a) We will draw the situation.
b) Considering the theory, the vertex of a parabola is represented in (h,k) where h is the x-coordinate, in this case, if our parabola has an amplitude of x=320feet then h=320/2=160feet, and k is the highest y-coordinate 80 feet. So the equation is:
[tex]\begin{gathered} y=a(x-h)^2+k \\ y=a(x-160)^2+80 \end{gathered}[/tex]We have to know the value for a, so we will use a point to replace it in the equation and with that, we will know the value, so in x=0 the y-value is 0 too so:
[tex]\begin{gathered} 0=a(0-160)^2+80 \\ a(-160)^2=-80 \\ a=-\frac{80}{25600}=-\frac{1}{320} \end{gathered}[/tex]So the equation is:
[tex]y=-\frac{1}{320}(x-160)^2+80[/tex]c. To know the answer to this question we will have to replace x=280feet with the y value we will know if at the moment the ball can go over the tree or if it would crash into it.
[tex]\begin{gathered} y=-\frac{1}{320}(280-160)^2+80 \\ y=35feet \end{gathered}[/tex]At that point, the ball would be 35 feet up so if it could pass over the tree 5 feet higher.
