Question

A cannon fires a cannonball as shown
in the figure. 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.
A) y-800=(-1/160)(x-4000)^2
B) y-800=(-1/120)(x-4000)^2
C) y-4000=(-1/160)(x-800)^2
D) y-4000=(-1/120)(x-800)^2

Answer 1

From your description, the vertex must be (800, 3200)
so its equation must be
y = a(x-800)^2 + 3200
but (1600,0) also lies on it, so
0 = a(1600-800)^2 + 3200
a = -3200/(800)^2 = - 1/200

the equation would be y = (-1/200)(x - 800)^2 + 3200

or in the form they use:
y - 3200 = (-1/200)(x - 800)^2

none of the choices match that.