Assume that t = 0 when θ = θ_{2}. Then,

θ = θ_{2} cos *ω*t

Given a seconds pendulum *ω* = 2π

At time t_{1}, let θ = θ_{0}/2

cos 2πt_{1} = 1/2 t_{1}=1/6

θ = θ_{0}2π sin 2πt [θ = dθ/dt]

At t_{1}= 1/6

θ = -θ_{0}2π sin 2π/6 = - √3πθ_{0}

Thus the linear velocity is u = –√3πθ_{0}l perpendicular to the string.

The vertical component is u_{y} = -√3πθ_{0}l sinθ_{0} and the horizontal component is u_{y} = -√3πθ_{0}l cosθ_{0}

At the time it snaps, the vertical height is H' = H+l(1 - cos(θ_{0}/2))

Let the time required for fall be t, then H' u_{y}t + gt^{2} (notice g is also in the negative direction)

Or, (1/2)gt^{2} + √3πθ_{0}l sinθ_{0}t - H' = o

Neglecting terms of order θ^{2}_{0} and heigher

t≃ √2H'/g.

Now H' ≃ H + l(1–1) = H∴ t≃ √2H/g

The distance travelled in the x direction is u_{x}^{t} to the left of where it snapped.

X = √3πθ_{0}lcosθ_{0}√2H/g

To order of θ_{0}.

X =√3πθ_{0}l√2H/g = √(6H/g)θ_{0}l.

At the time of snapping, the bob was l sinθ_{0} ≃ lθ_{0} distance from A.

Thus, the distance from A is

lθ_{0} - √(6H/g)lθ_{0} = lθ_{0} (1 - √(6H/g).