its displacement is given in vector form as follows
[tex]\vec d_1= 4 \hat j[/tex]
[tex]\vec d_2 = 2 cos45 \hat i + 2 sin45 \hat j[/tex]
[tex]\vec d_2 = 1.41 \hat i + 1.41 \hat j[/tex]
[tex]\vec d_3 = -1 sin30 \hat i - 1 cos30 \hat j[/tex]
[tex]\vec d_3 = -0.5 \hat i - 0.866 \hat j[/tex]
now the total displacement due to all three is given as
[tex]d = d_1 + d_2 + d_3[/tex]
[tex]d = (1.41 - 0.5)\hat i + (4 + 1.41 -0.866)\hat j[/tex]
[tex]d = 0.91 \hat i + 4.54 \hat j[/tex]
so magnitude of displacement is
[tex]d = \sqrt{0.91^2 + 4.54^2} = 4.63 m[/tex]
direction is given as
[tex]\theta = tan^{-1}\frac{4.54}{0.91}[/tex]
[tex]\theta = 78.7 degree[/tex]
so the ball will be displaced by 4.63 m at angle 78.7 degree North of East