Question
A ball was kicked from the ground with a velocity of 64 feet per second and followed the path modeled by the function: h(t) = −16t^2 + 64t
.(3 points)
The vertex of the rock's path indicates the rock took Response area seconds to reach its maximum height of Response area feet. Looking at the graph of this function, we can see the domain in context to this situation is 0 to 4 because the ball Response area.
.(3 points)
The vertex of the rock's path indicates the rock took Response area seconds to reach its maximum height of Response area feet. Looking at the graph of this function, we can see the domain in context to this situation is 0 to 4 because the ball Response area.
Answers
GPT-4o mini
To analyze the function \(h(t) = -16t^2 + 64t\), we can identify the vertex, which represents the maximum height of the ball.
### Finding the Vertex:
The formula to find the time \(t\) at which the maximum height occurs (the vertex) for a quadratic function in the form \(h(t) = at^2 + bt + c\) is given by:
\[
t = -\frac{b}{2a}
\]
In our function, \(a = -16\) and \(b = 64\).
Substituting in these values:
\[
t = -\frac{64}{2(-16)} = -\frac{64}{-32} = 2 \text{ seconds}
\]
### Finding the Maximum Height:
Now, we can find the maximum height by substituting \(t = 2\) back into the function \(h(t)\):
\[
h(2) = -16(2^2) + 64(2) = -16(4) + 128 = -64 + 128 = 64 \text{ feet}
\]
### Conclusion:
1. The vertex of the ball's path indicates the ball took **2 seconds** to reach its maximum height of **64 feet**.
### Domain Context:
The domain of the context is determined by the time the ball is in the air. Since we know the maximum height occurs at 2 seconds, we should also check when the ball hits the ground again. This occurs when \(h(t) = 0\).
Setting \(h(t) = 0\):
\[
-16t^2 + 64t = 0
\]
Factoring out \(16t\):
\[
16t(-t + 4) = 0
\]
This gives us two solutions:
1. \(t = 0\) (when it was kicked)
2. \(t = 4\) (when it hits the ground)
Thus, the domain of this function in the context is from 0 to 4 seconds because the ball is in the air during this interval.
**Final Responses:**
- The ball took **2 seconds** to reach its maximum height of **64 feet**.
- The ball is in the air from **0 to 4 seconds**.
### Finding the Vertex:
The formula to find the time \(t\) at which the maximum height occurs (the vertex) for a quadratic function in the form \(h(t) = at^2 + bt + c\) is given by:
\[
t = -\frac{b}{2a}
\]
In our function, \(a = -16\) and \(b = 64\).
Substituting in these values:
\[
t = -\frac{64}{2(-16)} = -\frac{64}{-32} = 2 \text{ seconds}
\]
### Finding the Maximum Height:
Now, we can find the maximum height by substituting \(t = 2\) back into the function \(h(t)\):
\[
h(2) = -16(2^2) + 64(2) = -16(4) + 128 = -64 + 128 = 64 \text{ feet}
\]
### Conclusion:
1. The vertex of the ball's path indicates the ball took **2 seconds** to reach its maximum height of **64 feet**.
### Domain Context:
The domain of the context is determined by the time the ball is in the air. Since we know the maximum height occurs at 2 seconds, we should also check when the ball hits the ground again. This occurs when \(h(t) = 0\).
Setting \(h(t) = 0\):
\[
-16t^2 + 64t = 0
\]
Factoring out \(16t\):
\[
16t(-t + 4) = 0
\]
This gives us two solutions:
1. \(t = 0\) (when it was kicked)
2. \(t = 4\) (when it hits the ground)
Thus, the domain of this function in the context is from 0 to 4 seconds because the ball is in the air during this interval.
**Final Responses:**
- The ball took **2 seconds** to reach its maximum height of **64 feet**.
- The ball is in the air from **0 to 4 seconds**.