For this function $\displaystyle h(t)=10-4t^2$ how do you determine what time interval the ball will be more than 40 feet in the air?
I am assuming t is time. As said max height is 50 at t=5. Due to symmetry of the curve it should be [5-a, 5+a].
20t-2t^2=40 gives you the solution.
I think it is: [5-sqrt(5), 5+sqrt(5)].
-O
P.S. I knew that plot of an object is a parabolla, f(x)=ax^2+bx+c, where x is the horizontal range. BUT I am not sure if it is still a parobolla when it is written with respect to time: f(t). Would that trajectory happen on the Jupiter?
If I understand you correctly you have to solve
$\displaystyle 20t-2t^2 \geq 50~\implies~2t^2-20t+50\leq 0~\implies~t^2-10t+25 \leq 0~\implies~(t-5)^2\leq 0$
The LHS of the inequality is a square which never will be negative (or smaller than zero - that's the same). At t = 5 the LHS will be zero.
Thus the answer is: At t = 5 the ball reaches a height of 50'.