1. Initial rolling motion on the horizontal plane

Since the sphere rolls without slipping with speed $v_0$,

$\omega_0 = \frac{v_0}{a}$

For a solid sphere, the moment of inertia about its center is

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

So the initial total kinetic energy is

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

Substitute $I$ and $\omega_0$:

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

$K_i = \frac{1}{2}mv_0^2 + \frac{1}{5}mv_0^2$

$K_i = \frac{7}{10}mv_0^2$

2. As the sphere climbs the incline

Because it continues to roll without slipping, static friction does no net work, so mechanical energy is conserved.

At the highest point reached, the sphere momentarily comes to rest:

$v=0, \quad \omega=0$

Hence all the initial kinetic energy converts into gravitational potential energy.

If the sphere moves a distance $s$ along the incline, the vertical rise is

$h = s\sin\theta$

Thus the gain in potential energy is

$U = mgh = mg(s\sin\theta)$

3. Apply conservation of energy

$\frac{7}{10}mv_0^2 = mg s\sin\theta$

Cancel $m$:

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

So,

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

4. Match with the options

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

This corresponds to Option B.

5. Comparison with stored correct answer

Stored correct answer: B

Derived answer: B

They match.