Answer:
[tex]0 \le t \le 3[/tex]
Step-by-step explanation:
[tex]h(t)[/tex] is a quadratic function centered around the origin, with zeros at (-3, 0) and (3, 0) (which you can find by solving [tex]h(t) = -16t^2 + 144 = 0[/tex], i.e. [tex]t^2 = 9[/tex], so [tex]t = 3[/tex] and [tex]t = -3[/tex]).
Given that the function curves downward at its peak, [tex]h(t)[/tex] is positive for [tex]-3 \le t \le 3[/tex]. If the value of the function is negative, the ball would be under the ground, which does not make sense. Also, a negative value of time does not make sense in the context of this problem. So we restrict t to be positive. This means that the only valid domain where t is positive and the value of the function is at least 0 is [tex]0 \le t \le 3[/tex]
