In the given figure, particle A moves along the line y = 30 m with a constant velocity $\overrightarrow{\mathrm{v}}$ of magnitude 3.0 m/s and parallel to the x-axis. At the instant particle A passes the y-axis, particle B leaves the origin with a zero initial speed and a constant acceleration $\overrightarrow{\mathrm{a}}$ of magnitude 0.40 m/s². What angle $\mathrm{\theta }$ (in degrees) between $\overrightarrow{\mathrm{a}}$ and the positive direction of the y-axis would result in a collision? 

Collision between particles A and B requires two things.
$\mathrm{y}=\frac{1}{2}{\mathrm{a}}_{\mathrm{y}}{\mathrm{t}}^2$

$\begin{array}{l}\Rightarrow 30\mathrm{m}=\frac{1}{2}\left[\left(0.40\mathrm{m}/{\mathrm{s}}^2\right)\cos \mathrm{\theta }\right]{\mathrm{t}}^2 .....\left(\mathrm{i}\right)\\ \mathrm{vt}=\frac{1}{2}{\mathrm{a}}_{\mathrm{x}}{\mathrm{t}}^2\Rightarrow (3.0\mathrm{m}/\mathrm{s})\mathrm{t}=\frac{1}{2}\left[\left(0.40\mathrm{m}/{\mathrm{s}}^2\right)\sin \mathrm{\theta }\right]{\mathrm{t}}^2\end{array}$

We eliminate a factor of t in the last relationship and formally solve 
for time: $\mathrm{t}=\frac{2\mathrm{v}}{{\mathrm{a}}_{\mathrm{x}}}=\frac{2(3.0\mathrm{m}/\mathrm{s})}{\left(0.40\mathrm{m}/{\mathrm{s}}^2\right)\sin \mathrm{\theta }}$

This is then plugged into Eq. (i) to produce
$30\mathrm{m}=\frac{1}{2}\left[\left(0.40\mathrm{m}/{\mathrm{s}}^2\right)\cos \mathrm{\theta }\right]{\left(\frac{2(3.0\mathrm{m}/\mathrm{s})}{\left(0.40\mathrm{m}/{\mathrm{s}}^2\right)\sin \mathrm{\theta }}\right)}^2$

which, with the use of ${\sin }^2\mathrm{\theta }=1-{\cos }^2\mathrm{\theta },$ simplifies to 

$30=\frac{9.0}{0.20}\frac{\cos \mathrm{\theta }}{1-{\cos }^2\mathrm{\theta }}\Rightarrow 1-{\cos }^2\mathrm{\theta }\frac{9.0}{(0.20)\left(30\right)}\cos \mathrm{\theta }$

We use the quadratic formula (choosing the positive root) to solve for cos $\mathrm{\theta }$:

$\cos \mathrm{\theta }=\frac{-1.5+\sqrt{1{.5}^2-4(1.0)(-1.0)}}{2}=\frac{1}{2}$

which yields $\mathrm{\theta }={\cos }^{-1}\left|\frac{1}{2}\right|={60}^{\circ }$.