well, we know the height is f(t), and we know that f(t) = -16t² + 96t, with "t" being the seconds.
so, when will it reach 128ft? namely, what is "t" when f(t) = 128?
[tex]\bf \stackrel{f(t)}{128}=-16t^2+96t\implies 16t^2-96t+128=0\\\\\\ 16(t^2-6t+8)=0
\\\\\\
t^2-6t+8=0\implies (t-2)(t-4)=0\implies t=
\begin{cases}
2\\
4
\end{cases}[/tex]
well, so it reaches it twice, check the picture below.
at 2 seconds on its way up, and at 4 seconds on its way down.
