soccer ball is kicked upward from a height of 3 ft with an initial velocity of 64 ft/s. How high will it go? Use – 32 ft/s² for the acceleration caused by gra esistance. nswer Ke Maximum height: ft
soccer ball is kicked upward from a height of 3 ft with an initial velocity of 64 ft/s. How high will it go? Use – 32 ft/s² for the acceleration caused by gra esistance. nswer Ke Maximum height: ft
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:**
A soccer ball is kicked upward from a height of 3 ft with an initial velocity of 64 ft/s. How high will it go? Use –32 ft/s² for the acceleration caused by gravity. Ignore air resistance.
**Answer:**
- **Maximum height:** [Blank Box] ft
**Explanation:**
To find the maximum height, use the kinematic equation for vertical motion:
\[ h = h_0 + v_0t + \frac{1}{2}at^2 \]
Where:
- \( h_0 \) is the initial height (3 ft)
- \( v_0 \) is the initial velocity (64 ft/s)
- \( a \) is the acceleration due to gravity (–32 ft/s²)
- \( t \) is the time at maximum height
Set the final velocity to 0 at maximum height and solve for time \( t \). Then substitute \( t \) back into the equation to solve for \( h \).
This equation will help you determine how high the ball will go above the ground.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7a6f7f12-582e-4d04-b850-47821b485a9b%2F0076c5c8-1abf-4f0b-984b-9648d7b1fd63%2F3ap9gj4_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem:**
A soccer ball is kicked upward from a height of 3 ft with an initial velocity of 64 ft/s. How high will it go? Use –32 ft/s² for the acceleration caused by gravity. Ignore air resistance.
**Answer:**
- **Maximum height:** [Blank Box] ft
**Explanation:**
To find the maximum height, use the kinematic equation for vertical motion:
\[ h = h_0 + v_0t + \frac{1}{2}at^2 \]
Where:
- \( h_0 \) is the initial height (3 ft)
- \( v_0 \) is the initial velocity (64 ft/s)
- \( a \) is the acceleration due to gravity (–32 ft/s²)
- \( t \) is the time at maximum height
Set the final velocity to 0 at maximum height and solve for time \( t \). Then substitute \( t \) back into the equation to solve for \( h \).
This equation will help you determine how high the ball will go above the ground.
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