.A frustum has been mounted with its axis vertical. It has a height h and radii of its upper and lower cross sections are R and r respectively. A particle is projected with horizontal velocity v0 along its upper brim. The particle spirals down the inner surface and leaves the lower face at point B. The inner wall of the frustum is smooth. 
The vertical component of velocity of the particle as it leaves the frustum at B is $V_v$ and the minimum value of h for which the particle will never come out of the frustum is $h_{min}$ (Take $\mathrm{r}=\frac{\mathrm{R}}{2}$ for solving this part of the problem). Then

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Forces on the particle are – normal reaction of the frustum wall and weight of the particle. These forces do not produce any torque about the central vertical axis. Hence, angular momentum of the particle about the central vertical axis is conserved 
$$\begin{gathered} \text{ Let }{\mathrm{v}}_{\mathrm{H}}=\text{ horizontal component of velocity at }\mathrm{B} \\ {\mathrm{mv}}_{\mathrm{H}}\mathrm{r}={\mathrm{mv}}_0\mathrm{R}\Rightarrow {\mathrm{v}}_{\mathrm{H}}={\mathrm{v}}_0\frac{\mathrm{R}}{\mathrm{r}} \end{gathered}$$
Energy conservation gives : 
$$\begin{gathered} \frac{1}{2}{\mathrm{mv}}^2=\frac{1}{2}{\mathrm{mv}}_0^2+\mathrm{mgh}\therefore {\mathrm{v}}^2={\mathrm{v}}_0^2+2\mathrm{gh} \\ {\mathrm{v}}_{\mathrm{H}}^2+{\mathrm{v}}_{\mathrm{v}}^2={\mathrm{v}}_0^2+2\mathrm{gh}\therefore {\mathrm{v}}_{\mathrm{v}}^2={\mathrm{v}}_0^2+2\mathrm{gh}-{\left({\mathrm{v}}_0\frac{\mathrm{R}}{\mathrm{r}}\right)}^2=2\mathrm{gh}-{\mathrm{v}}_0^2\left(\frac{{\mathrm{R}}^2}{{\mathrm{r}}^2}-1\right) \\ \Rightarrow {\mathrm{v}}_{\mathrm{V}}=\sqrt{2\mathrm{gh}-{\mathrm{v}}_0^2\left(\frac{{\mathrm{R}}^2}{{\mathrm{r}}^2}-1\right)} \end{gathered}$$
(b) The particle will not leave the frustum if vertical component of its velocity becomes zero before reaching the 
bottom
$$\begin{gathered} \Rightarrow 2\mathrm{gh}={\mathrm{v}}_0^2\left(\frac{{\mathrm{R}}^2}{{\mathrm{r}}^2}-1\right) \\ \Rightarrow 2\mathrm{gh}=3{\mathrm{v}}_0^2\Rightarrow \mathrm{h}=\frac{3{\left(V_0\right)}^2}{2g} \end{gathered}$$