Answer
$v_x = -50~m/s$
$v_y = -87~m/s$
Work Step by Step
The vector is directed at an angle which is $60^{\circ}$ below the negative x-axis in this coordinate system. Therefore,
$v_x = -v~cos(\theta)$
$v_x = -(100~m/s)~cos(60^{\circ})$
$v_x = -50~m/s$
$v_y = -v~sin(\theta)$
$v_y = -(100~m/s)~sin(60^{\circ})$
$v_y = -87~m/s$