Problem 38 - Algebra and Trigonometry

Question

Path of a Cannonball A cannon fires a cannonball as shown in the figure. The path of the cannonball is a parabola with vertex at the highest point of the path. If the cannon ball lands 1600 ft from the cannon and the highest point it reaches is 3200 ft above the ground, find an equation for the path of the cannonball. Place the origin at the location of the cannon.

Step-by-Step Solution

Answer

The equation is \(y = -\frac{1}{2}(x-800)^2 + 3200\).

Step 1: Understanding the Parabolic Path

The path is a parabola because the cannonball follows a projectile motion. The vertex is the highest point, which is directly above the origin as the cannonball lands at 1600 ft away horizontally.

Step 2: Setting up the Vertex Format

The vertex form of a parabola is given by \( y = a(x-h)^2 + k \), where \((h, k)\) is the vertex. In this problem, the vertex is at \((800, 3200)\) because the highest point of 3200 ft occurs halfway between the launch and the landing.

Step 3: Using Symmetry to Place Vertex

Since the symmetry of the parabola means the cannonball lands equidistant from the vertex, and the parabola opens downwards, the points on the horizontal axis will be at 0 ft (cannon) and 1600 ft (landing).

Step 4: Determining the Parabola's Equation

Using the vertex form equation \( y = a(x-800)^2 + 3200 \), we plug in the landing point \((1600, 0)\) to solve for \(a\).

Step 5: Substitute to Find 'a'

Substitute \(x = 1600\) and \(y = 0\) into the equation: \(0 = a(1600-800)^2 + 3200\). Then, solve for \(a\) to get \(a = -\frac{1}{2}\).

Step 6: Writing the Final Equation

Rewrite the parabola with the calculated \(a\) value: \(y = -\frac{1}{2}(x-800)^2 + 3200\). This is the equation representing the cannonball's path.

Key Concepts

Parabola

A parabola is a smooth, U-shaped curve that arises in various physical phenomena, such as the path of projectiles in motion. In projectile motion, forces like gravity act on the object, creating a specific type of trajectory—a parabola. This curvature is symmetric concerning its highest point, known as the vertex. The cannonball's path is modeled as a parabola because it follows this natural projective trajectory. Understanding the parabolic shape helps in predicting where the cannonball will land.

• Parabolas are symmetric, meaning they curve equally on both sides of the vertex.
• The highest or lowest point on a parabola is called the vertex.
• This symmetry is crucial for calculating distances and heights in projectile motion.

Recognizing the patterns in parabolas simplifies real-world calculations in contexts like the mentioned cannonball path.

Vertex Form

The vertex form of a quadratic equation is a powerful tool for analyzing parabolas. This form is expressed as \( y = a(x-h)^2 + k \), where \((h, k)\) is the vertex of the parabola. This format makes it easy to see the parabola's highest or lowest point and its direction. For the cannonball's path:

• The vertex is at \((800, 3200)\), indicating that the highest point of the parabola is 3200 ft above ground and 800 ft horizontally from the cannon.
• The parameter \(a\) determines the parabola's direction and width. Here, a negative \(a\) signifies an opening downwards.
• This downward opening represents the descent of the projectile after reaching its peak.

The vertex form simplifies finding the equation by clarifying these key features directly.

Symmetry

Symmetry is a defining characteristic of parabolic paths, especially in projectile motion. This concept means that the parabola is mirrored equally on both sides of the vertex. In terms of the cannonball, symmetry reveals:

• The paths taken to and from the peak are equidistant, allowing the landing point to be straightforwardly calculated.
• This symmetry helps us predict that the horizontal distance to the landing is 1600 ft, the midpoint being the vertex at 800 ft.
• This feature is pivotal in physics and engineering, where understanding symmetry aids in determining trajectories and points of impact.

This balanced nature of parabolas provides a reliable framework for analyzing motion paths.

Quadratic Equation

A quadratic equation takes the form \( ax^2 + bx + c = 0 \). However, when dealing with parabolas, especially in projectile motion, it's often easier to manipulate them using the vertex form. Quadratic equations characterize parabolas and encompass all possible solutions:

• The standard form is perfect for theoretical calculations and transformations, like deriving the path of a cannonball.
• In our cannonball scenario, transitioning from the standard form to the vertex form helped us pinpoint the vertex, maximizing calculation convenience.
• Solving for \(a\) using known points like the landing spot allowed us to solidify and confirm the projectile's trajectory equation.

Understanding quadratic equations in these contexts is essential for interpreting and predicting physical phenomena and their graphical representations.