A heavy but uniform rope of length L is suspended from a ceiling....

Let the mass of the rope be $M$ and its length be $L$. The linear mass density is $\mu = \dfrac{M}{L}$.

Consider a point at a distance $x$ from the bottom of the rope. The tension $T$ at this point is due to the weight of the rope below it.

$T = (\mu x) g$

The speed of the transverse pulse at a distance $x$ from the bottom is given by:

$v = \sqrt{\dfrac{T}{\mu}} = \sqrt{\dfrac{\mu x g}{\mu}} = \sqrt{gx}$

Since $v = \dfrac{dx}{dt}$, we have:

$\dfrac{dx}{dt} = \sqrt{gx}$

$dt = \dfrac{dx}{\sqrt{gx}}$

Integrating both sides to find the total time $t$ taken by the pulse to travel from the bottom ($x = 0$) to the ceiling ($x = L$):

$t = \int_{0}^{L} \dfrac{dx}{\sqrt{gx}} = \dfrac{1}{\sqrt{g}} \int_{0}^{L} x^{-1/2} dx$

$t = \dfrac{1}{\sqrt{g}} \left[ 2\sqrt{x} \right]_{0}^{L} = 2\sqrt{\dfrac{L}{g}} = \sqrt{\dfrac{4L}{g}}$