A is a fixed point at a height H above a perfectly inelastic smooth horizontal plane. A light inextensible string of length$L(>H)$ has one end attached to A and other to a heavy particle. The particle is held at the level of A with string just taut and released from rest. Find the height of the particle above the plane when it is next instantaneously at rest.

In perfectly inelastic collision with the horizontal surface the vertical velocity component after the impact becomes zero and the component parallel to the surface will remain unchanged. Similarly when the string becomes taut again, the component of velocity $\left(v\text{'}\right)$ parallel to the plane  will remain unchanged.

$$\begin{gathered} \cos \theta =\frac{H}{L} \\ If v_{c }be the velocity of the bob just before impact , then conserving energy \\ \frac{1}{2 }m{v_c}^2=mgH \Rightarrow v_c=\sqrt{2gH} \end{gathered}$$

$$\begin{gathered} v\text{'}=v_c \cos \theta \\ Therefore velocity of the bob perpendicular to the string =v\text{'} \cos \theta =v_c{\cos }^2\theta \\ \Rightarrow v_c{\cos }^2\theta =\left(\sqrt{2gH}\right)\frac{H^2}{L^2}=v\left(\text{ say }\right) \\ \therefore h=\frac{v^2}{2g}=\frac{\left(2gH\right)\frac{H^4}{L^4}}{2g}=\frac{H^5}{L^4} \end{gathered}$$t