College Algebra, Chapter 1, 1.3, Section 1.3, Problem 96

Suppose that an object is dropped from a height $96 ft$ tall above the ground. Then its height after $t$ seconds is given by $h = -16t^2 + 96$, where $h$ is measured in feet.

a.) How long will it take to fall half the distance to ground level?


$
\begin{equation}
\begin{aligned}

h =& -16t^2 + 96
&& \text{Model}
\\
\\
\frac{96}{2} =& -16t^2 + 96
&& h = \frac{96}{2} \text{ half the distance to ground level}
\\
\\
16t^2 =& 96 - \frac{96}{2}
&& \text{Add } 16t^2 \text{ and subtract } \frac{96}{2}
\\
\\
t^2 =& \frac{48}{16} = 3
&& \text{Solve for } t
\\
\\
t =& \pm \sqrt{3}
&& \text{Take the square root}
\\
\\
t =& \sqrt{3} \text{ and } t = \sqrt{3}
&& \text{Choose } t > 0
\\
\\
t =& \sqrt{3} \text{ seconds}
&&

\end{aligned}
\end{equation}
$


b.) How long will it take to fall to ground level?


$
\begin{equation}
\begin{aligned}

h =& -16t^2 + 96h = -16t^2 + 96
&& \text{Model}
\\
\\
0 =& -16t^2 + 96
&& h = 0 \text{ ground level}
\\
\\
16t^2 =& 96
&& \text{Add } 16t^2
\\
\\
t^2 =& \frac{96}{16}
&& \text{Divide } 16
\\
\\
t^2 =& 6
&& \text{Simplify}
\\
\\
t =& \pm \sqrt{6}
&& \text{Take the square root}
\\
\\
t =& \sqrt{6} \text{ and } t = - \sqrt{6}
&& \text{Choose } t > 0


\end{aligned}
\end{equation}
$