Zero Gravity. As part of their
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![### Zero Gravity
As part of their training, astronauts rode the “vomit comet,” NASA’s reduced-gravity KC 135A aircraft that performed parabolic flights to simulate weightlessness. The plane started at an altitude of 20,000 feet and made a steep climb at 528 with the horizon for 20–25 seconds and then dove at that same angle back down, repeatedly. The equation governing the altitude of the flight is
\[ A(x) = -0.0003x^2 + 9.3x - 46,075 \]
where \( A(x) \) is the altitude and \( x \) is the horizontal distance in feet.
1. **What is the maximum altitude the plane attains?**
2. **Over what horizontal distance is the entire maneuver performed?** (Assume the starting and ending altitude is 20,000 feet.)
#### Step-by-step Analysis:
1. **Determine the Maximum Altitude:**
- We need to find the vertex of the parabola described by \( A(x) = -0.0003x^2 + 9.3x - 46,075 \). The vertex form of a quadratic equation \( ax^2 + bx + c \) gives the x-coordinate of the vertex as \( x = -\frac{b}{2a} \).
- Plugging in the values \( a = -0.0003 \) and \( b = 9.3 \):
\[
x = -\frac{9.3}{2 \times -0.0003} = \frac{9.3}{0.0006} = 15,500 \, \text{feet}
\]
- To find the maximum altitude, substitute \( x = 15,500 \) back into the equation:
\[
A(15,500) = -0.0003(15,500)^2 + 9.3(15,500) - 46,075
\]
\[
A(15,500) = -0.0003(240,250,000) + 143,850 - 46,075
\]
\[
A(15,500) = -72,075 + 143,850 - 46,075
\]
\[
A(15,500) = 25,700 \,](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F180e35fb-9b09-4fc5-8a6a-511bee286828%2F1187c388-a4a5-46fe-88ed-4727942c51f5%2Fm0p9oxf_processed.jpeg&w=3840&q=75)
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