Answer
$\text{C}$
Work Step by Step
The graph of $h(t)=-16t^2+96t$ is a parabola that opens downward.
Thus, the maximum height of the ball is at the vertex.
The vertex of a parabola is at $\left(\frac{-b}{2a}, f\left(\frac{-b}{2a}\right)\right)$.
The given function has $a=-16$ and $b=96$.
Thus, its vertex is at:
\begin{align*}
\left(\frac{-96}{2(-16)}, h\left(\frac{-96}{2(-16)}\right)\right)&=\left(\frac{-96}{-32}, h\left(\frac{-96}{32}\right)\right)\\\\
&=\left(3, h(3)\right)\\\\
&=\left(3, -16\cdot 3^2+96\cdot 3\right) \\\\
&=\left(3, 144\right)
\end{align*}
Therefore, the maximum height that the ball will reach is $144\text{ ft}$.
