Answer:
+2.5 m/s and -2.5 m/s
Explanation:
There are two possibilities:
1) The velocity lies in the IV quadrant, with y being negative and x being positive. In this case, the magnitude of the horizontal component is
[tex]v_x = v cos 60^{\circ} = (5.0 m/s) cos 60^{\circ} =2.5 m/s[/tex]
2) The velocity lies in the III quadrant, with y being negative and x being negative as well. In this case, the magnitude of the horizontal component is
[tex]v_x = -v cos 60^{\circ} = -(5.0 m/s) cos 60^{\circ} =-2.5 m/s[/tex]
