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=0.62J C. 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.