Answer
$3.2\text{m}/\sec $
Work Step by Step
According to the provided statement,
$s=4.9{{t}^{2}}+{{v}_{0}}t$
Subtract $4.9{{t}^{2}}$ on both sides,
$\begin{align}
& s-4.9{{t}^{2}}=4.9{{t}^{2}}+{{v}_{0}}t-4.9{{t}^{2}} \\
& s-4.9{{t}^{2}}={{v}_{0}}t
\end{align}$
Carry out- solve for ${{v}_{0}}$,
Divide by $t$ on both sides,
$\begin{align}
& \frac{s-4.9{{t}^{2}}}{t}=\frac{{{v}_{0}}t}{t} \\
& \frac{s-4.9{{t}^{2}}}{t}={{v}_{0}} \\
\end{align}$
Substitute the values $s=91.2$ and $t=4$,
$\begin{align}
& \frac{s-4.9{{t}^{2}}}{t}={{v}_{0}} \\
& \frac{\left( 91.2 \right)-4.9{{\left( 4 \right)}^{2}}}{\left( 4 \right)}={{v}_{0}} \\
& 3.2={{v}_{0}}
\end{align}$
Thus, the initial velocity is $3.2\text{m}/\sec $.
