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+64th(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. 4842hits the ground after 4 seconds.0reaches a maximum height at 4 feet.64
Answers
GPT-4o mini
To analyze the quadratic function \( h(t) = -16t^2 + 64t \), we need to determine several key components:
1. **Finding the vertex**: The vertex of a quadratic equation in the format \( h(t) = at^2 + bt + c \) can be found using the formula \( t = -\frac{b}{2a} \). Here, \( a = -16 \) and \( b = 64 \).
\[
t = -\frac{64}{2 \times -16} = \frac{64}{32} = 2 \text{ seconds}
\]
2. **Finding the maximum height**: To find the height at \( t = 2 \) seconds, substitute \( t = 2 \) back into the function.
\[
h(2) = -16(2^2) + 64(2) = -16(4) + 128 = -64 + 128 = 64 \text{ feet}
\]
3. **Finding when the ball hits the ground**: To determine when the ball hits the ground, we set \( h(t) = 0 \):
\[
-16t^2 + 64t = 0
\]
Factoring out \( 16t \):
\[
16t(-t + 4) = 0
\]
This gives solutions:
\[
t = 0 \text{ seconds (launch)} \quad \text{or} \quad t = 4 \text{ seconds (hits ground)}
\]
Based on these calculations, we can fill in the responses:
- 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 after 4 seconds**.
To summarize:
- **First response** (time to reach max height): **2**
- **Second response** (max height): **64**
- **Third response** (what happens at 4 seconds): **hits the ground after 4 seconds**.
1. **Finding the vertex**: The vertex of a quadratic equation in the format \( h(t) = at^2 + bt + c \) can be found using the formula \( t = -\frac{b}{2a} \). Here, \( a = -16 \) and \( b = 64 \).
\[
t = -\frac{64}{2 \times -16} = \frac{64}{32} = 2 \text{ seconds}
\]
2. **Finding the maximum height**: To find the height at \( t = 2 \) seconds, substitute \( t = 2 \) back into the function.
\[
h(2) = -16(2^2) + 64(2) = -16(4) + 128 = -64 + 128 = 64 \text{ feet}
\]
3. **Finding when the ball hits the ground**: To determine when the ball hits the ground, we set \( h(t) = 0 \):
\[
-16t^2 + 64t = 0
\]
Factoring out \( 16t \):
\[
16t(-t + 4) = 0
\]
This gives solutions:
\[
t = 0 \text{ seconds (launch)} \quad \text{or} \quad t = 4 \text{ seconds (hits ground)}
\]
Based on these calculations, we can fill in the responses:
- 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 after 4 seconds**.
To summarize:
- **First response** (time to reach max height): **2**
- **Second response** (max height): **64**
- **Third response** (what happens at 4 seconds): **hits the ground after 4 seconds**.