Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![## Problem Statement: Height of a Launched Ball
The height (in feet) of a ball launched in the air *t* seconds after it is launched is given by the function:
\[ f(t) = -16t^2 + 280t + 22 \]
### Question
The height of the ball when its velocity is 0 ft/sec is ________ feet.
If necessary, round to two decimal places. If there is more than one correct answer, separate them with commas.
---
To solve this problem:
1. **Find the velocity function**: The velocity function is the first derivative of the height function \( f(t) \).
\[ v(t) = f'(t) \]
Starting from:
\[ f(t) = -16t^2 + 280t + 22 \]
We find the derivative \( f'(t) \):
\[ f'(t) = \frac{d}{dt} (-16t^2 + 280t + 22) = -32t + 280 \]
2. **Set the velocity to 0** and solve for \( t \):
\[ 0 = -32t + 280 \]
\[ 32t = 280 \]
\[ t = \frac{280}{32} = 8.75 \text{ seconds} \]
3. **Substitute \( t = 8.75 \) back into the original height function** to find the height at this time:
\[ f(8.75) = -16(8.75)^2 + 280(8.75) + 22 \]
Calculate:
\[ f(8.75) = -16(76.5625) + 2450 + 22 \]
\[ f(8.75) = -1225 + 2450 + 22 \]
\[ f(8.75) = 1247 \text { feet} \]
Thus, the height of the ball when its velocity is 0 ft/sec is **1247 feet**.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F12d36c83-f991-4c22-88cf-2adf2ca3b459%2Fda2947c8-d989-4b68-b37a-706cf35acfbb%2Ff49r1cl_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Problem Statement: Height of a Launched Ball
The height (in feet) of a ball launched in the air *t* seconds after it is launched is given by the function:
\[ f(t) = -16t^2 + 280t + 22 \]
### Question
The height of the ball when its velocity is 0 ft/sec is ________ feet.
If necessary, round to two decimal places. If there is more than one correct answer, separate them with commas.
---
To solve this problem:
1. **Find the velocity function**: The velocity function is the first derivative of the height function \( f(t) \).
\[ v(t) = f'(t) \]
Starting from:
\[ f(t) = -16t^2 + 280t + 22 \]
We find the derivative \( f'(t) \):
\[ f'(t) = \frac{d}{dt} (-16t^2 + 280t + 22) = -32t + 280 \]
2. **Set the velocity to 0** and solve for \( t \):
\[ 0 = -32t + 280 \]
\[ 32t = 280 \]
\[ t = \frac{280}{32} = 8.75 \text{ seconds} \]
3. **Substitute \( t = 8.75 \) back into the original height function** to find the height at this time:
\[ f(8.75) = -16(8.75)^2 + 280(8.75) + 22 \]
Calculate:
\[ f(8.75) = -16(76.5625) + 2450 + 22 \]
\[ f(8.75) = -1225 + 2450 + 22 \]
\[ f(8.75) = 1247 \text { feet} \]
Thus, the height of the ball when its velocity is 0 ft/sec is **1247 feet**.
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