Answer
$\theta=49.5^{\mathrm{o}}$
Work Step by Step
(b)
At $t=4.00s$
$\left[\begin{array}{lll}
v_{x} & =\frac{3}{2}(4.00)^{2}+5.00 & =(29.0\mathrm{m}/\mathrm{s})\\
v_{y} & =2.00(4.00)^{2}+2.00 & =(34.0\mathrm{m}/\mathrm{s})
\end{array}\right]$
$\vec{v}=(29.0\mathrm{m}/\mathrm{s})\hat{\mathrm{i}}+(34.0\mathrm{m}/\mathrm{s})\hat{\mathrm{j}}$.
The angle of the motion, counterclockwise to +x,
is in the first quadrant.
$\displaystyle \theta=\tan^{-1}[\frac{34.0\mathrm{m}/\mathrm{s}}{29.0\mathrm{m}/\mathrm{s}}]=49.5^{\mathrm{o}}$
$\theta=49.5^{\mathrm{o}}$
