If a ball is thrown vertically upward from the roof of 48 foot building with a velocity of 32 ft/sec, its height after t seconds is s(t)=48+32t–16t2. What is the maximum height the ball reaches?

Answer 1

s'(t) = 32 - 32t
= 0 for a max/min
t = 1
then s(1) = 48 + 32 - 16 = 64

max height is 64 above the ground, or 16 feet above the roof line