Using calculus...
dh/dt=-32t+32
When velocity, dh/dt, is equal to zero, the object is at its maximum height.
dh/dt=0 only when 32t=32, t=1 second, so the maximum height is h(1)
h(1)=-16+32=16ft
Using algebra
The maximum height will occur at the midpoint between the two zeros of the height function because of a quadratics symmetry...
h(t)=0 when
-16t^2+32t=0
-16t(t-2)=0
so t=0 and 2
The midpoint is t=1
And of course this will give you the same as we found earlier. The maximum height is h(1) ft
