To analyze the problem, we start with the function given for the height of the shot put:
\[ h(t) = -16t^2 + 48t + 6 \]
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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.
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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