A particle starting from rest at the highest point slides down the outside of a smooth vertical circular track of radius 0.3 m. When it leaves the track its vertical fall is $$h$$ and the linear velocity is $$v$$. The angle made by the radius at that position of the particle with the vertical is $$\theta$$. Now consider the following observations. $$(g = 10\ \text{m/s}^2)$$ (I) $$h = 0.1$$ m and $$\cos\theta = \frac{2}{3}$$. (II) $$h = 0.2$$ m and $$\cos\theta = \frac{1}{3}$$. (III) $$v = \sqrt{2}\ \text{ms}^{-1}$$. (IV) After leaving the circular track the particle will describe a parabolic path. Therefore,