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I don't understand why there is an additional $0.441$ added to the final distance... I do understand that the distance travelled by $A$ downwards is equal to the distance moved by $B$ upwards but why the $0.441$? I solved it until the point where you find $s_1$ an multiply it by $2$ but afterwards it makes no sense.. can someone please explain

bob
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wjkrkdfjkn
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1 Answers1

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When $A$ hits the ground $B$ continues to move upwards. The string is now loose, so $B$ moves freely as a particle under gravity. The $0.441$ is the extra distance it moves until it comes to instantaneous rest.

David Quinn
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