Answer
$\theta = -85.2^{\circ}$
Work Step by Step
In part (b), we found that at $t = 2.00~s$, the velocity is $~~v = (19.0~m/s)\hat{i}-(224~m/s)\hat{j}$
We can find the angle of the velocity relative to the positive x axis:
$tan~\theta = \frac{-224~m/s}{19.0~m/s}$
$\theta = tan^{-1}~\frac{-224~m/s}{19.0~m/s}$
$\theta = -85.2^{\circ}$
Note that the angle between the positive direction of the x axis and a line tangent to the particle’s path is equal to the angle of the velocity vector relative to the positive direction of the x axis at a given time $t$.
