A rocket is fired upward from some initial distance above the ground. Its height (in feet), h, above the ground t seconds after it is fired is given by h(t) 16t² + 160t + 896. What is the rocket's maximum height? 12 feet = How long does it take for the rocket to reach its maximum height? 5 seconds After it is fired, the rocket reaches the ground at t= seconds
A rocket is fired upward from some initial distance above the ground. Its height (in feet), h, above the ground t seconds after it is fired is given by h(t) 16t² + 160t + 896. What is the rocket's maximum height? 12 feet = How long does it take for the rocket to reach its maximum height? 5 seconds After it is fired, the rocket reaches the ground at t= seconds
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
how do I find the seconds it takes to fall to the ground? thank you!
![### Rocket Motion Analysis
A rocket is launched vertically upward from an initial height above the ground. The height \( h \) (in feet) of the rocket above the ground \( t \) seconds after launch is modeled by the quadratic equation:
\[ h(t) = -16t^2 + 160t + 896. \]
This equation provides crucial information about the rocket’s trajectory over time.
#### 1. Determining the Rocket's Maximum Height
To find the maximum height the rocket reaches, we need to identify the vertex of the parabolic equation \( h(t) \). The maximum or minimum value of a quadratic function \( ax^2 + bx + c \) occurs at \( t = -\frac{b}{2a} \).
For the given equation:
\[ a = -16, \, b = 160 \]
The time \( t \) at which the rocket reaches its maximum height is:
\[ t = -\frac{b}{2a} = -\frac{160}{2(-16)} = 5 \text{ seconds} \]
Now, substituting \( t = 5 \) back into the height function to find the maximum height:
\[ h(5) = -16(5)^2 + 160(5) + 896 = -16(25) + 800 + 896 \]
\[ h(5) = -400 + 800 + 896 = 1296 \text{ feet} \]
Therefore, the **rocket's maximum height** is **1296 feet**.
#### 2. Time to Reach Maximum Height
As calculated above, the rocket reaches its maximum height at:
\[ t = 5 \text{ seconds} \]
#### 3. Time to Hit the Ground
To determine when the rocket returns to the ground, we set the height equation to zero and solve for \( t \):
\[ 0 = -16t^2 + 160t + 896 \]
This can be solved using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ a = -16, \, b = 160, \, c = 896 \]
\[ t = \frac{-160 \pm \sqrt{160^2 - 4(-16)(896)}}{2(-16)} \]
\[ t = \frac{-160 \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F90d1e96e-989a-4849-aeb9-585c1e797a4d%2Fcfed9cf9-1455-4bf3-a067-74c3f015d7d1%2Fj7ujf38_processed.png&w=3840&q=75)
Transcribed Image Text:### Rocket Motion Analysis
A rocket is launched vertically upward from an initial height above the ground. The height \( h \) (in feet) of the rocket above the ground \( t \) seconds after launch is modeled by the quadratic equation:
\[ h(t) = -16t^2 + 160t + 896. \]
This equation provides crucial information about the rocket’s trajectory over time.
#### 1. Determining the Rocket's Maximum Height
To find the maximum height the rocket reaches, we need to identify the vertex of the parabolic equation \( h(t) \). The maximum or minimum value of a quadratic function \( ax^2 + bx + c \) occurs at \( t = -\frac{b}{2a} \).
For the given equation:
\[ a = -16, \, b = 160 \]
The time \( t \) at which the rocket reaches its maximum height is:
\[ t = -\frac{b}{2a} = -\frac{160}{2(-16)} = 5 \text{ seconds} \]
Now, substituting \( t = 5 \) back into the height function to find the maximum height:
\[ h(5) = -16(5)^2 + 160(5) + 896 = -16(25) + 800 + 896 \]
\[ h(5) = -400 + 800 + 896 = 1296 \text{ feet} \]
Therefore, the **rocket's maximum height** is **1296 feet**.
#### 2. Time to Reach Maximum Height
As calculated above, the rocket reaches its maximum height at:
\[ t = 5 \text{ seconds} \]
#### 3. Time to Hit the Ground
To determine when the rocket returns to the ground, we set the height equation to zero and solve for \( t \):
\[ 0 = -16t^2 + 160t + 896 \]
This can be solved using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ a = -16, \, b = 160, \, c = 896 \]
\[ t = \frac{-160 \pm \sqrt{160^2 - 4(-16)(896)}}{2(-16)} \]
\[ t = \frac{-160 \
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 2 steps with 2 images
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
