An arrow is shot upward with an initial velocity of 64ft/s. The height of the arrow h(t), in terms of the time the arrow is released t, is h(t)=-16t^2+64t.What is the maximum height that the arrow reaches???

this equation for h(t) is a parabola that opens down

in the equation a = -16    b=  64    c=0

Max height is at the vertex which is for x = -b/2a = -64/(2(-16)) = -64/-32 = 2

h(2)= -16*4 + 64(2) = -64 + 128 = 64

max height is 64 feet.