Question

A ball is at rest on an inclined plane, as shown below. When it is released, what will its velocity be at the bottomof the ramp?

Answer 1

`\begin{gathered} KE=PE \\ (mv^2)/(2)=mgh \\ (v^2)/(2)=gh \\ v=√(2gh) \\ v=√(2\cdot9.81m/s^2\cdot.55)m \\ v=3.28m/s \end{gathered}`