An object is launched upward at 64 ft/sec from a platform that is 80 feet high. what is the object's maximum height if the equation of height (h) in terms of time (t) of the object is given by h(t) = -16t2 + 64t + 80?

The object's maximum height is the same as the y-coordinate of the vertex.To find that, first, finds the x- coordinate(t).You can find it by:
first identifying a,b and c.Multiply the b by negative 1 or simply change its sign and divide it by 2a.Afterward, plug the answer in the equation to find y, your answer.
(ax+bx²+c=-16t²+64t+80). Therefore, a=-6, b= 64 and c=80
-64/2(-16)=-64/-32=2
-16(2)²+64(2)+80=144 ft
So the maximum height of the object, when launched 64ft/sec from a platform that is 80 ft high, is 144 ft
Hope this helps(there are other methods)