Calculating Gravitational Potential Energy of a Child on a Swing

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To calculate the gravitational potential energy of a child on a swing at a 30.9° angle with the vertical, the height (h) must be determined using the formula h = R * (1 - cos(30.9°)), where R is the length of the swing's ropes (2.14 m). The gravitational potential energy (U) can then be calculated using U = mgh, where m is the child's mass (30.3 N divided by gravitational acceleration). Previous attempts to derive the energy using sine functions were incorrect, as the correct approach involves using the cosine function to find the height above the ground. Thus, the potential energy is given by U = mg(R * (1 - cos(30.9°))). This method accurately reflects the increase in height and potential energy as the swing moves.
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A 30.3 N child is in a swing that is attached to ropes 2.14 m long. What is the gravitational potential energy when the ropes make a 30.9° angle with the vertical?

So far I've tried 6 different answers ranging between 0-64.8 J and no luck. Need a clue to get me on the right track. Thanks in advance.
 
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You know E_{gravitational potential}=mgh, where h is height off the ground and m is mass. I believe you must assume that when the ropes make an angle of 0 with the vertical h=0. Does that help?
 
Pretend that its like those incline questions.
 
I did assume that when the angle is zero E=0 and when the angle is 90 E=mgh=64.8, so I thought the answer would be (sin30.9)(mgh), but that answer was incorrect. I also tried (sin59.1)(mgh) which was also incorrect, I am not sure where to go from here.
 
Well I think when the kid swings up to make a 30.9^\circ with the vertical, his height above the ground is increased. By how much? The answer is

R \cos (30.9^\circ)

So the kid's U=mg(R\cos (30.8^\circ))
 
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