Massive Chain Lagrangian
$$L=\frac{1}{4} \mu g \left(-h^2-2 h L+L^2\right)+\frac{1}{4} \mu g(h+L) \dot{h}^2$$
$$\ddot{h}=-g-\frac{\dot{h}^2}{2(L+h)}$$
Conclusion: A massive chain's end in this initial configuration falls faster than a ball dropped from the same height!