An object is projected vertically upward from the top of a building with an initial velocity of 112 ft/sec. Its distance s(t) in feet above the ground after t seconds is given by the equation

Question

s(t) = −16t^2 + 112t + 100.
(a) Find its maximum distance above the ground.
s(t) = ft

Answer 1

Find the derivative of the function:

s(t) = -32t^2+112
0 = -32t^2 + 112
t=sqrt(112/32)

now once you find t plug it back into the original equation to find the max distance above the ground

Answer 2

usually college algebra students haven't learned about derivatives. However, s(t) is just a parabola, with its vertex at

t = 112/32

so, plug that into s(t) to find the max height

Note that you got your derivative wrong, anyway.