A uniform solid sphere is projected with speed 7 m s^-1 along a rough horizontal floor with no initial spin. After it begins to roll without…

While sliding, friction acts about the contact point, so angular momentum about any point on the floor along the line of contact is conserved. For a sphere, $mv_0R=mvR+\tfrac25mR^2\cdot\dfrac vR$, which gives $v_0=\tfrac75v$ and $v=\tfrac57v_0=5\ \mathrm{m\,s^{-1}}$. This is independent of the friction coefficient. On the incline there is no slipping, so mechanical energy is conserved: $\tfrac12mv^2+\tfrac12\left(\tfrac25mR^2\right)\dfrac{v^2}{R^2}=mgh$, i.e. $\tfrac{7}{10}v^2=gh$. Hence $h=\dfrac{0.7\times25}{10}=1.75\ \mathrm{m}$ and $4h=7$. Note that the answer does not depend on the friction coefficient or on the radius, because only the moment of inertia factor $2/5$ of the sphere matters in the angular momentum balance.