A particle in projected with velocity `sqrt(3gL)` at point A (lowest point of the circle) in the vertical plane. Find the maximum height about horizontal level of point A if the string slacks at the point B as shown in figure.

Explanation

As tension at B, `T=0`
`:. mg cos theta=(mv_B^2)/(L) :. v_B=sqrt(gLcos theta)` (i)
Now by equation of energy between A and B.
`0+1/2m3gL=1/2mv_B^2+mgL(1+costheta)` (ii)
From Eqs. (i) and (ii), `cos theta=1/3`
Therefore, height attended by particle after point B where the string slacks is
`h^'=(v_B^2sin^2theta)/(2g)=(gLcostheta(1-cos^2theta))/(2g)=(4L)/(27)`
Therefore, maximum height about point A is given by.
`H_(max)=L+Lcostheta+h^'=L+L/3+(4L)/(27)=(40L)/(27)`