What is the maximum angle measured from the vertical that the pendulum will swing, after the projectile impact

What is , the maximum angle measured from the vertical that the pendulum will swing, after the projectile impacts the pendulum?

A pendulum consists of a slender rod, AB, of weight = W_s 8.80lb and a wooden sphere of weight W_s = 20.6 .(Figure 1) The length of the rod is = 5.60 ft and the radius of the sphere is = 0.500 ft. A projectile of weight = 0.500 lb strikes the center of the sphere at a velocity of = 741 ft/sand becomes embedded in the center of the sphere. What is _omega_, the angular velocity of the pendulum, immediately after the projectile strikes the sphere?

I have recieved: 2.57 rad/s, now my question is how to find the What is , the maximum angle measured from the vertical that the pendulum will swing, after the projectile impact?

Initial kinetic energy i have recieved is 90.21 ft * lb

Projectile's final potential energy:
V_p=W_p(d_1+R)(1-cos(theta))

Sphere's final potential energy:

V_s=W_s(d_1+R)(1-cos(theta))

Rod's final potential energy

V_r=W_r(d_1/d_2)(1-cos(theta))

I'm lost at this point, what do i have to do, to get the maximum angle ?

Help greately appreciated.

m1v(L+R) = (m1+m2)(L+R)^2 omega + 1/3 m3 L^2 omega so
2260 = 785.1 omega + 109 omega and omega = 2.53 rad/s close enough!

all rotational energy = 1/2 I omega^2 = 1/2 (785.1 + 109) 2.53^2 = 2862 lb.ft^2/s^2

This goes into potential energy = g (1 - costheta)( 8.80 x 5.6/2 + 21.1 x 6.1) = 4907(1 - costheta)

so 1- costheta = 0.583 and costheta = 0.417 then theta = 65 degrees.
Sounds about right for a VERY fast moving incident ball.