Answer
The component of the cannonball's velocity that is parallel to the ground is 86.6 m/s
Work Step by Step
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.
