Answer
The magnitude of the velocity of the ball relative to Juan is 7.39 m/s.
The direction of the ball's velocity relative to Juan is $77.7^{\circ}$ east of north.
Work Step by Step
Let $J$ be Juan.
Let $E$ be the earth.
Let $B$ be the ball.
$v_{B/E} = v_{B/J} + v_{J/E}$
$v_{B/J} = v_{B/E} - v_{J/E}$
We can find the east component of $v_{B/J}$.
$v_{B/J} = v_{B/E} - v_{J/E}$
$v_{B/J} = (12.0~m/s)~sin(37.0^{\circ}) - 0$
$v_{B/J} = 7.22~m/s$ (toward the east)
We can find the north component of $v_{B/J}$.
$v_{B/J} = v_{B/E} - v_{J/E}$
$v_{B/J} = (12.0~m/s)~cos(37.0^{\circ}) - 8.00~m/s$
$v_{B/J} = 1.58~m/s$ (toward the north)
We can find the magnitude of the velocity of the ball relative to Juan.
$v_{B/J} = \sqrt{(7.22~m/s)^2+(1.58~m/s)^2}$
$v_{B/J} = 7.39~m/s$
The magnitude of the velocity of the ball relative to Juan is 7.39 m/s.
We can find the angle $\theta$ east of north.
$tan(\theta) = \frac{7.22~m/s}{1.58~m/s}$
$\theta = tan^{-1}(\frac{7.22}{1.58})$
$\theta = 77.7^{\circ}$
The direction of the ball's velocity relative to Juan is $77.7^{\circ}$ east of north.
