A sphere of radius $$a$$ and mass $$m$$ rolls along a horizontal plane with constant speed $$v_0$$. It encounters an inclined plane at angle $$\theta$$ and climbs upward. Assuming that it rolls without slipping, how far up the sphere will travel?

$$I = \frac{2}{5}ma^2$$

$$K_i = \frac{1}{2}mv_0^2 + \frac{1}{2}I\omega_0^2$$

$$K_i = \frac{1}{2}mv_0^2 + \frac{1}{2}\left(\frac{2}{5}ma^2\right)\left(\frac{v_0}{a}\right)^2 = \frac{1}{2}mv_0^2 + \frac{1}{5}mv_0^2 = \frac{7}{10}mv_0^2$$

$$K_i = mgh \implies \frac{7}{10}mv_0^2 = mgh \implies h = \frac{7v_0^2}{10g}$$

Distance traveled along the inclined plane ($$s$$):

$$h = s\sin\theta \implies s = \frac{h}{\sin\theta} = \frac{7v_0^2}{10g\sin\theta}$$