Velocity of the solid cylinder, v = 5 m/s

Angle of inclination, θ = 30°

Height reached by the cylinder =h

Energy of the cylinder at point A:

$\begin{array}{l}K{E}_{rot}=K{E}_{\text{ trans }}\\ \frac{1}{2}I{\omega }^2=\frac{1}{2}m{v}^2\end{array}$
Energy of the cylinder at point B = mgh

Using the law of conservation of energy, 

$\frac{1}{2}I{\omega }^2+\frac{1}{2}m{v}^2=mgh$
$\begin{array}{l}\Rightarrow \frac{1}{2}\left(\frac{1}{2}{\mathrm{mr}}^2\right){\mathrm{\omega }}^2+\frac{1}{2}{\mathrm{mv}}^2=\mathrm{mgh}\\ =\frac{1}{4}{\mathrm{mr}}^2{\mathrm{\omega }}^2+\frac{1}{2}{\mathrm{mv}}^2=\mathrm{mgh}\\ \frac{1}{4}{\mathrm{v}}^2+\frac{1}{2}{\mathrm{v}}^2=\mathrm{gh} \left(as, v=r\omega \right)\\ \frac{3}{4}{\mathrm{v}}^2=\mathrm{gh}\\ \Rightarrow \mathrm{h}=\frac{3}{4}\frac{{\mathrm{v}}^2}{\mathrm{g}}\\ \\ =\frac{3}{4}\times \frac{5\times 5}{9.8}=1.91\mathrm{m}\end{array}$
In ΔABC:

$\begin{array}{l}\sin \theta =\frac{BC}{AB}\\ \sin {30}^{\circ }=\frac{h}{AB}\\ AB=\frac{1.91}{0.5}=3.82m\end{array}$

Hence, the cylinder will go 3.82 m up the inclined plane.

(b) For a radius of gyration K, the velocity of the cylinder at the instance when it rolls back to the bottom is,

$\begin{array}{l}\mathrm{v}={\left(\frac{2\mathrm{gh}}{1+\frac{{\mathrm{K}}^2}{{\mathrm{R}}^2}}\right)}^{\frac{1}{2}}\\ \Rightarrow \mathrm{v}={\left(\frac{2\mathrm{gABsin}\mathrm{\theta }}{1+\frac{{\mathrm{K}}^2}{{\mathrm{R}}^2}}\right)}^{\frac{1}{2}}\\ \text{ For the solid cylinder, }{\mathrm{K}}^2=\frac{{\mathrm{R}}^2}{2}\end{array}$
$\begin{array}{l}\Rightarrow \mathrm{v}={\left(\frac{2\mathrm{gABsin}\mathrm{\theta }}{1+\frac{1}{2}}\right)}^{\frac{1}{2}}\\ ={\left(\frac{4}{3}\mathrm{gABsin}\mathrm{\theta }\right)}^{\frac{1}{2}}\end{array}$

The time taken to return to the bottom is,

$\begin{array}{l}\mathrm{t}=\frac{\mathrm{AB}}{\mathrm{v}}\\ =\frac{\mathrm{AB}}{{\left(\frac{4}{3}\mathrm{gABsin}\mathrm{\theta }\right)}^{\frac{1}{2}}}={\left(\frac{3\mathrm{AB}}{4\mathrm{gsin}\mathrm{\theta }}\right)}^{\frac{1}{2}}\\ ={\left(\frac{11.46}{19.6}\right)}^{\frac{1}{2}}=0.764\mathrm{s}\end{array}$

Therefore, the total time taken by the cylinder to return to the bottom$=2\times 0.764 =1.53 s$