Question

A rock is thrown from one side of a river to another. The function h(t) = -16t^2 + 80t + 30 gives the height in inches of the rock t seconds after it has been thrown.

Answer 1

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:

1. **Quadratic Function**: This is a quadratic function of the form \( h(t) = at^2 + bt + c \) where:
- \( a = -16 \)
- \( b = 80 \)
- \( c = 30 \)

2. **Parabola Orientation**: Since \( a < 0 \), the parabola opens downwards, indicating that there will be a maximum height achieved before the rock falls back down.

3. **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 \( h(t) \):
\[
h(2.5) = -16(2.5)^2 + 80(2.5) + 30
\]
\[
= -16(6.25) + 200 + 30
\]
\[
= -100 + 200 + 30 = 130 \text{ inches}
\]

4. **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!