If a ball is thrown vertically upward from the roof of 64 foot building with a velocity of 64 ft/sec, its height after t seconds is s(t)=64+64t–16t2 . What is the maximum height the ball reaches? What is the velocity of the ball when it hits the ground (height 0 )?

Answer 1

Differentiate to find the max height.

Velocity is 64 - 32t.

Max is at s'(t) = 0, when velocity is zero
s'(t)= 64 - 32t = 0

so it peaks at t=2. From that you can get the height.

You now need to find s(t) = 0, so solve 64+64t–16t2 = 0

That will give you the time, and from that the velocity function 64 - 32t will give you the answer to the second part