Formal Circular Motion...
A particle hanging by a light string of length 'L' is projected horizontally from its lowest point with velocity \sqrt{7gL/2} . The string slackens after swinging through _ _ _ __ degrees.
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1 solution
let the angle be x, then at the highest point- mgcosx=mv^2/r v=sqrt(rgcosx) now use third eqn. of motion putting v=sqrt(rgcosx) u=sqrt(7gl/2) a=-g s=l(1=cosx) solve for x to find x to be 120