Question

An object is launched upward at 64 ft/sec from a platform that is 80 feet high. What is the objects maximum height if the equation of height (h) in terms of time (t) of the object is given by h(t) = -16t2 + 64t + 80?

Answer 1

Max height is the vertex
convert to vertex form (y=a(x-h)^2+k) by completeing the square

h(t)=-16(t²-4t)+80
h(t)=-16(t²-4t+4-4)+80
h(t)=-16((t-2)²-4)+80
h(t)=-16(t-2)²+64+80
h(t)=-16(t-2)²+144

vertex is (2,144)
at t=2, the height is 144

max height is 144ft