If $V = gt + V_0$ and $S = \frac {1}{2}gt^2 + V_0t$, then $t$ equals: 
$\textbf{(A)}\ \frac{2S}{V+V_0}\qquad \textbf{(B)}\ \frac{2S}{V-V_0}\qquad \textbf{(C)}\ \frac{2S}{V_0-V}\qquad \textbf{(D)}\ \frac{2S}{V}\qquad \textbf{(E)}\ 2S-V$
Solve for $t$ in the first equation:
\[V - V_0 = gt\]
\[t = \frac{V - V_0}{g}\]
Then use the second equation to solve for $g:$
\[2(S - V_0 t) = gt^2\]
\[g = \frac{2(S - V_0 t)}{t^2}\]
Plug in $g$ in $t$ to get $\boxed{\textbf{(A)}}.$
-coolmath34