Velocity & Energy Homework: Find Speed at Low/High Points

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The discussion focuses on calculating the speed of a ball attached to a string at its lowest and highest points during a swing. The conservation of energy principle is applied, using the equations for kinetic energy (K) and gravitational potential energy (U). The initial potential energy is calculated as mgL, while the final kinetic energy is expressed as 1/2mv^2. The user attempts to solve for speed but arrives at an incorrect value of 50.28 m/s, indicating a misunderstanding or miscalculation in applying the energy conservation equations. Further clarification or correction is needed to resolve the discrepancy in the solution.
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Homework Statement


The string in Fig. 8-37 is L = 129 cm long, has a ball attached to one end, and is fixed at its other end. The distance d from the fixed end to a fixed peg at point P is 78 cm. When the initially stationary ball is released with the string horizontal as shown, it will swing along the dashed arc. What is its speed when it reaches (a) its lowest point and (b) its highest point after the string catches on the peg?

Homework Equations


K=1/2mv^2
U=mgh where in this case, h=L

The Attempt at a Solution


I know I need to use the conservation of energy (U(i) + K(i) = U(f) + K(f)) and then K(i)=0 and U(f)=0. So then you can use the equation mgL = 1/2mv^2 and the m's cancel out so youre left with gL = 1/2v^2. I took this to mean that I would have 9.8(129) = 1/2v^2. Solving for v, I got 50.28 m/s. The online program says that's wrong so I'm at a loss.
 

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I'm still at a loss for this one
 
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