The bob of a stationary pendulum is given a sharp hit to impart it a horizontal speed of $\sqrt{3} \mathrm{gl}$. Find the angle -rotated by the string before it becomes slack.
$\frac{1}{2} \mathrm{~m} \mathrm{v}^{2}-\frac{1}{2} \mathrm{mu}^{2}=-\mathrm{mgh}$
$v^{2}=3 g l-2 g l(1+\cos \theta) \ldots \ldots \ldots 1$
and
$\left(m v^{2}\right) / l=m g \cos \theta$
$\mathrm{v}^{2}=\mathrm{gl} \cos \theta \ldots \ldots \ldots 2$
from equation 1 and 2
$3 g l-2 g l-2 g l \cos \theta=g l \cos \theta$
$\theta=\cos ^{-1}\left(\frac{1}{3}\right)$
Thus,
Angle $=180^{\circ}-\cos ^{-1}\left(\frac{\frac{1}{3}}{3}\right)$
Angle $=\cos ^{-1}(-1 / 3)$
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