Center of mass Help!! Physics and Engineering?

[muata2]

In the 1968 Olympic Games, University of Oregon jumper Dick Fosbury introduced a new technique of high jumping called the "Fosbury flop." It contributed to raising the world record by about 30 cm and is presently used by nearly every world-class jumper. In this technique, the jumper goes over the bar face up while arching her back as much as possible, as shown below. This action places her center of mass outside her body, below her back. As her body goes over the bar, her center of mass passes below the bar. Because a given energy input implies a certain elevation for her center of mass, the action of arching her back means her body is higher than if her back were straight. As a model, consider the jumper as a thin, uniform rod of length L. When the rod is straight, its center of mass is at its center. Now bend the rod in a circular arc so that it subtends an angle of 94.0° at the center of the arc, as shown in Figure (b) below. In this configuration, how far outside the rod is the center of mass? The answer should be a multiple of L. (....*L).
I cant seem to figure it out, I have no clue what I am doing wrong. Thank you very much in advanced please provide a real number answer with work. Thank you!!!

[orson19]

Let
? = angle of arc = 94º
a = radius of arc
L = length of arc = a?
? = angle along the arc, measured from y-axis, in +x direction
y = a cos?

Not being shown the figure, I will set the y-axis vertically, positive upward, and the x-axis horizontal, with the arc in the xy-plane, centered at the origin, and with its midpoint on the +y-axis. By symmetry, the center of mass (CM) will lie on the y-axis, and it will be:

CM(y) = (1/?) ?[?=-½?,½?] y d?
= (2a/?) ?[?=0,½?] cos? d?
= (2L/?²) sin? [?=0,½?]
= (2L/?²) sin(½?)

And its distance below the arc is
a - CM(y) = L/? - (2L/?²) sin½?
= [1/? - (2/?²) sin½?] L
= (0.0660953...)L