A small bob of mass 0.20 kg hangs at rest from a fixed point O on a light, inextensible string of…

Step 1. Where the string goes slack. Let the string make an angle $\alpha$ above the horizontal. Along the string, towards O: $T + mg\sin\alpha = \dfrac{mv^2}{L}$. The string goes slack when $T = 0$, that is when $v^2 = gL\sin\alpha$.

The bob has then risen $L(1 + \sin\alpha)$, so energy gives $u^2 = v^2 + 2gL(1 + \sin\alpha)$. Together:

So $\alpha = 30^\circ$, and $v = \sqrt{gL\sin\alpha} = \sqrt{5}\ \text{m/s} = 2.24\ \text{m/s}$.

Step 2. Free flight. Put O at the origin. The slack point is $(L\cos\alpha,\ L\sin\alpha)$, and the velocity there is along the circle: $v(-\sin\alpha,\ \cos\alpha)$. As a projectile,

The string is taut again when $x^2 + y^2 = L^2$. On expanding, the terms in $t$ cancel, and so do the terms in $t^2$, because $v^2 = gL\sin\alpha$. What is left is

This is negative until $t = \dfrac{4v\cos\alpha}{g} = 0.775\ \text{s}$, so the bob flies inside the circle until then. At that moment

$x = L\cos\alpha(1 - 4\sin^2\alpha) = 0$ and $y = L\sin\alpha(1 - 4\cos^2\alpha) = -1.0\ \text{m}$: the lowest point of the circle.

Step 3. Velocity just before the jerk. Horizontally: $v\sin\alpha = 1.118\ \text{m/s}$, back towards the side it started from. Vertically: $v\cos\alpha - gt = -3v\cos\alpha$, that is $5.81\ \text{m/s}$ downwards. The speed squared is $1.25 + 33.75 = 35\ \text{m}^2/\text{s}^2 = u^2$, as energy says, since the bob is back at its starting height.

Step 4. The jerk. At the lowest point the string is vertical. The sudden pull wipes out the vertical part of the velocity, which is along the string, and leaves the horizontal part, $1.118\ \text{m/s}$.

Energy before: $\tfrac12 \times 0.20\ \text{kg} \times 35\ \text{m}^2/\text{s}^2 = 3.50\ \text{J}$. After: $\tfrac12 \times 0.20\ \text{kg} \times 1.25\ \text{m}^2/\text{s}^2 = 0.125\ \text{J}$. So $3.375\ \text{J}$ is lost.

Step 5. The swing. The bob rises $h = \dfrac{(1.118\ \text{m/s})^2}{2 \times 10\ \text{m/s}^2} = 0.0625\ \text{m}$. Then $\cos\theta = 1 - \dfrac{h}{L} = 0.9375$, so $\theta = 20.4^\circ$.