A uniform inextensible string of length $L=6m$ and linear mass density $\mu$ is suspended vertically and a transverse pulse is produced at the lower end. At the same time a point object is released from rest from the top of the string and falls freely. How far (in meter) from the top does the object meet the pulse?

Tension in the string at a height y from the lower and = Weight of the lower part (having y length =  $\mu yg$
Wave speed at this point  $V=\sqrt{\frac{T}{\mu }}$
$$\begin{gathered} \Rightarrow V=\sqrt{\frac{\mu yg}{\mu }}=\sqrt{yg} \\ \Rightarrow \frac{dy}{dt}=\sqrt{yg} \end{gathered}$$

On integration we get,
$y=\frac{g{t}^2}{4}$
• Distance travelled by the body in time $t=\frac{1}{2}g{t}^2$ When the pulse & the body are at same level 
$$\begin{gathered} \frac{g{t}^2}{4}+\frac{1}{2}g{t}^2=L \\ \Rightarrow t=\sqrt{\frac{4L}{3g}} \end{gathered}$$
Therefore, distance from the top 
$=\frac{1}{2}g{t}^2=\frac{2L}{3}$