---### Basketball Projectile Motion Problem**Problem Statement:**Xavier shoots a basketball in which the height, in feet, is modeled by the equation:\[ h(t) = -4t^2 + 10t + 18 \]where $ t $ is time, in seconds. What is the maximum height of the basketball?---In this problem, we are given a quadratic equation that models the height of a basketball as a function of time. To determine the maximum height of the basketball, we need to find the vertex of the parabola represented by the given quadratic function. The general form of a quadratic equation is:\[ h(t) = at^2 + bt + c \]For our given equation:\[ h(t) = -4t^2 + 10t + 18 \]- $ a = -4 $- $ b = 10 $- $ c = 18 $The formula to find the time $ t $ at which the maximum height occurs for a parabola $ ax^2 + bx + c $ is:\[ t = \frac{-b}{2a} \]Substituting the given values:\[ t = \frac{-10}{2(-4)} = \frac{-10}{-8} = 1.25 \]Now, we find the height at $ t = 1.25 $:\[ h(1.25) = -4(1.25)^2 + 10(1.25) + 18 \]\[ h(1.25) = -4(1.25 \times 1.25) + 12.5 + 18 \]\[ h(1.25) = -4(1.5625) + 12.5 + 18 \]\[ h(1.25) = -6.25 + 12.5 + 18 \]\[ h(1.25) = 24.25 \]Therefore, the maximum height of the basketball is **24.25 feet**.---### Explanation of Method:To find the maximum height of the basketball using the quadratic equation, we:1. Identified the coefficients $ a $, $ b $, and $ c $ from the given equation.2. Used the vertex formula for the time $ t $ at which the maximum height occurs.3. Substituted this value of \(

Name: ---### Basketball Projectile Motion Problem**Problem Statement:**Xavi
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Description: ---### Basketball Projectile Motion Problem**Problem Statement:**Xavier shoots a basketball in which the height, in feet, is modeled by the equation:\[ h(t) = -4t^2 + 10t + 18 \]where $ t $ is time, in seconds. What is the maximum height of the ba

Answered: Image /qna-images/answer/382fe189-e678-4859-b0d1-6a6959fce9df.jpg

Answered: , h(t) = -4t² + 10t + 18, w %3D

Math

Algebra

, h(t) = -4t² + 10t + 18, w %3D

Big Ideas Math A Bridge To Success Algebra 1: Student Edition 2015

1st Edition

ISBN:9781680331141

Author:HOUGHTON MIFFLIN HARCOURT

Publisher:HOUGHTON MIFFLIN HARCOURT

Chapter5: Solving Systems Of Linear Equations

Section: Chapter Questions

Problem 4CA

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Question

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### Basketball Projectile Motion Problem

**Problem Statement:**

Xavier shoots a basketball in which the height, in feet, is modeled by the equation:

\[ h(t) = -4t^2 + 10t + 18 \]

where $ t $ is time, in seconds. What is the maximum height of the basketball?

---

In this problem, we are given a quadratic equation that models the height of a basketball as a function of time. To determine the maximum height of the basketball, we need to find the vertex of the parabola represented by the given quadratic function.

The general form of a quadratic equation is:

\[ h(t) = at^2 + bt + c \]

For our given equation:

\[ h(t) = -4t^2 + 10t + 18 \]

- $ a = -4 $
- $ b = 10 $
- $ c = 18 $

The formula to find the time $ t $ at which the maximum height occurs for a parabola $ ax^2 + bx + c $ is:

\[ t = \frac{-b}{2a} \]

Substituting the given values:

\[ t = \frac{-10}{2(-4)} = \frac{-10}{-8} = 1.25 \]

Now, we find the height at $ t = 1.25 $:

\[ h(1.25) = -4(1.25)^2 + 10(1.25) + 18 \]
\[ h(1.25) = -4(1.25 \times 1.25) + 12.5 + 18 \]
\[ h(1.25) = -4(1.5625) + 12.5 + 18 \]
\[ h(1.25) = -6.25 + 12.5 + 18 \]
\[ h(1.25) = 24.25 \]

Therefore, the maximum height of the basketball is **24.25 feet**.

---

### Explanation of Method:

To find the maximum height of the basketball using the quadratic equation, we:
1. Identified the coefficients $ a $, $ b $, and $ c $ from the given equation.
2. Used the vertex formula for the time $ t $ at which the maximum height occurs.
3. Substituted this value of \(

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