Answer
The magnitude of the velocity of the ball relative to the ground is 3.01 m/s. The direction is $56.3^{\circ}$ east of north.
Work Step by Step
Let $M$ be Mia.
Let $B$ be the ball.
Let $G$ be the ground.
$v_{B/G} = v_{B/M} + v_{M/G}$
We can find the east component of $v_{B/G}$.
$v_{B/G} = v_{B/M} + v_{M/G}$
$v_{B/G} = (5.00~m/s)~sin(30.0^{\circ}) + 0$
$v_{B/G} = 2.50~m/s$ (toward the east)
We can find the north component of $v_{B/G}$.
$v_{B/G} = v_{B/M} + v_{M/G}$
$v_{B/G} = -(5.00~m/s)~cos(30.0^{\circ}) + (6.00~m/s)$
$v_{B/G} = 1.67~m/s$ (toward the north)
We can find the magnitude of the velocity of the ball relative to the ground.
$v_{B/G} = \sqrt{(2.50~m/s)^2+(1.67~m/s)^2}$
$v_{B/G} = 3.01~m/s$
We can find the angle $\theta$ east of north.
$tan(\theta) = \frac{2.50~m/s}{1.67~m/s}$
$\theta = tan^{-1}(\frac{2.50}{1.67}) = 56.3^{\circ}$
The magnitude of the velocity of the ball relative to the ground is 3.01 m/s. The direction is $56.3^{\circ}$ east of north.
