Answer
$v_{x}=161\ m/s.$
Work Step by Step
We use the discussion of part (a).
The horizontal component of velocity remains unchanged,
( from release to just before impact)
We have found $ v_{0}=202m/s$, $\theta=-37^{o}$
$v_{x}=v_{0}\cos\theta_{0}$
$=(202m/s)\cos(-37.0^{o})$
$=161m/s.$
