Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

The horizontal component of the velocity is parallel to the ground. Therefore; $v_x = v~cos(\theta)$ $v_x = (100~m/s)~cos(30^{\circ})$ $v_x = 86.6~m/s$ The component of the cannonball's velocity that is parallel to the ground is 86.6 m/s.