College Algebra, Chapter 1, 1.3, Section 1.3, Problem 98

How fast would a ball have to be thrown upward to reach a maximum height of $100 ft$? Use the discriminant of the equation $16t^2 - v_0 t + h = 0$.



$
\begin{equation}
\begin{aligned}

16t^2 - v_0 t + h =& 0
&& \text{Model}
\\
\\
16t^2 - v_0 t + 100 =& 0
&& \text{Substitute the given}

\end{aligned}
\end{equation}
$


Since the time it takes for the ball to reach the maximum height will only happen once, the equation has only one exact solution.

So the discriminant,


$
\begin{equation}
\begin{aligned}

b^2 - 4ac =& 0
&&
\\
\\
(-v_0)^2 - 4(16)(100) =& 0
&&
\\
\\
v_0^2 - 6400 =& 0
&& \text{Add } 6400
\\
\\
v_0^2 =& 6400
&& \text{Take the square root}
\\
\\
v_0 =& \pm \sqrt{6400}
&& \text{Solve for } v_0
\\
\\
v_0 =& 80 \text{ and } v_0 = -80
&& \text{Choose } v_0 > 0
\\
\\
v_0 =& 80 ft/s
&&

\end{aligned}
\end{equation}
$