A uniform solid sphere rolls without slipping on a horizontal surface and then up a rough incline of angle 30^°. When it reaches the foot of the…

For rolling without slipping the deceleration up the incline is $a=\frac{g\sin\theta}{1+k^2/R^2}=\frac{5}{1+2/5}=\frac{25}{7}\ \mathrm{m\,s^{-2}}$ for a solid sphere. Hence $v^2=v_0^2-2as=70-2\cdot\frac{25}{7}\cdot 4.2=70-30=40\ \mathrm{m^2s^{-2}}$. After leaving the edge only the vertical component $v\sin30^\circ$ matters for the rise, so $h=\frac{v^2\sin^2 30^\circ}{2g}=\frac{40\times 0.25}{20}=0.50\ \mathrm{m}$. A frictionless block (no rotational energy) would give $v^2=28$ and $h=0.35$ m; using the solid-cylinder factor $2/3$ gives $0.53$ m; forgetting the $\sin^2$ factor gives $2.0$ m. Rotation of the sphere is irrelevant after take-off, since no torque acts on it. Note that friction does no work in pure rolling, so mechanical energy is conserved and the sum of translational and rotational kinetic energy fixes the speed at the top. The quantity to be careful with is the factor for the moment of inertia, $\frac25MR^2$, which makes the effective inertia $\frac75m$.