Calculating Impact Speed of a Wrecking Ball: A Quick Energy Question

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The discussion revolves around calculating the impact speed of a wrecking ball used for demolition. The ball, weighing 600 kg and suspended by a 37-meter cable at a 22-degree angle, strikes the building when the cable is vertical. The initial calculations led to a speed of 3.6 m/s, while the book states the correct speed is 7.3 m/s. The error was identified in the formula used; the correct equation is v = √(2gL(1 - cos(θ)). Following this corrected formula yields the expected result.
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a wreacking ball is used to demolish a building. The 600-kg ball starts from rest, with it's 37-m-long cable making a 22 degrees angle with the vertical. it strikes the building when the cable is vertical. What is the speed of the ball on impact?

it doesn't seem like a hard question, but i can't figure it out.

U_o =mgL(1-cos(\theta))
K_0 = 0

U_f = 0
K_f = 1/2mv^2

mgL(1-cos(\theta)) = 1/2mv^2
V= \sqrt{\frac{gL(1-cos(\theta))}{2}}

plugging in known values, i get 3.6 but the book gets 7.3 m/s. am i missing something?
 
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ProBasket said:
a wreacking ball is used to demolish a building. The 600-kg ball starts from rest, with it's 37-m-long cable making a 22 degrees angle with the vertical. it strikes the building when the cable is vertical. What is the speed of the ball on impact?

it doesn't seem like a hard question, but i can't figure it out.

U_o =mgL(1-cos(\theta))
K_0 = 0

U_f = 0
K_f = 1/2mv^2

mgL(1-cos(\theta)) = 1/2mv^2
V= \sqrt{\frac{gL(1-cos(\theta))}{2}}

plugging in known values, i get 3.6 but the book gets 7.3 m/s. am i missing something?
Your final equation should be:

\blacktriangleright \blacktriangleright \ \ \ \ \color{red} v \ = \ \sqrt{2gL(1 - \cos(\theta))}

Use this correct version, and you'll obtain the book answer.


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oops your right, thanks for the help
 
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