A solid, uniform density sphere with mass $M$, radius $R$, and moment of inertia $I=\frac{2}{5} M R^{2}$ rolls without slipping towards a ramp with translational speed $v$. How high $(h)$ above its initial height will it get when it comes to rest? Assume the acceleration due to gravity has magnitude g.
$\frac{5 v^{2}}{8 R}$
$\frac{3 v^{2}}{2 g R}$
$\frac{7 v^{2}}{10 g}$
$\frac{3 v}{10 g}$

Step 1 ：We begin by considering the initial kinetic energy of the sphere, which is due to both its translational and rotational motion. The total initial kinetic energy is given by \(\frac{1}{2} M v^{2} + \frac{1}{2} I \omega^{2}\).

Step 2 ：Since the sphere is rolling without slipping, we have \(v = R \omega\), so we can substitute this into the kinetic energy to get \(\frac{1}{2} M v^{2} + \frac{1}{2} \left(\frac{2}{5} M R^{2}\right) \left(\frac{v}{R}\right)^{2}\).

Step 3 ：Simplifying this expression gives us \(\frac{1}{2} M v^{2} + \frac{1}{5} M v^{2} = \frac{7}{10} M v^{2}\) as the total initial kinetic energy.

Step 4 ：When the sphere comes to rest at the top of the ramp, all of its kinetic energy has been converted into gravitational potential energy, which is given by \(M g h\).

Step 5 ：Setting the initial kinetic energy equal to the final potential energy gives us \(\frac{7}{10} M v^{2} = M g h\).

Step 6 ：Solving this equation for \(h\) gives us \(h = \frac{7 v^{2}}{10 g}\).

Step 7 ：So, the sphere will rise to a height of \(\boxed{\frac{7 v^{2}}{10 g}}\) above its initial height.