A person of 60 kg is holding on a rope of 3m while standing on a the ledge of a building of height 7m. The rope is fixed to a point at roughly eye level 3 m from ledge. The person walks off the building and is swung in a vertical circle. If the person lets go at approximately the same height he starts (C), what is the tangential speed at this point given that the distance from (C) and the ground is 6.5m. Assume no air resistance, no elasticity in rope Relevant equations: Ek=(mv^2)/2 Eg=mgh c=2∏r v=d/t The attempt at a solution for releasing rope at (C): Energy before = Energy after Ek+Eg=Ek'+Eg' Ek=0 so, Eg=Ek'+Eg' Ek'=Eg-Eg' (mv^2)/2=mg(h-h') v=√[2g(h-h')] =√2x9.8(7-6.5) =3.1m/s Are these equations accurate under these conditions or centripetal acceleration missing? Also, what equations are necessary to deal with changing speed in circular motion?