Kinetic and Potential Energy calculation

[Blubla]

A 215 g particle is released from rest at point A inside a smooth hemispherical bowl of radius 30.0 cm. Point B to C have Height= 2/3(R)
Calculate the following: A. The gravitational potential energy at A relative to B
B. The particle's kinetic energy at B
c. Th particle's speed at B
D. The potential energy and kinetic energy at C

I solved A,B,C and D(The potential Energy but not the kinetic energy)

Here is my work:

A. Changed 215 g into Kg so it will be 0.215 kg
30.0 cm into m so it will be 0.30m
PE=mgh
PE=0.215*9.81*0.30
PE=0.633J
B. KE= 1/2mv^2
KE=1/2*0.215*2.4^2
KE=0.62JC. V^2f=2gh
V^2f=2*9.81*030
V^2f=5.9 Apply square root
Vf= 2.4m/s

D. PE=mgh
PE=0.30*9.8(2/3*0.30)
PE=0.422

Can someone please help me find the KE at C, the answer is 0.211J according to the book. Well, what I got is 0.43J..
Thanks in advance.

 

[Phlogistonian]

You missed the point of the problem. Use conservation of energy.

 

[phyteacher]

That is true, you need to use the conservation of energy between C and B.
mgR= KEc + mg(2/3)R
KEc=1/3mgR=0.211J