QuestionSeptember 6, 2026

Amaya is standing 30 ft from a volleybal I net. The net is 8 ft high. Amaya serves the ball. The path of the ball is modeled by the equation y=-0.02(x-18)^2+12 , where x is the ball's horizontal distance in feet from Amaya's position and y is the distance in feet from the ground to the ball. a. How far away is the ball from Amaya when it is at its maximum height?Explain. b. Describe how you would find the ball's height when it crosses the net at x=30 a. How far away is the ball from Amaya when it is at its maximum height?Explain. The ball is square ft away from Amaya when it is at its maximum height. Use the square of the vertex.

Amaya is standing 30 ft from a volleybal I net. The net is 8 ft high. Amaya serves the ball. The path of the ball is modeled by the equation y=-0.02(x-18)^2+12 , where x is the ball's horizontal distance in feet from Amaya's position and y is the distance in feet from the ground to the ball. a. How far away is the ball from Amaya when it is at its maximum height?Explain. b. Describe how you would find the ball's height when it crosses the net at x=30 a. How far away is the ball from Amaya when it is at its maximum height?Explain. The ball is square ft away from Amaya when it is at its maximum height. Use the square of the vertex.
Amaya is standing 30 ft from a volleybal I net. The
net is 8 ft high. Amaya serves the ball. The path of
the ball is modeled by the equation
y=-0.02(x-18)^2+12 , where x is the ball's horizontal
distance in feet from Amaya's position and y is the
distance in feet from the ground to the ball.
a. How far away is the ball from Amaya when it is at
its maximum height?Explain.
b. Describe how you would find the ball's height
when it crosses the net at x=30
a. How far away is the ball from Amaya when it is at
its maximum height?Explain.
The ball is square  ft away from Amaya when it is at its
maximum height. Use the square  of the vertex.

Solution
4.1(168 votes)

Answer

18 ft Explanation 1. Identify the Vertex of the Parabola The equation y = -0.02(x-18)^2 + 12 is in vertex form y = a(x-h)^2 + k, where (h, k) is the vertex. Here, h = 18 and k = 12. 2. Determine the Horizontal Distance at Maximum Height The maximum height occurs at the vertex, which is at x = 18. This means the ball is 18 ft away from Amaya when it reaches its maximum height.

Explanation

1. Identify the Vertex of the Parabola<br /> The equation $y = -0.02(x-18)^2 + 12$ is in vertex form $y = a(x-h)^2 + k$, where $(h, k)$ is the vertex. Here, $h = 18$ and $k = 12$.<br /><br />2. Determine the Horizontal Distance at Maximum Height<br /> The maximum height occurs at the vertex, which is at $x = 18$. This means the ball is 18 ft away from Amaya when it reaches its maximum height.
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