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
The function modeling the height \( h(t) \) of the ball kicked is given by:
\[
h(t) = -16t^2 + 64t
\]
To find the vertex of this quadratic function, which indicates the maximum height of the ball, we can use the vertex formula for a parabola in the standard form \( ax^2 + bx + c \). The time \( t \) at which the vertex occurs is given by:
\[
t = -\frac{b}{2a}
\]
Here, \( a = -16 \) and \( b = 64 \). Plugging these values into the formula gives:
\[
t = -\frac{64}{2(-16)} = \frac{64}{32} = 2 \text{ seconds}
\]
Next, we need to find the maximum height of the ball by substituting \( t = 2 \) back into the height equation:
\[
h(2) = -16(2)^2 + 64(2) = -16(4) + 128 = -64 + 128 = 64 \text{ feet}
\]
Thus, the vertex indicates that the ball took **2 seconds** to reach its maximum height of **64 feet**.
Now, regarding the domain, the ball is kicked from the ground and returns to the ground when it reaches a height of 0. To find the time when the ball hits the ground, we solve for \( t \) when \( h(t) = 0 \):
\[
0 = -16t^2 + 64t
\]
Factoring out \( -16t \):
\[
-16t(t - 4) = 0
\]
This gives us two solutions:
\[
t = 0 \quad \text{and} \quad t = 4
\]
This means the ball is in the air from **0 seconds** to **4 seconds**. Therefore, the domain of this function, in the context of this problem, is indeed from **0 to 4 seconds**.
To summarize:
- The maximum height is reached at **2 seconds**.
- The maximum height of the ball is **64 feet**.
- The domain of the function (the time the ball is in the air) is from **0 to 4 seconds**.
\[
h(t) = -16t^2 + 64t
\]
To find the vertex of this quadratic function, which indicates the maximum height of the ball, we can use the vertex formula for a parabola in the standard form \( ax^2 + bx + c \). The time \( t \) at which the vertex occurs is given by:
\[
t = -\frac{b}{2a}
\]
Here, \( a = -16 \) and \( b = 64 \). Plugging these values into the formula gives:
\[
t = -\frac{64}{2(-16)} = \frac{64}{32} = 2 \text{ seconds}
\]
Next, we need to find the maximum height of the ball by substituting \( t = 2 \) back into the height equation:
\[
h(2) = -16(2)^2 + 64(2) = -16(4) + 128 = -64 + 128 = 64 \text{ feet}
\]
Thus, the vertex indicates that the ball took **2 seconds** to reach its maximum height of **64 feet**.
Now, regarding the domain, the ball is kicked from the ground and returns to the ground when it reaches a height of 0. To find the time when the ball hits the ground, we solve for \( t \) when \( h(t) = 0 \):
\[
0 = -16t^2 + 64t
\]
Factoring out \( -16t \):
\[
-16t(t - 4) = 0
\]
This gives us two solutions:
\[
t = 0 \quad \text{and} \quad t = 4
\]
This means the ball is in the air from **0 seconds** to **4 seconds**. Therefore, the domain of this function, in the context of this problem, is indeed from **0 to 4 seconds**.
To summarize:
- The maximum height is reached at **2 seconds**.
- The maximum height of the ball is **64 feet**.
- The domain of the function (the time the ball is in the air) is from **0 to 4 seconds**.