Given that, $\theta = 30^{\circ}$Speed of C.M. of cylinder at the bottom,$v = 5\ \mathrm{m}\,\mathrm{s}^{-1}$(a) As cylinder goes up, it attains potential energy at the expense of its kinetic energy of translational and rotational motion. Suppose that the cylinder goes up to the height h on the inclined plane.According to the principle of conservation of energy$\frac{1}{2} m v^{2} + \frac{1}{2} I \omega^{2} = m g h$ $\therefore \frac{1}{2} m v^{2} + \frac{1}{4} m v^{2} = m g h$$\Rightarrow \frac{3}{4} m v^{2} = m g h$$h = \frac{3 v^{2}}{4 g} = \frac{3 \times 5^{2}}{4 \times 9.8} = 1.913\ \text{m}$Suppose that the cylinder covers a distance S along the inclined plane, when it goes up to the height h on the plane. If θ is the inclination of the plane, then $\sin \theta = \frac{h}{S} \Rightarrow S = \frac{h}{\sin \theta} = \frac{1.913}{\sin 30^{\circ}}$$\Rightarrow h = \frac{1.913}{\frac{1}{2}} = 3.826\ \text{m}$(b) Time taken by the cylinder to return to the bottom, $T = 2t$,where t = Time of ascending or descending.Acceleration of a body rolling down an inclined plane is given by$a = \frac{g \sin \theta}{1 + \frac{k^{2}}{r^{2}}}$where k is radius of gyration of rolling body$a = \frac{2}{3} g \sin \theta$$\text{Now}, S = u t + \frac{1}{2} a t^{2}$Since $u = 0,$ we get$t = \sqrt{ \frac{2S}{a} } = \sqrt{ \frac{2 \times 3.826}{ \frac{2}{3} g \sin \theta } }$;$t = \sqrt{ \frac{2 \times 3.826 \times 3}{2 \times 9.8 \times \sin 30^{\circ}} }$ $= 1.53\ \text{s}$$T = 2t = 2 \times 1.53 = 3.065 \approx 3.0\ \text{s}$