A small object of uniform density rolls up a rough curved surface with an initial translational velo

Question

A small object of uniform density rolls up a rough curved surface with an initial translational velocity $v$. It reaches a maximum vertical height of exactly $h = \frac{7v^2}{10g}$ with respect to its initial position. The geometric shape of the object is most likely a

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(d) : The initial kinetic energy of the rolling object is completely converted into potential energy at the maximum height $h$. By the law of conservation of mechanical energy: $\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = mgh$. Substituting $\omega = v/R$ and $I = mK^2$, the total kinetic energy is $\frac{1}{2}mv^2(1 + \frac{K^2}{R^2})$. Equating this to $mg(\frac{7v^2}{10g})$ gives $\frac{1}{2}(1 + \frac{K^2}{R^2}) = \frac{7}{10} \Rightarrow 1 + \frac{K^2}{R^2} = 1.4 \Rightarrow \frac{K^2}{R^2} = 0.4 = \frac{2}{5}$. The shape factor $2/5$ corresponds definitively to a solid sphere.