A mosquito is moving with a velocity $$\vec{v} = 0.5t^2\hat{i} + 3t\hat{j} + 9\hat{k}$$ m s$$^{-1}$$ and accelerating in uniform conditions. What will be the direction of mosquitoes after 2 s?

The velocity of the mosquito at any time $$t$$ is given by $$\vec{v} = 0.5t^2\hat{i} + 3t\hat{j} + 9\hat{k}$$ m/s.

At $$t = 2$$ s, the velocity components are: $$v_x = 0.5 \times 4 = 2$$ m/s, $$v_y = 3 \times 2 = 6$$ m/s, and $$v_z = 9$$ m/s.

The direction of the mosquito is along its velocity vector $$\vec{v} = 2\hat{i} + 6\hat{j} + 9\hat{k}$$. To find the angle this vector makes with the $$y$$-axis, we compute the components perpendicular and parallel to $$\hat{j}$$.

The component along $$y$$-axis is $$v_y = 6$$. The component perpendicular to the $$y$$-axis lies in the $$xz$$-plane and has magnitude $$\sqrt{v_x^2 + v_z^2} = \sqrt{4 + 81} = \sqrt{85}$$.

The angle from the $$y$$-axis is therefore $$\theta = \tan^{-1}\left(\frac{\sqrt{85}}{6}\right)$$.