A ball is thrown vertically upward with an initial velocity of 96 feet per second. The distance s (in feet) of the ball from the ground after t seconds is S(t)=96t-16t^2. At what time t will the ball strike the ground? For what time t is the ball more than 96 feet above the ground?

Answer 1

final position: s(t)=0
0=96t-16t^2=16t(6-t)

so t=6 it hits the ground.

Answer 2

for your last part, you want
-16t^2 + 96t > 96

let's see when -16t^2 + 96t = 96
t^2 - 6t +6 = 0
t = (6 ± √12)/2
= 3 ± √3
= 1.268 or 4.732

so between these two times the ball is above 96 ft
so the ball is above 96 ft for 3.464 seconds