formula is a highly effective tool for finding the roots of any quadratic equation of the form \(ax^2 + bx + c = 0\). Roots are the values of \(x\) that make the equation equal zero, which often represent where a parabola intersects the \(x\)-axis, or in real-world applications, where an object reaches ground level.

The formula is: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). This allows you to solve any quadratic equation efficiently without needing to manually factor it. In our exercise, applying this formula to \(y = -0.005x^2 + x + 5\) helped us find the point \(x\) when the ball hits the ground.

• Quadratic formula finds roots of \(ax^2 + bx + c = 0\).
• Critical for determining intercepts where the curve crosses the \(x\)-axis.
• Formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) provides solutions directly.
• Important in applications involving distances and intersections.