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)=−16t2+64t .(3 points) Put responses in the correct input to answer the question. Select a response, navigate to the desired input and insert the response. Responses can be selected and inserted using the space bar, enter key, left mouse button or touchpad. Responses can also be moved by dragging with a mouse. 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. The option "48" (4 of 7) has been selected. Press tab to choose a response area, and spacebar to insert it. Press escape to cancel.
Answers
GPT-4o mini
To answer the question, we first need to analyze the function given for the ball's height, \( h(t) = -16t^2 + 64t \).
1. **Finding the vertex**: The vertex of a quadratic function in the form \( h(t) = at^2 + bt + c \) can be found using the formula \( t = -\frac{b}{2a} \).
- Here, \( a = -16 \) and \( b = 64 \).
- Therefore, \( t = -\frac{64}{2(-16)} = \frac{64}{32} = 2 \) seconds.
2. **Finding the maximum height**: To find the maximum height at \( t = 2 \), we substitute \( t = 2 \) back into the function:
\[
h(2) = -16(2)^2 + 64(2) = -16(4) + 128 = -64 + 128 = 64 \text{ feet}.
\]
Thus, the responses are:
- The ball took **2** seconds to reach its maximum height of **64** feet.
3. **Finding the domain**: The domain of the function in this context is from the time the ball is kicked until it hits the ground. We need to find when \( h(t) = 0 \):
\(
-16t^2 + 64t = 0
\)
\(
t(-16t + 64) = 0
\)
The solutions are \( t = 0 \) and \( t = 4 \). Therefore, the domain in context to this situation is **0 to 4** because the ball **hits the ground**.
So the final responses are:
- The vertex of the rock's path indicates the rock took **2** seconds to reach its maximum height of **64** 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 **hits the ground**.
1. **Finding the vertex**: The vertex of a quadratic function in the form \( h(t) = at^2 + bt + c \) can be found using the formula \( t = -\frac{b}{2a} \).
- Here, \( a = -16 \) and \( b = 64 \).
- Therefore, \( t = -\frac{64}{2(-16)} = \frac{64}{32} = 2 \) seconds.
2. **Finding the maximum height**: To find the maximum height at \( t = 2 \), we substitute \( t = 2 \) back into the function:
\[
h(2) = -16(2)^2 + 64(2) = -16(4) + 128 = -64 + 128 = 64 \text{ feet}.
\]
Thus, the responses are:
- The ball took **2** seconds to reach its maximum height of **64** feet.
3. **Finding the domain**: The domain of the function in this context is from the time the ball is kicked until it hits the ground. We need to find when \( h(t) = 0 \):
\(
-16t^2 + 64t = 0
\)
\(
t(-16t + 64) = 0
\)
The solutions are \( t = 0 \) and \( t = 4 \). Therefore, the domain in context to this situation is **0 to 4** because the ball **hits the ground**.
So the final responses are:
- The vertex of the rock's path indicates the rock took **2** seconds to reach its maximum height of **64** 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 **hits the ground**.