Explanation:
So this will be an upside down parabola....the leading coefficient (for x^2 ) will be negative ...
Vertex at 60,90 <=====given
Vertex form y = a (x-h) ^2 + k
y = a ( x -60)^2 + 90 to find 'a' substitute in a point on the parabola...I'll use 0,0
0 = a ( 0-60)^2 + 90 shows a = - 1/40
so the equation is y = -1/40 ( x -60)^2 + 90
( or expanded to y= -1/40 x^2 + 3x )
Solve for 'x' when y = 6 ft ( to keep from hitting your head)
6 = -1/40x^2 +3x
0 = -1/40 x^2 + 3x - 6 Use Quadratic Formula to find x = ~ 2 feet