## University Physics with Modern Physics (14th Edition)

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.
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.