Answer
(a) $v = 31.0~m/s$
(b) $\theta_x = 58.1^{\circ}$
$\theta_y = 31.9^{\circ}$
Work Step by Step
(a) We can find the magnitude of the velocity vector.
$v= \sqrt{v_x^2+v_y^2}$
$v= \sqrt{(16.4~m/s)^2+(-26.3~m/s)^2}$
$v = 31.0~m/s$
(b) We can find the angle $\theta_x$ below the +x-axis.
$tan(\theta_x) = \frac{26.3}{16.4}$
$\theta_x = arctan(\frac{26.3}{16.4})$
$\theta_x = 58.1^{\circ}$
We can find the angle $\theta_y$ with the -y-axis.
$\theta_y = 90.0^{\circ} - \theta_x$
$\theta_y = 90.0^{\circ} - 58.1^{\circ}$
$\theta_y = 31.9^{\circ}$
