Problem 98 - College Algebra with Modeling and Visualization

Question

When a projectile is shot into the air, it attains a maximum height and then falls back to the ground. Suppose that \(x=0\) corresponds to the time when the projectile's height is maximum. If air resistance is ignored, its height \(h\) above the ground at any time \(x\) may be modeled by \(h(x)=-16 x^{2}+h_{\max }\) where \(h_{\max }\) is the projectile's maximum height above the ground. Height is measured in feet and time in seconds. Let \(h_{\max }=400\) feet. (a) Evaluate \(h(-2)\) and \(h(2) .\) Interpret these results. (b) Evaluate \(h(-5)\) and \(h(5) .\) Interpret these results. (c) Graph \(h\) for \(-5 \leq x \leq 5 .\) Is \(h\) even or odd? (d) How do \(h(x)\) and \(h(-x)\) compare when \(-5 \leq x \leq 5 ?\) What does this result indicate?

Step-by-Step Solution

Answer

(a) 336 feet; (b) 0 feet; (c) The graph is even; (d) \(h(x) = h(-x)\) indicates symmetry (even function).

Step 1: Understanding the Height Formula

The height of the projectile is given by the equation \(h(x) = -16x^2 + h_{\max}\). The given maximum height \( h_{\max} = 400 \) feet. So, the height equation simplifies to \( h(x) = -16x^2 + 400 \).

Step 2: Calculating Height at \( x = -2 \) and \( x = 2 \)

Let's substitute \( x = -2 \) and \( x = 2 \) into the height equation:\[ h(-2) = -16(-2)^2 + 400 = -16(4) + 400 = -64 + 400 = 336 \text{ feet} \]\[ h(2) = -16(2)^2 + 400 = -16(4) + 400 = -64 + 400 = 336 \text{ feet} \]Both \(h(-2)\) and \(h(2)\) equal 336 feet, which means the heights at 2 seconds before and after reaching the maximum height are the same.

Step 3: Calculating Height at \( x = -5 \) and \( x = 5 \)

Substitute \( x = -5 \) and \( x = 5 \) into the height equation to find the heights:\[ h(-5) = -16(-5)^2 + 400 = -16(25) + 400 = -400 + 400 = 0 \text{ feet} \]\[ h(5) = -16(5)^2 + 400 = -16(25) + 400 = -400 + 400 = 0 \text{ feet} \]Both \( h(-5) \) and \( h(5) \) are 0 feet, which indicates that the projectile hits the ground at 5 seconds before and after reaching the maximum height.

Step 4: Graphing the Height Equation

The equation for height \( h(x) = -16x^2 + 400 \) is a downward-opening parabola. To graph this for \(-5 \leq x \leq 5\):- The maximum point at \( x = 0 \) is 400 feet.- At \( x = -2 \) and \( x = 2 \), \( h = 336 \) feet.- At \( x = -5 \) and \( x = 5 \), \( h = 0 \) feet.The graph is symmetric about the y-axis, suggesting it is an even function.

Step 5: Determining Evenness or Oddness

A function is even if \( f(x) = f(-x) \). In this case, we observe that:- \( h(-2) = 336 \) and \( h(2) = 336 \)- \( h(-5) = 0 \) and \( h(5) = 0 \)This symmetry around the y-axis shows that \( h(x) \) is an even function.

Step 6: Comparing \(h(x)\) and \(h(-x)\)

For the given domain \(-5 \leq x \leq 5\), the function \(h\) satisfies \( h(x) = h(-x) \). This implies that for every point on the graph at \( x \), there is an identical point at \( -x \). This symmetry indicates that \( h \) is exactly even over the interval.

Key Concepts

Projectile Motion

When we talk about projectile motion, we are discussing how objects move through the air when they are launched. Think of throwing a ball in the air or shooting an arrow. Once these objects are sent on their way, they follow a specific path called a trajectory. This path is usually curved, looking like a hill or an arch. In math and physics, this trajectory can be described with a formula.

Some key points about projectile motion include:

• The motion occurs in two dimensions: horizontal (side to side) and vertical (up and down).
• The movement is affected by gravity, pulling the object downwards.
• Air resistance is often ignored to simplify calculations.

In our exercise, without considering air resistance, we use a mathematical formula to track the height of the projectile over time. This simplifies understanding how high or low an object gets after being launched.

Quadratic Function

Quadratic functions are a type of mathematical formula that can help us understand parabolic motion, like the path of our projectile. They have the standard form: \[ ax^2 + bx + c \]
To grasp projectile motion, specifically how high an object will go or how fast it might fall back to earth, one can use these functions. In our scenario:

• The formula used is quadratic: \( h(x) = -16x^2 + h_{\max} \).
• The term \(-16x^2\) represents the effect of gravity.
• The constant \( h_{\max} \) is the peak height the object reaches.

There are no linear terms in our equation, which means changes occur symmetrically around the peak height. This makes it easier to predict and calculate where the object will be at different points in time.

Symmetry in Mathematics

Symmetry in mathematics refers to how similar the two halves of something are. Imagine a picture that looks exactly the same on the left and right, that's symmetric. In math, functions can show symmetry, too.

For our exercise, we observe symmetry around the y-axis:

• The function \( h(x) = -16x^2 + 400 \) is even.
• An even function satisfies \( f(x) = f(-x) \), meaning that values of the function at positive and negative \(x\) around zero are the same.
• Symmetry ensures that the projectile's path up is a mirror image of its path back down.

This is why the object reaches the same height after 2 seconds as it did 2 seconds before reaching the maximum. Symmetry around the maximum height is crucial for calculating the time it takes for a projectile to reach the ground.

Maximum Height Calculation

The maximum height of a projectile is the highest point it reaches in its path. In a parabolic motion, this is at the very top of the curve. To find the maximum height, we are interested in when the rate of climb shifts to descent — a crucial turning point.

To determine the maximum height:

• Identify where the vertex of the parabolic function \( ax^2 + bx + c \) lies.
• In our simplified function \(-16x^2 + h_{\max}\), with no linear term, achieving the maximum is straightforward.
• At \( x = 0 \), the projectile is at its peak, achieving \( h_{\max} = 400 \) feet in our scenario.

Calculating when and what maximum height is crucial for predicting when a projectile will hit the ground again. This applies to sports, engineering, and any field where understanding the path of moving objects is vital.