An object is projected vertically upward from the top of a building with an initial velocity of 144 ft/sec. Its distance s(t) in feet above the ground after t seconds is given by the function

Question

s(t) = −16t2 + 144t + 120.
A) Find its maximum distance above the ground. (in feet)
B) Find the height of the building. (in feet)

Answer 1

v(t) = -32t + 144
at max height v(t) = 0
32t = 144
t = 4.5
s(4.5) = -16(4.5)^2 + 144(4.5) + 120 = 444 ft

b) height of building --- t = 0
height = 120