Here since friction is not present in this case so we will use mechanical energy conservation law
As per this mechanical energy conservation law sum of potential energy and kinetic energy will always remains constant
To find the speed at point B we will have
[tex]\frac{1}{2}mv_1^2 + mgh_1= \frac{1}{2}mv_2^2 + mgh_2[/tex]
[tex]0 + mg*10= \frac{1}{2}mv_2^2 + mg*4[/tex]
[tex]mg*(10 - 4) = \frac{1}{2}mv_2^2 [/tex]
[tex]2*g*(10 - 4) = v_2^2 [/tex]
[tex]v = 10.95 m/s[/tex]
Now similarly for speed at point C
[tex]\frac{1}{2}mv_1^2 + mgh_1= \frac{1}{2}mv_2^2 + mgh_2[/tex]
[tex]0 + mg*10= \frac{1}{2}mv_2^2 + 0[/tex]
[tex]mg*(10) = \frac{1}{2}mv_2^2 [/tex]
[tex]2*g*(10) = v_2^2 [/tex]
[tex]v = 14.14 m/s[/tex]
