Answer
$\theta = 70.5^{\circ}$
Work Step by Step
At the horizontal where the rod was released:
$K_C + U_C = K_B +U_B$
$\frac{1}{2}mv_C + mgh_C = \frac{1}{2}mv_B^2 + mgh_B$
$v_C$ and h_B are equal to zero
$h_c = l$
$(\frac{1}{2}m\times 0) + mgl= \frac{1}{2}mv_B^2 + (mg\times 0)$
$ mgl = \frac{1}{2}mv_B^2 $
$ gl = \frac{1}{2}v_B^2 $
$ 2gl = v_B^2 $
$v_{_B} = \sqrt { 2gl}$
Finding the angle between the rod and the vertical:
$\theta = cos^{-1}(\frac{mg}{T})$
$\theta = cos^{-1}(\frac{mg}{mv_B^2l^{-1}+mg})$
$\theta = cos^{-1}(\frac{mg}{m(\sqrt {2gl})^2l^{-1}+mg})$
$\theta = cos^{-1}(\frac{mg}{m(2gl)l^{-1}+mg})$
$\theta = cos^{-1}(\frac{mg}{(2mg)+mg})$
$\theta = cos^{-1}(\frac{mg}{3mg})$
$\theta = cos^{-1}(\frac{1}{3})$
$\theta = 70.5^{\circ}$
