If a cannonball is shot directly upward with a velocity of 160 ft per second, its height above the ground after t seconds is given by s(t) = 160t - 16t. Find the velocity and the acceleration after t seconds. What is the maximum height the cannonball reaches? When does it hit the ground? The velocity after t seconds is v(t) = ft/sec. The acceleration after t seconds is a(t) = ft/sec?. %3D The cannonball reaches a maximum height of ft. choic The cannonball hits the ground after tD sec.

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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### Transcript of Educational Exercise

**Problem 8**  
a. Find the time at which the concentration is a maximum.  
b. Find the maximum concentration.

- a. The concentration will be maximized at \( x = 1 \) hour(s).  
- b. The maximum concentration is \( 2.5 \% \).

**Problem 9**  
If a cannonball is shot directly upward with a velocity of 160 ft per second, its height above the ground after \( t \) seconds is given by \( s(t) = 160t - 16t^2 \). Find the velocity and the acceleration after \( t \) seconds. What is the maximum height the cannonball reaches? When does it hit the ground?

- The velocity after \( t \) seconds is \( v(t) = \_\_\_\_\_\_ \) ft/sec.  
- The acceleration after \( t \) seconds is \( a(t) = \_\_\_\_\_\_ \) ft/sec\(^2\).  
- The cannonball reaches a maximum height of \_\_\_\_\_\_ ft.  
- The cannonball hits the ground after \( t = \_\_\_\_\_\_ \) sec. 

(Additional instructions or diagrams are not provided in the text provided.)
Transcribed Image Text:### Transcript of Educational Exercise **Problem 8** a. Find the time at which the concentration is a maximum. b. Find the maximum concentration. - a. The concentration will be maximized at \( x = 1 \) hour(s). - b. The maximum concentration is \( 2.5 \% \). **Problem 9** If a cannonball is shot directly upward with a velocity of 160 ft per second, its height above the ground after \( t \) seconds is given by \( s(t) = 160t - 16t^2 \). Find the velocity and the acceleration after \( t \) seconds. What is the maximum height the cannonball reaches? When does it hit the ground? - The velocity after \( t \) seconds is \( v(t) = \_\_\_\_\_\_ \) ft/sec. - The acceleration after \( t \) seconds is \( a(t) = \_\_\_\_\_\_ \) ft/sec\(^2\). - The cannonball reaches a maximum height of \_\_\_\_\_\_ ft. - The cannonball hits the ground after \( t = \_\_\_\_\_\_ \) sec. (Additional instructions or diagrams are not provided in the text provided.)
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