In the above case the torque due to weight of the rod, here use the concept of translation work done and rotational kinetic energy, 

$\Rightarrow W=K.E.\Rightarrow mgl=\frac{1}{2}I{\omega }^2$, now the moment of inertia of the rod when the axis of rotation passes through one end of the rod, $I=\frac{1\times m{l}^2}{3}$,
$\Rightarrow mgl=\frac{1}{2}\times \frac{m{l}^2}{3}{\omega }^2$
$\Rightarrow \omega =\sqrt{\frac{6\mathrm{g}}{l}}$, now use the relation between linear velocity and angular velocity, 
$\Rightarrow v=l\omega =\sqrt{3.6g}=6 \mathrm{m} {\mathrm{s}}^{-1}$