h(t)=-16t^2+80t. whats the maximum height that the rocket will reach?

　
This is a parabola, so put it in vertex form:

h(t) = −16t² + 80t
h(t) = −16 (t²−5t+25/4) + 16(25/4)
h(t) = −16 (t−5/2)² + 100

Maximum height = 100 ft, when t = 2.5 sec

v =-32t +80=, v=0
0=-32t+80
32t=80 then t 80/32=5/2 (max height at t=5/2)
h(5/2)=-16(5/2)^2+80(5/2)=-16(25)/4 +400/2
=-4(25)+200=-100+200=100

You need to find the vertex, either by putting the equation into vertex form, or by finding the point where the first derivative equals 0.

h(t) = -16t² + 80t

h'(t) = -32t + 80 = 0
32t = 80
t = 80/32 = 2.5

max height = h(2.5) = 100 ft

H(t) = at^2 + bt + c
the maximum height = H(-b/(2a))