Problem 62 - Algebra for College Students

Question

Solve each problem. If a soccer ball is kicked straight up with an initial velocity of 32 feet per second, then its height above the earth is a function of time given by \(s(t)=-16 t^{2}+32 t .\) Graph this function for \(0 \leq t \leq 2\). What is the maximum height reached by this ball?

Step-by-Step Solution

Answer

Maximum height is 16 feet. Occurs at t = 1 second.

Step 1: Understand the Problem

We are given the height function of a soccer ball kicked vertically: \[s(t) = -16t^{2} + 32t\]We need to graph this function for the interval \[0 \leq t \leq 2\]and determine the ball's maximum height.

Step 2: Find the Vertex of the Parabola

The function \[s(t) = -16t^{2} + 32t\]is a downward-opening parabola. The vertex represents the maximum height. The vertex of \[at^{2}+bt+c\]occur at \[t = -\frac{b}{2a}\].Here, \[a = -16\], \[b = 32\].So, \[t = -\frac{32}{2 \cdot (-16)} = 1\].

Step 3: Calculate the Maximum Height

Substitute \[t = 1\]back into the function: \[s(1) = -16(1)^{2} + 32(1) = -16 + 32 = 16\].Thus, the maximum height reached is 16 feet.

Step 4: Graph the Function

To graph \[s(t) = -16t^{2} + 32t\]on the interval \[0 \leq t \leq 2\],calculate the height at key points.- When \[t = 0\]: \[s(0) = -16(0)^{2} + 32(0) = 0\].- When \[t = 2\]: \[s(2) = -16(2)^{2} + 32(2) = -64 + 64 = 0\].Plot these points and the vertex point (1, 16) on the graph. The graph will be a parabola opening downwards, starting from (0, 0), reaching the peak at (1, 16), and returning to (2, 0).

Key Concepts

Quadratic Functions

Quadratic functions are essential in algebra, as they represent parabolas when graphed. A quadratic function is typically written in the form \[ ax^{2} + bx + c \]. The coefficients \(a\), \(b\), and \(c\) determine the shape and position of the parabola.
In this exercise, our quadratic function is \[ s(t) = -16t^{2} + 32t \].
Here:

• \(a = -16\)
• \(b = 32\)

Quadratic functions can open upwards or downwards.
If \(a > 0\), the parabola opens upwards, forming a U-shape.
If \(a < 0\), it opens downwards, forming an inverted U-shape.
In our case, since \(a = -16\), the parabola opens downwards. This downward-opening parabola indicates that the height of the soccer ball increases to a maximum point and then decreases.

Vertex of a Parabola

The vertex of a parabola in the quadratic function \[ ax^{2} + bx + c \] is crucial.
It represents the highest or lowest point of the parabola.
For a quadratic function \[ at^{2} + bt + c \], the vertex's x-coordinate (or in this case, t-coordinate) is found using the formula: \[ t = -\frac{b}{2a}\].
Plugging in the values from our exercise:
- \(a = -16\)
- \(b = 32\)
We get: \[ t = -\frac{32}{2 \cdot (-16)} = 1\].
This means that the vertex occurs at \( t = 1 \) second.
To find the height at this vertex, substitute \( t \) back into the original function: \[ s(1) = -16(1)^{2} + 32(1) = -16 + 32 = 16 \].
Therefore, the maximum height reached by the soccer ball is 16 feet.

Graphing Functions

Graphing a quadratic function helps visualize its behavior.
To graph \[ s(t) = -16t^{2} + 32t \], consider the interval \[ 0 \leq t \leq 2 \].
Begin by finding key points:

• When \(t = 0\): \[ s(0) = -16(0)^{2} + 32(0) = 0 \].
• When \(t = 2\): \[ s(2) = -16(2)^{2} + 32(2) = -64 + 64 = 0 \].
• The vertex at \( t = 1 \): \[ s(1) = 16 \].

Now plot these points:

• \( (0, 0) \)
• \( (1, 16) \)
• \( (2, 0) \)

Connecting these points will give you a downward-opening parabola.
Starting at \( (0, 0) \), reaching the maximum at \( (1, 16) \), and returning to \( (2, 0) \). The graph shows how the ball rises, peaks, and then falls back down.

Maximum Height Calculation

In parabolic motion, calculating the maximum height is often needed.
For a quadratic function \[ s(t) = at^{2} + bt + c \] that opens downwards, the vertex gives the maximum height.
From the previous sections, we found the time at which the vertex occurs:

• \( t= -\frac{b}{2a} = 1\) second

Plug this time into the function to find the height: \[ s(1) = -16(1)^{2} + 32(1) = 16 \].
Therefore, the maximum height the soccer ball reaches is 16 feet.
Ensure to always identify the vertex when dealing with quadratic motion problems, as it simplifies finding the maximum or minimum values.
This helps in understanding the physical behavior of objects in projectile motion.