Answer
(a) The ball will strike the ground after 6 seconds.
(b) The ball is more than 128 feet high for 2 seconds.
Work Step by Step
(a) set s(t) to zero and solve for t:
$0=96t-16t^2$
$0=16(6t-t^2)$
$0=6t-t^2$
$0=t(6-t)$
$t_1=0$
$t_2=6$
The first value means that the ball is on the ground right before being thrown, so the second value is the correct solution: the ball will strike the ground after 6 seconds.
(b) Set s(t) to 128 and move it to the right side:
$128-128=96t-16t^2-128$
$0=-16t^2+96t-128$
Now, we'll solve for t:
$0=-16(t^2-6t+8)$
$0=t^2-6t+8$
$0=(t-2)(t-4)$
$t_1=2$
$t_2=4$
This means that the ball crosses the 128 feet mark after 2 and 4 seconds. That means that the ball is more than 128 feet high for 2 seconds.
