If a ball is thrown vertically upward with a velocity of 80 ft./s, then its height after t seconds is s = 80t-16t^2.
a) What is the velocity of the ball when it is 96 feet above the ground on its way up? On its way down?
I already found in another part of this question that the maximum height is 100 ft. However I think for this I need to set up an inequality, but of what?
80-32t>96
-96 -96
-16(2t+1) > 0
t > (-1/2) ???????
80-32t<96
t < (-1/2) ???????
I have tried some things but I cant' seem to get any further...