Amaya is standing 30 ft from a volleyball 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 nothing ft away from Amaya when it is at its maximum height. Use the ▼ y-coordinate x-coordinate of the vertex.

a) 30feet

b) 9.12feet

Step-by-step explanation:

a) Given 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

y is the distance in feet from the ground to the ball.

The ball will be at the maximum height when dy/dx = 0

Differentiate the function:

dy/dx = 2(-0.02)(x-18)^(2-1)+0

dy/dx = -0.04(x-18)

Equate the resulting expression to zero and find x as shown;

-0.04(x-18) = 0

Open the parentheses

-0.04x+0.04(18) = 0

-0.04x+0.72 =0

-0.04x = -0.72

x = -0.72/-0.04

x = 18ft

Hence the ball is 18ft away from Amaya when it is at its maximum height

b) To get the balls height y when x = 30, we will substitute x = 30 into the equation given and calculate the value of y as shown:

y=-0.02(x-18)²+12

y = -0.02(30-18)²+12

y= -0.02(12)²+12

y = -0.02(144)+12

y = -2.88+12

y = 9.12 feet

Hence the Ball's height is 9.12feet when it crosses the net at x = 30.

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