Answer
Just before impact, the angle $\theta$ below the horizontal of the velocity vector is $63.4^{\circ}$
Work Step by Step
We can find the time of flight:
$h = \frac{1}{2}gt^2$
$t = \sqrt{\frac{2h}{g}}$
We can find an expression for $v_y$ just before impact:
$v_y = v_{0y}+at$
$v_y = 0+gt$
$v_y = g~(\sqrt{\frac{2h}{g}})$
$v_y = \sqrt{2gh}$
We can find an expression for $v_x$:
$h = v_x~t$
$v_x = \frac{h}{t}$
$v_x = \frac{h}{\sqrt{\frac{2h}{g}}}$
$v_x = \frac{\sqrt{2gh}}{2}$
We can find the angle $\theta$ below the horizontal of the velocity vector just before impact:
$tan~\theta = \frac{v_y}{v_x}$
$tan~\theta = \frac{\sqrt{2gh}}{(\frac{\sqrt{2gh}}{2})}$
$tan~\theta = 2$
$\theta = tan^{-1}~(2)$
$\theta = 63.4^{\circ}$
Just before impact, the angle $\theta$ below the horizontal of the velocity vector is $63.4^{\circ}$
