h(x)=-16(t-2)^2+64

Answer 1

this looks a lot like the parabolic equation describing the height of an object with seemingly initial height 64 but really zero, and initial velocity not =0, gravity is -32 ft/sec^2

you need to expand to see the initial height & velocity

h= -16(t^2 -4t + 4) + 64 = -16t^2 +64t initial velocity = 64, initial height =0

for max height take the derivative and set =0

h'(x) = 32(t-2) = 0

32t-64 =0

t = 2 seconds at max height

max height = 64 feet

at time t=0 the height = 0

at t =4 seconds, it's back down to the ground