There are many ways to solve this...however without calculus or derivations from physics, the simplest is to find the midpoint of the two zeros of the function:
(*note, that this will occur for -b/(2a), where a and b are from the general quadratic ax^2+bx+c. So the time when the ball is at its maximum height is -36/(2*-16), -36/-32, 1.125 seconds. And the maximum/minimum height will be (4ac-b^2)/(4a), 25.25 ft. In general the vertex, minimum/maximum point of a parabola will always be (-b/(2a), (4ac-b^2)/(4a) if you care to commit such to memory.)
So back to the midpoint of the zeros. Simply use the quadratic formula to find when h=0
t=(-b±√(b^2-4ac))/(2a)
t=(-36±√1616)/-32 regardless of what is under the radical the midpoint for t is simply
t=-36/-32=1.125 seconds
You then can plug this value in for t in the equation h(t) to get the maximum height of 25.25 ft.
