Consider the diagram,

Let us assume $t=0$ when $\theta=\theta_0$, then $\theta=\theta_0 \cos \omega t$

Given a seconds pendulum $\omega=2 \pi \Rightarrow \theta=\theta_0 \cos 2 \pi t\quad \text{... (i)}$

At time $t_1$ let $\theta=\theta_0 / 2$

$$\begin{aligned} \therefore \quad \cos 2 \pi t_1 & =1 / 2 \Rightarrow t_1=\frac{1}{6} \quad\left[\because \cos 2 \pi t_1=\cos \frac{\pi}{3} \Rightarrow 2 \pi t_1=\frac{\pi}{3}\right] \\ \frac{d \theta}{d t} & =-\left(\theta_0 2 \pi\right) \sin 2 \pi t \quad \text{[from Eq. (i)]} \end{aligned}$$

$$\begin{aligned} \text { At } \quad t & =t_1=\frac{1}{6} \\ \frac{d \theta}{d t} & =-\theta_0 2 \pi \sin \frac{2 \pi}{6}=-\sqrt{3} \pi \theta_0 \end{aligned}$$

Negative sign shows that it is going left.

Thus, the linear velocity is

$$u=-\sqrt{3} \pi \theta_0 l \text { perpendicular to the string. }$$

The vertical component is

$$u_y=-\sqrt{3} \pi \theta_0 l \sin \left(\theta_0 / 2\right)$$

and the horizontal component is

$$u_x=-\sqrt{3} \pi \theta_0 l \cos \left(\theta_0 / 2\right)$$

At the time it snaps, the vertical height is

$$H^{\prime}=H+l\left(1-\cos \left(\theta_0 / 2\right)\right)\quad \text{... (ii)}$$

$$\begin{aligned} &\text { Let the time required for fall be } t \text {, then }\\ &H^{\prime}=u_y t+(1 / 2) g t^2 \quad \text { (notice } g \text { also in the negative direction) } \end{aligned}$$

or $$\frac{1}{2} g t^2+\sqrt{3} \pi \theta_0 l \sin \frac{\theta_o}{2} t-H^{\prime}=0$$

$\therefore$ $$t=\frac{-\sqrt{3} \pi \theta_0 l \sin \frac{\theta_0}{2} \pm \sqrt{3 \pi^2 \theta_0^2 l^2 \sin ^2 \frac{\theta_0}{2}+2 g H^{\prime}}}{g}$$

$$=\frac{-\sqrt{3} \pi l \frac{\theta_0^2}{2} \pm \sqrt{3 \pi^2\left(\frac{\theta_0{ }^4}{4}\right) l^2+2 g H^{\prime}}}{g}\left[\because \sin \frac{\theta_0}{2} \approx \frac{\theta_0}{2} \text { for small angle }\right]$$

$$\begin{aligned} &\text { Given that } \theta_0 \text { is small, hence neglecting terms of order } \theta_0^2 \text { and higher }\\ &t=\sqrt{\frac{2 H^{\prime}}{g}} \quad \text{[from Eq. (iii)]} \end{aligned}$$

Now, $$H^{\prime} \approx H+l(1-1) \quad\left[\therefore \cos \theta_0 / 2 \approx 1\right]$$

$$ \begin{aligned} &\begin{aligned} & =H \quad \text { [from Eq. (ii)] }\\ \Rightarrow \quad t & =\sqrt{\frac{2 H}{g}} \end{aligned}\\ \end{aligned}$$

$$\begin{aligned} &\text { The distance travelled in the } x \text {-direction is } u_x t \text { to the left of where the bob is snapped }\\ &\begin{gathered} X=U x t=\sqrt{3} \pi \theta_0 l \operatorname{Cos}\left(\frac{\theta_0}{2}\right) \sqrt{\frac{2 H}{g}} \mathrm{~s} \\ \text { As } \theta_0 \text { is small } \Rightarrow \cos \left(\frac{\theta_0}{2}\right) \approx 1 \\ X=\sqrt{3} \pi \theta_0 l \sqrt{\frac{2 H}{g}}=\sqrt{\frac{6 H}{g}} \theta_0 l \pi \end{gathered} \end{aligned}$$

At the time of snapping, the bob was at a horizontal distance of $l \sin \left(\theta_0 / 2\right) \approx l \frac{\theta_0}{2}$ from $A$.

Thus, the distance of bob from $A$ where it meets the ground is

$$\begin{aligned} & =\frac{l \theta_0}{2}-X=\frac{l \theta_0}{2}-\sqrt{\frac{6 H}{g}} \theta_0 l \pi \\ & =\theta_0 l\left(\frac{1}{2}-\pi \sqrt{\frac{6 H}{g}}\right) \end{aligned}$$