Question 970087
 h(t)=-16t^2+128t

maximum height occurs at the vertex.
t value at vertex is -b/2a
b=128
a=-16

vertex has t value of -128/-32   = 4 seconds to reach maximum height

h(4)=-16 (4^2) + (128) (4)
h=-256+512=256feet

4 seconds to return to ground, because launched from ground, and parabolas are symmetric.

112=-16(t^2) +128
Divide everything by 16
7=-1 (t^2) +8
-1=-1 (t^2)
1=1 t^2
t^2=1
t=(1) second OR 7 seconds, on the way down.  Note, t=-1 is a mathematical but not real world solution.

It is h(t), because height is a function of t.  That is by convention.

{{{graph(300,200,-1,8,0,400,-16x^2+128x)}}}