## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

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