To analyze the motion of the shirts launched from the cannon, we can look at the function given:
\[ h(t) = -16t^2 + 38t + 2 \]
where \( h(t) \) is the height in feet after \( t \) seconds, given that the cannon starts at a height of 2 feet.
1. Finding the intercepts:
- To find the ground level (when height = 0):
\[ 0 = -16t^2 + 38t + 2 \]
Using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = -16 \), \( b = 38 \), and \( c = 2 \).
Calculating the discriminant:
\[ b^2 - 4ac = 38^2 - 4(-16)(2) = 1444 + 128 = 1572 \]
Now using the quadratic formula:
\[ t = \frac{-38 \pm \sqrt{1572}}{2(-16)} \]
Calculating \( \sqrt{1572} \approx 39.6 \):
\[ t = \frac{-38 \pm 39.6}{-32} \]
Two possible solutions:
- \( t \approx \frac{-38 + 39.6}{-32} \Rightarrow t \approx 0.049 \) seconds
- \( t \approx \frac{-38 - 39.6}{-32} \Rightarrow t \approx 2.38 \) seconds
So the intercepts are approximately \( (0, 2) \) and \( (2.38, 0) \).
2. Finding the vertex:
The vertex of a parabola given by \( h(t) \) can be found using the formula \( t = -\frac{b}{2a} \):
\[ t = -\frac{38}{2 \times -16} = \frac{38}{32} = 1.19 \text{ seconds} \]
Now calculate the height at the vertex:
\[ h(1.19) = -16(1.19)^2 + 38(1.19) + 2 \] \[ = -16(1.4161) + 45.62 + 2 \] \[ \approx -22.66 + 45.62 + 2 \approx 24.96 \text{ feet} \]
So the vertex is around \( (1.19, 24.96) \).
3. Summary of findings:
- Intercepts: \( (0, 2) \) and \( (2.38, 0) \)
- Vertex: \( (1.19, 24.96) \)
4. Interpreting the results:
The shirts are launched from a height of 2 feet. After about 1.19 seconds, they reach their maximum height of approximately 24.96 feet. Then, they descend back to ground level after approximately 2.38 seconds.
Conclusion:
The correct response to fit these findings about the intercepts and the path of the shirts is:
(0, 2), (1.19, 24.96), and (2.38, 0); After being launched from the cannon at a height of 2 feet, the shirts reach the height of 24.96 feet after 1.19 seconds before returning to ground level after 2.38 seconds.
Since this option includes the important context of starting from 2 feet, maximally reaching 24.96 feet at 1.19 seconds, and returning to ground level at 2.38 seconds, it correctly summarizes the situation.