Calculus
posted by Jery on .
If a ball is thrown vertically upward from the roof of 32 foot building with a velocity of 112 ft/sec, its height after t seconds is s(t)=32+112tâ€“16t2 . What is the maximum height the ball reaches?What is the velocity of the ball when it hits the ground (height 0 )?

v(t) = s'(t) = 112  32t
at max v = 0
32t = 112
t = 3.5
then s(3.5) = ......
when it hits the ground, s(t) = 0
16t^2  112t  32 = 0
t^2  7t  2 = 0
solve using the quadratic formula,
sub the positive answer into v(t) 
Hi I have the same question. The first part is correct to find the maximum height, but while solving for the quadratic formula to find the velocity when it hits the ground, the answer is incorrect. (You would get 7.2749 and 0.27492 after solving the quadratic)

Nevermind! I figured out you're supposed the plug the positive value into the derivative v'(t).