# Cannonball Parabola Conics Problem

I found this problem in a math textbook and I was a little confused on how to solve it. Here is the problem:

A cannon fires a cannonball. The path of the cannonball is a parabola with vertex at the highest point of the path. If the cannonball lands 1600 ft from the cannon and the highest point it reaches is 3200 ft above the ground, find an equation for the path of the cannonball. Place the origin at the location of the cannon.

Since the cannonball's path is parabolic, we can assume the generic conic equation for parabolas, which can be applied to this question:

\begin{align} \ x^2 &= {\ 4py} \\\\ \end{align}

Further, the roots of the parabolic equation are (0, 0) and (0, 1600), respectively, since the cannon begins at the origin and comes down at the 1600 mark. Further, the middle of the x-values of 0 and 1600 is 800, which is where the maximum value of 800 would be. So, the vertex of the parabolic graph is placed at (800, 3200).

Lastly, since the vertex is not at (0, 0), the origin, its (h, k) values would be different since this is a parabolic graph that has a shift. The equation would then be: (I am substituting a for the p value because we are uncertain about it)

\begin{align} \ (x-800)^2 &= {\ a(y-3200)^2} \\\\ \end{align}

I am confused on how to go further with this problem. Am I supposed to plug in one of the roots in order to get the final equation?

• In the last line, the square on right hand side looks like a typo. Don't you mean $$(x-800)^2 = a(y-3200)$$ ? – ganeshie8 Mar 17 '15 at 5:30
• Right, sorry. That was a mistake. – Shrey Mar 17 '15 at 5:30
• you're almost done then ! simply plugin any of the known points on parabola and solve $a$ – ganeshie8 Mar 17 '15 at 5:31
• for example, the point $(0,0)$ is on parabola – ganeshie8 Mar 17 '15 at 5:32
• My answer was: (x-800)^2 = 1/16(y-3200) – Shrey Mar 17 '15 at 5:33