$\theta=30^{\circ}, v=5 \mathrm{~m} / \mathrm{s}$
Let $h$ be the height on the plane upto which the cylinder will go up.
$\therefore \quad$ From conservation of energy,
Total K.E $=$ P.E $\Rightarrow \frac{1}{2} m v^2+\frac{1}{2} I \omega^2=\mathrm{mgh}$
$$
\begin{aligned}
&\Rightarrow \frac{1}{2} m v^2+\frac{1}{2}\left(\frac{1}{2} m r^2\right) \omega^2=m g h\left[I_{\mathrm{cyl}}=\frac{1}{2} m r^2\right] \\
&\left.\Rightarrow \frac{3}{4} m v^2=m g h \text { [using } v=r \omega\right] \\
&\Rightarrow h=\frac{3 v^2}{4 g}=\frac{3 \times 5^2}{4 \times 9.8}=1.913 \mathrm{~m} .
\end{aligned}
$$
Let, $s=$ distance moved up by the cylinder on the inclined plane.


$$
\therefore \sin \theta=\frac{h}{s} \Rightarrow s=\frac{h}{\sin \theta}=\frac{1.913}{\sin 30^{\circ}}=3.826 \mathrm{~m}
$$
Time taken to return to the bottom $=t$
$$
=\sqrt{\frac{2 s\left(1+\frac{k^2}{r^2}\right)}{g \sin \theta}}=\sqrt{\frac{2 \times 3.826\left(1+\frac{1}{2}\right)}{9.8 \sin 30^{\circ}}}=1.53 \mathrm{~s} .
$$