We are standing on the top of a 704 feet tall building and launch a small object upward. The object's vertical position, measured in feet, after t seconds is h(t) = − 16t² + 320t +704. What is the highest point that the object reaches? feet
We are standing on the top of a 704 feet tall building and launch a small object upward. The object's vertical position, measured in feet, after t seconds is h(t) = − 16t² + 320t +704. What is the highest point that the object reaches? feet
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![### Problem Statement
We are standing on the top of a 704 feet tall building and launch a small object upward. The object's vertical position, measured in feet, after \( t \) seconds is \( h(t) = -16t^2 + 320t + 704 \). What is the highest point that the object reaches?
\[ \text{feet} \]
### Explanation
This question involves finding the vertex of a quadratic equation. The given equation \( h(t) = -16t^2 + 320t + 704 \) describes the height of the object as a function of time, \( t \).
To find the highest point, we need to determine the vertex of the parabola represented by the quadratic function. The formula to find the time at which the maximum height occurs (vertex of the parabola) is given by:
\[ t = -\frac{b}{2a} \]
where \( a \) and \( b \) are the coefficients from the quadratic equation \( at^2 + bt + c \).
For the given equation \( h(t) = -16t^2 + 320t + 704 \):
- \( a = -16 \)
- \( b = 320 \)
Plugging these values into the vertex formula:
\[ t = -\frac{320}{2(-16)} = \frac{320}{32} = 10 \]
Thus, the object reaches its maximum height at \( t = 10 \) seconds.
To find the maximum height, substitute \( t = 10 \) back into the original equation:
\[ h(10) = -16(10)^2 + 320(10) + 704 \]
\[ h(10) = -16(100) + 3200 + 704 \]
\[ h(10) = -1600 + 3200 + 704 \]
\[ h(10) = 1600 + 704 \]
\[ h(10) = 2304 \]
Therefore, the highest point that the object reaches is **2304 feet**.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8b3342fe-b335-4689-a0aa-7d956754a511%2F3d738220-352d-4ecb-b37c-6cf727f10111%2F3uhbhid_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
We are standing on the top of a 704 feet tall building and launch a small object upward. The object's vertical position, measured in feet, after \( t \) seconds is \( h(t) = -16t^2 + 320t + 704 \). What is the highest point that the object reaches?
\[ \text{feet} \]
### Explanation
This question involves finding the vertex of a quadratic equation. The given equation \( h(t) = -16t^2 + 320t + 704 \) describes the height of the object as a function of time, \( t \).
To find the highest point, we need to determine the vertex of the parabola represented by the quadratic function. The formula to find the time at which the maximum height occurs (vertex of the parabola) is given by:
\[ t = -\frac{b}{2a} \]
where \( a \) and \( b \) are the coefficients from the quadratic equation \( at^2 + bt + c \).
For the given equation \( h(t) = -16t^2 + 320t + 704 \):
- \( a = -16 \)
- \( b = 320 \)
Plugging these values into the vertex formula:
\[ t = -\frac{320}{2(-16)} = \frac{320}{32} = 10 \]
Thus, the object reaches its maximum height at \( t = 10 \) seconds.
To find the maximum height, substitute \( t = 10 \) back into the original equation:
\[ h(10) = -16(10)^2 + 320(10) + 704 \]
\[ h(10) = -16(100) + 3200 + 704 \]
\[ h(10) = -1600 + 3200 + 704 \]
\[ h(10) = 1600 + 704 \]
\[ h(10) = 2304 \]
Therefore, the highest point that the object reaches is **2304 feet**.
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