The function \( h(t) = -16t^2 + 80t + 30 \) describes the height of the rock over time, where \( h(t) \) is the height in inches and \( t \) is the time in seconds.
Key Features of the Function:
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Quadratic Function: This is a quadratic function of the form \( h(t) = at^2 + bt + c \) where:
- \( a = -16 \)
- \( b = 80 \)
- \( c = 30 \)
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Parabola Orientation: Since \( a < 0 \), the parabola opens downwards, indicating that there will be a maximum height achieved before the rock falls back down.
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Finding the Vertex (Maximum Height): The vertex of a quadratic function given by \( h(t) = at^2 + bt + c \) occurs at \( t = -\frac{b}{2a} \). \[ t = -\frac{80}{2 \cdot -16} = \frac{80}{32} = 2.5 \text{ seconds} \]
To find the maximum height, we plug \( t = 2.5 \) back into the function \( h(t) \): \[ h(2.5) = -16(2.5)^2 + 80(2.5) + 30 \] \[ = -16(6.25) + 200 + 30 \] \[ = -100 + 200 + 30 = 130 \text{ inches} \]
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Finding when the rock hits the ground: Set \( h(t) = 0 \) to find when the rock hits the ground: \[ -16t^2 + 80t + 30 = 0 \] We can use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- \( a = -16 \)
- \( b = 80 \)
- \( c = 30 \)
\[ t = \frac{-80 \pm \sqrt{80^2 - 4 \cdot (-16) \cdot 30}}{2 \cdot -16} \] \[ t = \frac{-80 \pm \sqrt{6400 + 1920}}{-32} \] \[ t = \frac{-80 \pm \sqrt{8320}}{-32} \] \[ t = \frac{-80 \pm 91.144}{-32} \]
Calculating the two possible solutions: \[ t_1 = \frac{-80 + 91.144}{-32} = \frac{11.144}{-32} \ (\text{not physically meaningful}) \] \[ t_2 = \frac{-80 - 91.144}{-32} = \frac{-171.144}{-32} \approx 5.34 \text{ seconds} \]
Summary:
- Maximum Height: 130 inches at \( t = 2.5 \) seconds.
- Time the rock hits the ground: approximately \( t = 5.34 \) seconds.
If you have specific questions or would like further analysis on this function, feel free to ask!