A body is rolling horizontally without slipping with speed \(v\). It then rolls up a hill to a maximum height \(h\). If \(h=3 v^{2} / 4 g\) what might the body be?

Let us start by converting the given formula \(h = \frac{3v^{2}}{4g}\) into the form \(h = \frac{v^{2}}{2g}\), since it can be matched with the term \(mgh\) in the energy equation of a rolling body. So, the factor that multiplies \(\frac{v^{2}}{2g}\) should help us to find the type of the body. Therefore, multiply the equation by 2/3 to get \(h = \frac{2}{3} \times \frac{v^{2}}{2g} = \frac{v^{2}}{3g}\).

Now let's compare this result with the energy equation for a rolling body, which is given by \(\frac{1}{2}mv^{2}+\frac{1}{2}Iw^{2}=mgh\). Here, the term \(\frac{1}{2}mv^{2}\) is the translational kinetic energy of the body and \(\frac{1}{2}Iw^{2}\) is the rotational kinetic energy. For a rolling body, \(w\) can be expressed as \(v/r\) where r is the radius of the body. So, \(\frac{1}{2}Iw^{2}\) becomes \(\frac{1}{2}I\times(\frac{v}{r})^{2}\). And for a rolling body, \(h = \frac{v^{2}}{2g}\) (where \(I = \frac{kr^{2}}{2}\) and k is the radii of gyration, r is the radius of the body) gives us the value of k. Hence, the rolling body ascends the hill with 100% of its initial kinetic energy converted to potential energy. Therefore, if \(h = \frac{v^{2}}{3g}\) (from the given equation), to balance the energy equation, the moment of inertia \(I\) should be \(\frac{2}{3}mr^{2}\) which corresponds to a solid cylinder or disk.

Now, look up the list of rotational inertias, it is clear that the body whose moment of inertia matches the derived value of \(\frac{2}{3}mr^{2}\) is a solid disk or solid cylinder.