Question
A soccer ball was kicked from the ground with an initial velocity of 56 feet per second and followed the path modeled by the function: h(t)=−16t2+56t
.(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 ball's path indicates that the ball took Response areaseconds to reach its maximum height of Response area feet. Looking at the graph of this function, we can see that the domain in context to this situation is 0 to 3.5 because the ball Response area.
.(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 ball's path indicates that the ball took Response areaseconds to reach its maximum height of Response area feet. Looking at the graph of this function, we can see that the domain in context to this situation is 0 to 3.5 because the ball Response area.
Answers
GPT-4o mini
To solve this problem, we start with the function given for the height of the ball:
\[ h(t) = -16t^2 + 56t \]
### Finding the Vertex
The vertex of a parabola described by the function \( h(t) = at^2 + bt + c \) can be found using the formula for the time \( t \) at which the maximum height occurs:
\[
t = -\frac{b}{2a}
\]
In this case, \( a = -16 \) and \( b = 56 \):
\[
t = -\frac{56}{2 \times -16} = \frac{56}{32} = 1.75 \, \text{seconds}
\]
Next, we find the maximum height by substituting \( t = 1.75 \) back into the function:
\[
h(1.75) = -16(1.75)^2 + 56(1.75)
\]
Calculating \( (1.75)^2 \):
\[
(1.75)^2 = 3.0625
\]
Now substituting back into the height function:
\[
h(1.75) = -16(3.0625) + 56(1.75)
\]
\[
= -49 + 98 = 49 \, \text{feet}
\]
### Domain of the Function
The domain in the context of this problem describes the interval of time during which the ball is in the air. The maximum height occurs at \( t = 1.75 \) seconds, and we need to find when the ball hits the ground again.
Setting \( h(t) = 0 \):
\[
-16t^2 + 56t = 0
\]
Factoring out \( t \):
\[
t(-16t + 56) = 0
\]
This gives us \( t = 0 \) (the time when the ball is kicked) and solving for the other part:
\[
-16t + 56 = 0
\]
\[
16t = 56
\]
\[
t = 3.5 \, \text{seconds}
\]
Thus, the ball is in the air from \( t = 0 \) seconds to \( t = 3.5 \) seconds.
### Final Responses
Fill in the required responses based on the calculations we've made:
1. **Response area seconds to reach its maximum height**: **1.75**
2. **Response area feet**: **49**
3. **Response area**: **hits the ground.**
The final responses are:
- The vertex of the ball's path indicates that the ball took **1.75** seconds to reach its maximum height of **49** feet. Looking at the graph of this function, we can see that the domain in context to this situation is 0 to 3.5 because the ball **hits the ground**.
\[ h(t) = -16t^2 + 56t \]
### Finding the Vertex
The vertex of a parabola described by the function \( h(t) = at^2 + bt + c \) can be found using the formula for the time \( t \) at which the maximum height occurs:
\[
t = -\frac{b}{2a}
\]
In this case, \( a = -16 \) and \( b = 56 \):
\[
t = -\frac{56}{2 \times -16} = \frac{56}{32} = 1.75 \, \text{seconds}
\]
Next, we find the maximum height by substituting \( t = 1.75 \) back into the function:
\[
h(1.75) = -16(1.75)^2 + 56(1.75)
\]
Calculating \( (1.75)^2 \):
\[
(1.75)^2 = 3.0625
\]
Now substituting back into the height function:
\[
h(1.75) = -16(3.0625) + 56(1.75)
\]
\[
= -49 + 98 = 49 \, \text{feet}
\]
### Domain of the Function
The domain in the context of this problem describes the interval of time during which the ball is in the air. The maximum height occurs at \( t = 1.75 \) seconds, and we need to find when the ball hits the ground again.
Setting \( h(t) = 0 \):
\[
-16t^2 + 56t = 0
\]
Factoring out \( t \):
\[
t(-16t + 56) = 0
\]
This gives us \( t = 0 \) (the time when the ball is kicked) and solving for the other part:
\[
-16t + 56 = 0
\]
\[
16t = 56
\]
\[
t = 3.5 \, \text{seconds}
\]
Thus, the ball is in the air from \( t = 0 \) seconds to \( t = 3.5 \) seconds.
### Final Responses
Fill in the required responses based on the calculations we've made:
1. **Response area seconds to reach its maximum height**: **1.75**
2. **Response area feet**: **49**
3. **Response area**: **hits the ground.**
The final responses are:
- The vertex of the ball's path indicates that the ball took **1.75** seconds to reach its maximum height of **49** feet. Looking at the graph of this function, we can see that the domain in context to this situation is 0 to 3.5 because the ball **hits the ground**.