Aball is kicked upward with an initial velocity of 68 feet per second. the ball's height, h (in feet), from the ground is modeled by h= -16t2 + 68t, where t is measured in seconds. what is the practical domain in this situation?

The practical dominion is from t = 0, when the ball starts motion until the time when the ball hits the ground.

To find the time when the ball hits the ground solve the equation:

h(t) = 0

=> -16t^2 + 68t = 0

Factor: t(-16t + 68) = 0 => t = 0 and -16t + 68 = 0

=> 16t = 68

=> t = 68/16

=> t = 4.25

So, the practical domain is 0 ≤ t ≤ 4.25 s.

Which also may be represented by t ∈ [0, 4.25s]

The practical domain in this question is is the time between which the ball is kicked and when the ball comes down. This means that the practical domain are the two x-intercepts, or times t at which h=0.
We know the lower bound of the practical domain to be t=0, since this is the moment the ball is kicked, and is also the first point at which h=0. The ball is then launched into the air in a parabolic path, at comes back down to h=0 at some point t>0. This will point is the upper bound of our practical domain.
Let's find the second point where h=0.
h(t) = -16t² + 68 t               Set h(t) equal to 0
0 = -16t² + 68t         Add 16t² to both sides
16t² = 68t                Divide both sides by 4t
4t = 17                    Divide both sides by 4
t = 17/4 = 4.25
Therefore, the practical domain, or the domain in which the height is greater than or equal to zero, and serves as an accurate model of the position of the ball, is 0 ≤ t ≤ 17/4

4hours and 40 mins she would be 120 meters high