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Kayden is throwing a shot put at a track meet. He is 4.1 feet tall. The shot put leaves Kayden's hand at a height of 6 feet. It...Question
Kayden is throwing a shot put at a track meet. He is 4.1 feet tall. The shot put leaves Kayden's hand at a height of 6 feet. It moves with an upward velocity of 48 feet per second. Using the function h=−16t^2+48t+6, which of the following correctly identifies the start and the vertex, plus also explains its meaning? Round to the nearest hundredth if necessary.(1 point)
Responses
A: (0,6); (1.5,42); The shot put leaves Kayden's hand at a distance of 0 feet and a height of 6
feet and will reach a maximum height of 42 feet 1.5 seconds after being thrown.
B: (6,0); (42,1.5) The shot put is thrown at a height of 6 feet. It will reach a maximum height of 42 feet 1.5 seconds after being thrown.
C: (4.1,0); (42,1.5) The shot put is thrown from a distance of 0 feet at a height of 4.1 feet and will land after 42 seconds.
D: (0,4.1) ; (1.5,42) The shot put is thrown at a height of 0 feet and a distance of 4.1 feet and will reach a maximum height of 1.5 feet 42 seconds after being thrown
Responses
A: (0,6); (1.5,42); The shot put leaves Kayden's hand at a distance of 0 feet and a height of 6
feet and will reach a maximum height of 42 feet 1.5 seconds after being thrown.
B: (6,0); (42,1.5) The shot put is thrown at a height of 6 feet. It will reach a maximum height of 42 feet 1.5 seconds after being thrown.
C: (4.1,0); (42,1.5) The shot put is thrown from a distance of 0 feet at a height of 4.1 feet and will land after 42 seconds.
D: (0,4.1) ; (1.5,42) The shot put is thrown at a height of 0 feet and a distance of 4.1 feet and will reach a maximum height of 1.5 feet 42 seconds after being thrown
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To analyze the problem, we start with the function given for the height of the shot put:
\[ h(t) = -16t^2 + 48t + 6 \]
1. **Starting Point**:
- When \( t = 0 \):
\[
h(0) = -16(0)^2 + 48(0) + 6 = 6
\]
This means the shot put starts at a height of 6 feet when it leaves Kayden's hand.
2. **Finding the Vertex**:
The time \( t \) at which the maximum height occurs can be found using the formula for the vertex of a parabola:
\[
t = -\frac{b}{2a}
\]
where \( a = -16 \) and \( b = 48 \).
\[
t = -\frac{48}{2 \times -16} = \frac{48}{32} = 1.5 \, \text{seconds}
\]
Now, we plug \( t = 1.5 \) into the height equation to find the maximum height:
\[
h(1.5) = -16(1.5)^2 + 48(1.5) + 6
\]
\[
h(1.5) = -16(2.25) + 72 + 6
\]
\[
= -36 + 72 + 6 = 42
\]
Thus, the vertex is at \( (1.5, 42) \), indicating that the shot put reaches a maximum height of 42 feet at 1.5 seconds after being thrown.
Now we can summarize the information and evaluate the options:
- Starting point: \( (0, 6) \)
- Vertex: \( (1.5, 42) \)
The correct option is **A**:
- \( A: (0, 6); (1.5, 42); \)
- Meaning: The shot put leaves Kayden's hand at a distance of 0 feet (the point of release) and a height of 6 feet, and will reach a maximum height of 42 feet 1.5 seconds after being thrown.
**Answer: A**
\[ h(t) = -16t^2 + 48t + 6 \]
1. **Starting Point**:
- When \( t = 0 \):
\[
h(0) = -16(0)^2 + 48(0) + 6 = 6
\]
This means the shot put starts at a height of 6 feet when it leaves Kayden's hand.
2. **Finding the Vertex**:
The time \( t \) at which the maximum height occurs can be found using the formula for the vertex of a parabola:
\[
t = -\frac{b}{2a}
\]
where \( a = -16 \) and \( b = 48 \).
\[
t = -\frac{48}{2 \times -16} = \frac{48}{32} = 1.5 \, \text{seconds}
\]
Now, we plug \( t = 1.5 \) into the height equation to find the maximum height:
\[
h(1.5) = -16(1.5)^2 + 48(1.5) + 6
\]
\[
h(1.5) = -16(2.25) + 72 + 6
\]
\[
= -36 + 72 + 6 = 42
\]
Thus, the vertex is at \( (1.5, 42) \), indicating that the shot put reaches a maximum height of 42 feet at 1.5 seconds after being thrown.
Now we can summarize the information and evaluate the options:
- Starting point: \( (0, 6) \)
- Vertex: \( (1.5, 42) \)
The correct option is **A**:
- \( A: (0, 6); (1.5, 42); \)
- Meaning: The shot put leaves Kayden's hand at a distance of 0 feet (the point of release) and a height of 6 feet, and will reach a maximum height of 42 feet 1.5 seconds after being thrown.
**Answer: A**
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