The velocity of a rolling object at the bottom of an inclined plane is given by:
\[ v = \sqrt{\frac{2gh}{1 + \frac{I}{mr^2}}}, \]
where:
For different objects:
Comparison:
Since \( \frac{2}{5} < \frac{1}{2} < 1 \), the solid sphere has the smallest value of \( \frac{I}{mr^2} \) and thus the largest velocity.
Conclusion: The solid sphere reaches the bottom with maximum velocity.