When a solid cylinder rolls up an incline, its total initial kinetic energy is converted into gravitational potential energy at the highest point. Total kinetic energy of a rolling solid cylinder: $K = K_{trans} + K_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$ For a solid cylinder, moment of inertia $I = \frac{1}{2}mr^2$. In pure rolling, $\omega = \frac{v}{r}$. $K = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{1}{2}mr^2\right)\left(\frac{v}{r}\right)^2 = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$ Let the distance travelled along the incline be $s$. The vertical height reached is $h = s \sin 30^\circ$. By conservation of mechanical energy: $mgh = \frac{3}{4}mv^2$ $mgs \sin 30^\circ = \frac{3}{4}mv^2$ $s = \frac{3v^2}{4g \sin 30^\circ}$ Substitute the given values ($v = 4 \text{ m/s}$, $g = 10 \text{ m/s}^2$, $\sin 30^\circ = 0.5$): $s = \frac{3 \times (4)^2}{4 \times 10 \times 0.5} = \frac{3 \times 16}{20} = \frac{48}{20} = 2.4 \text{ m}$.