Assume that t = 0 when θ = θ2. Then,
θ = θ2 cos ωt
Given a seconds pendulum ω = 2π
At time t1, let θ = θ0/2
cos 2πt1 = 1/2 t1=1/6
θ = θ02π sin 2πt [θ = dθ/dt]
At t1= 1/6
θ = -θ02π sin 2π/6 = - √3πθ0
Thus the linear velocity is u = –√3πθ0l perpendicular to the string.
The vertical component is uy = -√3πθ0l sinθ0 and the horizontal component is uy = -√3πθ0l cosθ0
At the time it snaps, the vertical height is H' = H+l(1 - cos(θ0/2))
Let the time required for fall be t, then H' uyt + gt2 (notice g is also in the negative direction)
Or, (1/2)gt2 + √3πθ0l sinθ0t - H' = o
Neglecting terms of order θ20 and heigher
Now H' ≃ H + l(1–1) = H∴ t≃ √2H/g
The distance travelled in the x direction is uxt to the left of where it snapped.
X = √3πθ0lcosθ0√2H/g
To order of θ0.
X =√3πθ0l√2H/g = √(6H/g)θ0l.
At the time of snapping, the bob was l sinθ0 ≃ lθ0 distance from A.
Thus, the distance from A is
lθ0 - √(6H/g)lθ0 = lθ0 (1 - √(6H/g).