Answer
(a) The ball will strike the ground after 5 seconds.
(b) The ball is more than 96 feet high for 1 second.
Work Step by Step
(a) set s(t) to zero and solve for t:
$0=80t-16t^2$
$0=16(5t-t^2)$
$0=5t-t^2$
$0=t(5-t)$
$t_1=0$
$t_2=5$
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 5 seconds.
(b) Set s(t) to 96 and move it to the right side:
$96-96=80t-16t^2-96$
$0=-16t^2+80t-96$
Now, we'll solve for t:
$0=-16(t^2-5t+6)$
$0=t^2-5t+6$
$0=(t-2)(t-3)$
$t_1=2$
$t_2=3$
This means that the ball crosses the 96 feet mark after 2 and 3 seconds. This means that the ball is more than 96 feet high for 1 second.
