Problem 42 - Precalculus Mathematics for Calculus

Question

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 cannonball lands \(1600 \mathrm{ft}\) from the cannon and the highest point it reaches is \(3200 \mathrm{ft}\) above the ground, find an equation for the path of the cannonball. Place the origin at the location of the cannon. CAN'T COPY THE GRAPH

Step-by-Step Solution

Answer

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

Step 1: Identify the Nature of the Problem

The problem involves finding the parabolic equation of a path followed by a projectile, such as a cannonball. We know the parabola's vertex and where it lands, which helps us set up the equation.

Step 2: Define the Standard Equation of a Parabola

The path of the projectile is a parabola, which generally can be defined as \(y = ax^2 + bx + c\). Since the parabola is vertical and symmetric with respect to the y-axis, it can be simplified to \(y = ax^2\) when the vertex is at the origin (0, 0).

Step 3: Position the Vertex at the Highest Point

The highest point the cannonball reaches is on the y-axis at (0, 3200). Because the parabola is symmetric about the y-axis, the vertex form of the equation is \(y = a(x - h)^2 + k\) with \(h = 0\) and \(k = 3200\), giving us \(y = ax^2 + 3200\).

Step 4: Use the Landing Point to Find 'a'

The cannonball lands 1600 ft from the cannon, meaning it lands at (1600, 0). Substitute these values into the equation \(0 = a(1600)^2 + 3200\) to solve for \(a\).

Step 5: Solve for 'a'

Rearrange the equation from Step 4: \(-3200 = a(1600)^2\). Simplify and solve for \(a\): \(a = \frac{-3200}{1600^2} = \frac{-3200}{2560000} = -\frac{1}{800}\).

Step 6: Write the Final Equation of the Parabola

Substitute \(a = -\frac{1}{800}\) back into the equation from Step 3: \(y = -\frac{1}{800}x^2 + 3200\). This is the equation of the path of the cannonball.

Key Concepts

Projectile Motion

When we talk about projectile motion, we're referring to objects that are thrown or propelled into the air and move along a curved path under the influence of gravity. The most common examples are things like cannonballs, arrows, or basketballs, all moving through an arc before landing. These paths are often parabolic in shape, which means they can be described using quadratic functions. Understanding projectile motion helps us determine key points such as where the object will land and what its maximum height will be.

The path of the projectile, such as the cannonball in our problem, is assumed to be under constant gravitational acceleration. This means the only force acting on the cannonball, once it is fired, is gravity. This results in a predictable parabolic motion that we can calculate using mathematical equations.

The key aspects of projectile motion include:

• Initial velocity and angle of projection
• Maximal height reached
• Total horizontal distance covered, known as the range
• The symmetry of the path due to gravity

All these factors help us to establish the precise path, like the one seen in our cannonball example, based on the initial launch conditions.

Vertex Form

The vertex form of a quadratic equation is quite useful when dealing with parabolas, especially if you know the highest or lowest point — the vertex. In this form, the equation is written as \[ y = a(x - h)^2 + k \]where

• \( (h, k) \) is the vertex of the parabola
• \( a \) determines the direction of the parabola (upward or downward) and how "wide" or "narrow" it is.

This form highlights the vertex directly, making it easier to understand and manipulate the parabola in equations. For instance, in the cannonball’s path, we place the vertex at \( h = 0 \) and \( k = 3200 \), indicating the parabola reaches its maximum height at this point. Utilizing the vertex form is an effective way to quickly identify these peaks in the projectile's path.

Quadratic Functions

A quadratic function is what defines nearly all parabola shapes in mathematics and physics. In its simplest form, it is expressed as \( y = ax^2 + bx + c \). This function graphs into a parabola, with many real-world applications, such as modeling trajectories, like our cannonball example.

The coefficient \( a \) tells us whether the parabola opens upwards (\( a > 0 \)) or downwards (\( a < 0 \)). The values of \( b \) and \( c \) help to place the parabola at the correct location on a coordinate axis. In our exercise, we simplify it to \( y = ax^2 \) because the parabola is vertically aligned and symmetric around the y-axis, allowing us to focus solely on the height of the vertex and the landing point to solve for \( a \).

As you determine the path for the cannonball, you'll see how the quadratic function molds the parabolic shape of the flight path.

Symmetry in Parabolas

Symmetry plays a fundamental role in understanding parabolas. A parabola is symmetrical about its axis, which means if you fold the graph along its vertical line of symmetry (which runs through the vertex), both sides will match perfectly. This property simplifies solving and understanding many problems, such as in projectile motion scenarios.

For the cannonball example, symmetry tells us that the projectile’s path to its maximum height will mirror its descent, creating a symmetric arc. The vertex, where the cannonball reaches its maximum height, acts as the midpoint of the distance it travels horizontally. This feature of symmetry allows us to use predictable equations to calculate where the cannonball lands based on its initial launch position and speed.

Recognizing symmetry can help quickly estimate where certain key events in the projectile motion, such as the maximum height and range, will occur without needing complicated calculations. By embracing symmetry, it aligns easily with the quadratic functions to solve real-world problems efficiently.