Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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![### Projectile Motion of a Thrown Ball
The height \( y \) (in feet) of a ball thrown by a child is given by the equation:
\[ y = -\frac{1}{12} x^2 + 6x + 3 \]
where \( x \) is the horizontal distance in feet from the point at which the ball is thrown.
#### Problem Statement:
(a) **Initial Height:**
- **Question:** How high is the ball when it leaves the child's hand?
- **Answer format:** \[\_\_\_\_\_\_\_\_\_\] feet
(b) **Maximum Height:**
- **Question:** What is the maximum height of the ball?
- **Answer format:** \[\_\_\_\_\_\_\_\_\_\] feet
(c) **Distance to Ground Impact:**
- **Question:** How far from the child does the ball strike the ground? Round your answers to the nearest 0.01.
- **Answer format:** \[\_\_\_\_\_\_\_\_\_\] feet
### Explanation of Equations and Method
1. **Determining the Initial Height:**
- Substitute \( x = 0 \) into the equation \( y = -\frac{1}{12} x^2 + 6x + 3 \).
- Calculation:
\[
y = -\frac{1}{12}(0)^2 + 6(0) + 3 = 3
\]
- Thus, the initial height when \( x = 0 \) is 3 feet.
2. **Determining the Maximum Height:**
- This can be found by determining the vertex of the parabolic equation \( y = -\frac{1}{12} x^2 + 6x + 3 \).
- The formula for the vertex \( x \) value in a quadratic equation \( ax^2 + bx + c \) is \( x = -\frac{b}{2a} \).
- For the given equation, \( a = -\frac{1}{12} \) and \( b = 6 \):
\[
x = -\frac{6}{2 \times -\frac{1}{12}} = 36
\]
- Substitute \( x = 36 \) back into the equation to find \( y \):
\[
y](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4f8a397f-4115-4363-8e32-37a2c2950f08%2F79cdac87-e17e-4291-a1da-1aa8cca51446%2Fvn55gcg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Projectile Motion of a Thrown Ball
The height \( y \) (in feet) of a ball thrown by a child is given by the equation:
\[ y = -\frac{1}{12} x^2 + 6x + 3 \]
where \( x \) is the horizontal distance in feet from the point at which the ball is thrown.
#### Problem Statement:
(a) **Initial Height:**
- **Question:** How high is the ball when it leaves the child's hand?
- **Answer format:** \[\_\_\_\_\_\_\_\_\_\] feet
(b) **Maximum Height:**
- **Question:** What is the maximum height of the ball?
- **Answer format:** \[\_\_\_\_\_\_\_\_\_\] feet
(c) **Distance to Ground Impact:**
- **Question:** How far from the child does the ball strike the ground? Round your answers to the nearest 0.01.
- **Answer format:** \[\_\_\_\_\_\_\_\_\_\] feet
### Explanation of Equations and Method
1. **Determining the Initial Height:**
- Substitute \( x = 0 \) into the equation \( y = -\frac{1}{12} x^2 + 6x + 3 \).
- Calculation:
\[
y = -\frac{1}{12}(0)^2 + 6(0) + 3 = 3
\]
- Thus, the initial height when \( x = 0 \) is 3 feet.
2. **Determining the Maximum Height:**
- This can be found by determining the vertex of the parabolic equation \( y = -\frac{1}{12} x^2 + 6x + 3 \).
- The formula for the vertex \( x \) value in a quadratic equation \( ax^2 + bx + c \) is \( x = -\frac{b}{2a} \).
- For the given equation, \( a = -\frac{1}{12} \) and \( b = 6 \):
\[
x = -\frac{6}{2 \times -\frac{1}{12}} = 36
\]
- Substitute \( x = 36 \) back into the equation to find \( y \):
\[
y
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