Question

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.

Answer 1

a) 30feet

b) 9.12feet

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.