Answer
$$10 \sqrt{6 g}~m/s\approx 76.68~m/s$$
Work Step by Step
We have the position and velocity functions:
\begin{array}{l}
s(t)=s_{0}+v_{0} t-\frac{1}{2} g t^{2} \\
v(t)=v_{0}-g t
\end{array}
then for $s_0=300$ and $v_0=0$, we have
$$s(t) =300- -\frac{1}{2} g t^{2} $$
at $s=0$, we get $t=\sqrt{\frac{600}{g}} $
Hence
\begin{aligned}
v\left(\sqrt{\frac{600}{g}}\right) &= -g \sqrt{\frac{600}{g}} \\
&=\frac{g}{\sqrt{g}} \sqrt{600} \\
&=10 \sqrt{6 g}
\end{aligned}
