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

Let $x$ be the distance of the pulse from the bottom of the rope at time $t$.

The tension in the rope at a distance $x$ from the bottom is $T = \mu g x$, where $\mu$ is the mass per unit length of the rope.

The speed of the transverse pulse is $v = \sqrt{\dfrac{T}{\mu}} = \sqrt{g x}$.

Since $v = \dfrac{dx}{dt}$, we have $\dfrac{dx}{dt} = \sqrt{g x}$.

Rearranging and integrating from $t = 0$ to $t$:

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

$2\sqrt{x} = \sqrt{g} t \Rightarrow x = \dfrac{g t^2}{4}$

At the same time $t$, the distance fallen by the particle dropped from the ceiling is $y = \dfrac{1}{2} g t^2$.

Comparing $x$ and $y$, we get $y = 2x$.

The particle and the pulse meet when the sum of their distances equals the total length of the rope $L$:

$x + y = L$

$x + 2x = L$

$3x = L \Rightarrow x = \dfrac{L}{3}$

Thus, they meet at a distance $\dfrac{L}{3}$ from the bottom.