An equation that models the path of Johnny’s ball is [tex]y = -\frac{40}9(x - 1.5)^2 + 10[/tex]
The maximum height is given as:
[tex]h_{max}= 10ft[/tex]
And the time spent is:
[tex]t = 3s[/tex]
So, the vertex of the ball would be:
[tex]Vertex = (t/2,h_{max})[/tex]
This gives
[tex]Vertex = (3/2,10)[/tex]
[tex]Vertex = (1.5,10)[/tex]
Rewrite properly as:
[tex](h,k) = (1.5,10)[/tex]
The ball is on the floor after 3 seconds.
So, another point on the graph is:
[tex](x,y) = (3,0)[/tex]
A quadratic equation is represented as:
[tex]y=a(x - h)^2 + k[/tex]
Substitute [tex](x,y) = (3,0)[/tex] and [tex](h,k) = (1.5,10)[/tex] in [tex]y=a(x - h)^2 + k[/tex]
[tex]0=a(3 - 1.5)^2 + 10[/tex]
[tex]0=a(1.5)^2 + 10[/tex]
Evaluate the exponent
[tex]0=2.25a + 10[/tex]
Subtract 10 from both sides
[tex]2.25a =- 10[/tex]
Divide both sides by 2.25
[tex]a = -\frac{10}{2.25}[/tex]
Multiply by 4/4
[tex]a = -\frac{40}{9}[/tex]
Substitute [tex]a = -\frac{40}{9}[/tex] and [tex](h,k) = (1.5,10)[/tex] in [tex]y=a(x - h)^2 + k[/tex]
[tex]y = -\frac{40}9(x - 1.5)^2 + 10[/tex]
See attachment for the graph that models the path
Read more about quadratic functions at:
https://brainly.com/question/11441586
