Beginning Algebra With Applications, Chapter 3, 3.2, Section 3.2, Problem 178

The distance $s$, in feet, than an object will fall in $t$ seconds is given by $s = 16t^2 + vt$, where $v$ is the initial downward velocity of the object in feet per second.

Find the initial velocity of an object that falls $144$ ft in $3$ s.

We solve for initial velocity,


$
\begin{equation}
\begin{aligned}

s =& 16t^2 + vt
&& \text{Given equation}
\\
\\
s - 16t^2 =& vt
&& \text{Subtract } 16t^2
\\
\\
\frac{s - 16t^2}{t} =& v
&& \text{Divide by } t
\\
\\
\frac{144 - 16(3)^2}{3} =& v
&& \text{Substitute } s = 144 \text{ and } t = 3
\\
\\
\frac{144 - 144}{3} =& v
&&
\\
\\
\frac{0}{3} =& v
&&
\\
\\
v =& 0 \text{ ft/s}
&&

\end{aligned}
\end{equation}
$


The initial velocity is ft/sec.