Answer
(a) $\vec{v_f}=(16.6m/s)\hat x-(17.2m/s)\hat y$
(b) $23.9\frac{m}{s},-46.0^{\circ}$
Work Step by Step
(a) We know that
$\vec {v_f}=\vec{v_{\circ}}+\vec a t$
$\implies \vec{v_f}=(16.6m/s)\hat x+(-9.81m/s^2)(1.75s)\hat y$
$\vec{v_f}=(16.6m/s)\hat x-(17.2m/s)\hat y$
(b) We can find the magnitude of the velocity as
$v=\sqrt{(16.6)^2+(-17.2)^2}$
$v=23.9m/s$
and the direction of velocity is given as
$\theta=tan^{-1}(\frac{-17.2}{16.6})$
$\theta=-46.0^{\circ}$
