Answer:
Explanation:
Given
Pivot is at h=0.76 L
Where L is the length of String
Conserving Energy at A and B
[tex]mgL=\frac{1}{2}mu^2[/tex]
where u=velocity at bottom
[tex]u=\sqrt{2gL}[/tex]
After coming at bottom the ball completes the circle with radius r=L-0.76 L
Suppose v is the velocity at the top
Conserving Energy at B and C
[tex]\frac{1}{2}mu^2=mg(2r)+\frac{1}{2}mv^2[/tex]
Eliminating m
[tex]u^2=4r+v^2[/tex]
[tex]v^2=u^2-4\cdot gr[/tex]
[tex]v^2=2gL-4g(L-0.76L)[/tex]
[tex]v^2=1.04gL[/tex]
[tex]v=\sqrt{1.04gL}[/tex]
