"The hope is on the table"

Suppose that the rope is $L$ m long. When $x$ m of rope hangs off the edge of the table, the centre of mass of the hanging section of rope is a distance $\tfrac12x$ m below the tabletop. Thus the kinetic energy of the rope is $\tfrac12m\dot{x}^2$ , while the gravitational potential energy of the rope is $\frac{x}{L}mg \big(-\tfrac12x\big) \; = \; -\frac{mgx^2}{2L}$ Conservation of energy tells us that $\begin{aligned} \tfrac12m\dot{x}^2 - \frac{mgx^2}{2L} & = \; -\tfrac18mgL \\ \dot{x} & = \; \sqrt{\tfrac{g}{L}\big(x^2 - \tfrac14L^2\big)} \end{aligned}$ and so the time for the rope to fall off the table is $T \; = \; \sqrt{\tfrac{L}{g}}\int_{\frac12L}^L \frac{dx}{\sqrt{x^2 - \tfrac14L^2}} \; = \; \sqrt{\tfrac{L}{g}}\cosh^{-1}2$ using the substitution $x = \tfrac12L\cosh u$ to evaluate this integral. With $L=0.1$ and $g=10$ , we obtain a time of $\boxed{0.1316957897}$ seconds.