How does the law of conservation apply under these condition

[Henrybar]

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 let's 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?

 

[Dr. Courtney]

Law of conservation of what quantity?

Physics has a bunch of conservation laws.

 

[Henrybar]

Law of conservation of what quantity?

Physics has a bunch of conservation laws.

Can you not read what I wrote beneath that part?

 

[CWatters]

(mv^2)/2=mg(h-h')
v=√[2g(h-h')]
=√2x9.8(7-6.5)
=3.1m/s

That was OK except the problem statement says...

..the person let's go at approximately the same height he starts (C),

Why not set h=h' ?

 

[Dr. Courtney]

Can you not read what I wrote beneath that part?

How successful do expect to be in college if you speak to faculty that way?

 

[CWatters]

Are these equations accurate under these conditions or centripetal acceleration missing?
Also, what equations are necessary to deal with changing speed in circular motion?

The equations you used are just fine for this purpose. The problem says to ignore air resistance so there is no loss of energy. Therefore you can apply conservation of energy and equate the change in KE to the change in PE.