A correct solution came from Bradley University student Alec Witty. Other correct solutions were submitted by Bill Webb, Paul Botham, Juan Carlos Marivela, Cee Ann Franklin.

Let r be the radius of the semicircle, x, h, d, k, f be as in the sketch on the left.  Let p = x + h be the height of the flag pole; the total length of the rope is 2r so when the flag has reached its highest point we have the equation p² + (2r)² = (2p)², and (1) p = 2r/Ö3.  We are given (2) ds/dt = 20 ft/sec from which we want to determine dh/dt.  Since s = rf it follows that (3) df/dt = (1/r)×(ds/dt).  Also sin(f/2) = k/(2r), so 2rsin(f/2) = k.  Furthermore, x + d = 2p = 4r/Ö3 so the equation k² + p² = d² yield a relationship between f and h, namely (2rsin(f/2))² + p² = (p + h)² .  Differentiating implicitly with respect to time yields  (4) 2(2rsin(f/2))×(2rcos(f/2))×1/2(df/dt) = 2(p + h)×(dh/dt).  When the flag is halfway up the pole (5) h = p/2 and (6) f/2 = arcsin(Ö15/6) (Check this!).  Put together (1) through (6), season to taste, to get dh/dt = 10Ö105/16 = 11.3855...

The second part is left to you. 

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