2 = 2(0)+7(0) + ²₂ D 6. Use a(t)=-32 feet per second squared as the acceleration due to gravity. A ball is thrown vertically upward from the ground with an initial velocity of 96 feet per second. How high will the ball go (max height)? a) 32 ft b) 64 ft c)24 ft 144 ft e) None of these Dal+C -324
2 = 2(0)+7(0) + ²₂ D 6. Use a(t)=-32 feet per second squared as the acceleration due to gravity. A ball is thrown vertically upward from the ground with an initial velocity of 96 feet per second. How high will the ball go (max height)? a) 32 ft b) 64 ft c)24 ft 144 ft e) None of these Dal+C -324
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section7.6: The Inverse Trigonometric Functions
Problem 94E
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Q6: what is the correct answer for this question, please show work clearly
![**Problem Statement:**
Consider the following physics problem:
Use \( a(t) = -32 \) feet per second squared as the acceleration due to gravity. A ball is thrown vertically upward from the ground with an initial velocity of 96 feet per second. How high will the ball go (max height)?
**Options:**
a) 32 ft
b) 64 ft
c) 24 ft
d) 144 ft
e) None of these
In the given problem, the correct answer (option d) is circled, indicating that the ball will reach a maximum height of 144 feet.
**Solution Consideration:**
1. The initial velocity \( v_0 \) is given as 96 feet per second.
2. The acceleration \( a(t) \) is the constant due to gravity, \( -32 \) feet per second squared.
3. To find the maximum height, we need to determine when the velocity \( v(t) \) becomes zero (i.e., at the peak of the ball's trajectory).
**Solution Steps:**
1. Integrate the acceleration function to find the velocity function:
\[ v(t) = \int a(t) \, dt = \int -32 \, dt = -32t + v_0 \]
where \( v_0 = 96 \).
So, \( v(t) = -32t + 96 \).
2. Set \( v(t) = 0 \) to find the time at which the maximum height is reached:
\[ 0 = -32t + 96 \]
\[ 32t = 96 \]
\[ t = 3 \text{ seconds} \]
3. Now, integrate the velocity function to find the height function \( s(t) \):
\[ s(t) = \int v(t) \, dt = \int (-32t + 96) \, dt = -16t^2 + 96t + s_0 \]
where \( s_0 = 0 \) (the initial height is 0).
So, \( s(t) = -16t^2 + 96t \).
4. Substitute \( t = 3 \) seconds to find the maximum height:
\[ s(3) = -16(3)^2 + 96(3) \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb1751e3a-a174-44cb-807a-026103c1f5d6%2F8a67d93e-5aa0-4c77-91b3-f56d12b518d2%2Fst0oy3p_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Consider the following physics problem:
Use \( a(t) = -32 \) feet per second squared as the acceleration due to gravity. A ball is thrown vertically upward from the ground with an initial velocity of 96 feet per second. How high will the ball go (max height)?
**Options:**
a) 32 ft
b) 64 ft
c) 24 ft
d) 144 ft
e) None of these
In the given problem, the correct answer (option d) is circled, indicating that the ball will reach a maximum height of 144 feet.
**Solution Consideration:**
1. The initial velocity \( v_0 \) is given as 96 feet per second.
2. The acceleration \( a(t) \) is the constant due to gravity, \( -32 \) feet per second squared.
3. To find the maximum height, we need to determine when the velocity \( v(t) \) becomes zero (i.e., at the peak of the ball's trajectory).
**Solution Steps:**
1. Integrate the acceleration function to find the velocity function:
\[ v(t) = \int a(t) \, dt = \int -32 \, dt = -32t + v_0 \]
where \( v_0 = 96 \).
So, \( v(t) = -32t + 96 \).
2. Set \( v(t) = 0 \) to find the time at which the maximum height is reached:
\[ 0 = -32t + 96 \]
\[ 32t = 96 \]
\[ t = 3 \text{ seconds} \]
3. Now, integrate the velocity function to find the height function \( s(t) \):
\[ s(t) = \int v(t) \, dt = \int (-32t + 96) \, dt = -16t^2 + 96t + s_0 \]
where \( s_0 = 0 \) (the initial height is 0).
So, \( s(t) = -16t^2 + 96t \).
4. Substitute \( t = 3 \) seconds to find the maximum height:
\[ s(3) = -16(3)^2 + 96(3) \]
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