Question

anaya is serving a volleyball. the net is 8 ft high. anaya serves the ball. the path of the ball is modeled by the equation y = - 0.02(x - 18)^2+12, where x is the balls horizontal distance in feet from the point of the serve and y is the distance in feet from the ground.

a. how far is the ball from anaya when it is at its maximum height? explain.

b. how far is the ball from anaya when it is at its maximum height? how many feet would you find the balls height when it crosses the net at x = 30? describe how you would find the balls height when it crosses the net at x = 30.

Explanation:

Step1: Identify the vertex - form of a parabola

The equation of the ball's path is $y=-0.02(x - 18)^2+12$, which is in vertex - form $y=a(x - h)^2 + k$, where $(h,k)$ is the vertex of the parabola. Here, $h = 18$ and $k = 12$.

Step2: Answer part (a)

The ball reaches its maximum height at the vertex of the parabolic path. The $x$ - value of the vertex gives the horizontal distance of the ball from Anaya when it is at its maximum height. So the ball is 18 feet from Anaya when it is at its maximum height.

Step3: Answer part (b)

We are given the equation $y=-0.02(x - 18)^2+12$ and $x = 30$. Substitute $x = 30$ into the equation:
\[

$$\begin{align*} y&=-0.02(30 - 18)^2+12\\ &=-0.02\times12^2+12\\ &=-0.02\times144 + 12\\ &=-2.88+12\\ &=9.12 \end{align*}$$

\]

Answer:

a. 18 feet. The ball's path is a parabola given by $y=-0.02(x - 18)^2+12$. The vertex of the parabola is at $(18,12)$, and the $x$ - coordinate of the vertex represents the horizontal distance from Anaya when the ball is at its maximum height.
b. 9.12 feet. Substitute $x = 30$ into the equation $y=-0.02(x - 18)^2+12$. First, calculate $(30 - 18)^2=144$, then $-0.02\times144=-2.88$, and finally $-2.88 + 12=9.12$.