$A$ simple pendulum of time period $1\,s$ and length $l$ is hung from a fixed support at $O$. The bob is at a distance $H$ vertically above $A$ on the ground. The amplitude is $\theta_0$. The string snaps at $\theta = \frac{\theta_0}{2}$. Find the time taken by the bob to hit the ground and the distance from $A$ where the bob hits the ground. Assume $\theta_0$ to be small,so that $\sin \theta_0 \approx \theta_0$ and $\cos \theta_0 \approx 1$.

(N/A) The angular frequency is $\omega = \frac{2\pi}{T} = 2\pi \text{ rad/s}$.
At $t=0$,the bob is at $\theta = \theta_0$. The angular position is $\theta(t) = \theta_0 \cos(\omega t)$.
At the moment the string snaps,$\theta = \frac{\theta_0}{2}$,so $\frac{\theta_0}{2} = \theta_0 \cos(2\pi t_1) \implies \cos(2\pi t_1) = \frac{1}{2} \implies 2\pi t_1 = \frac{\pi}{3} \implies t_1 = \frac{1}{6} \text{ s}$.
The velocity components at this instant are $v_x = l\omega \sin(\frac{\theta_0}{2})$ (horizontal) and $v_y = l\omega \cos(\frac{\theta_0}{2})$ (downward).
Since $\theta_0$ is small,$v_x \approx l\omega(\frac{\theta_0}{2})$ and $v_y \approx l\omega$.
The height of the bob from the ground at this instant is $h = H + l(1 - \cos(\frac{\theta_0}{2})) \approx H + l(1 - 1) = H$.
The vertical motion is $h = v_y t + \frac{1}{2}gt^2$,so $H = (l\omega)t + \frac{1}{2}gt^2$.
Solving for $t$ gives the time to hit the ground. The horizontal distance from $A$ is $x = l \sin(\frac{\theta_0}{2}) + v_x t = l(\frac{\theta_0}{2}) + (l\omega \frac{\theta_0}{2})t$.