A solid sphere of mass m and radius r is released from rest at a vertical height h=9.8 m on a rough incline and rolls down without slipping. At the…

Rolling without slipping dissipates no energy, so at the foot $mgh=\tfrac12 mv^2+\tfrac12 I\omega^2$ with $I=\tfrac25 mr^2$ and $\omega=v/r$. This gives $mgh=\tfrac12 mv^2\left(1+\tfrac25\right)=\tfrac{7}{10}mv^2$, so the translational kinetic energy is $\tfrac12 mv^2=\tfrac57\,mgh$ and the rotational part is $\tfrac27\,mgh$. On the horizontal rough stretch the sphere keeps rolling at constant $v$. On the smooth incline there is no friction torque, hence $\omega$ stays constant and the rotational energy $\tfrac27 mgh$ is locked in the spin and cannot be converted to potential energy. Only the translational energy is available: $mgH=\tfrac57 mgh$, so $H=\tfrac57(9.8)=7\ \mathrm{m}$. The common mistake is to give $H=h$ by assuming full conversion.