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9 September, 08:57

Ted Clubber Lang. A hook in boxing primarily involves horizontal flexion of the shoulder while maintaining a constant angle at the elbow. During this punch, the horizontal flexor muscles of the shoulder contract and shorten at an average speed of 75 cm/s. They move through an arc length of 5 cm during the hook, while the first moves through an arc length of 100 cm. What is the average speed of the first during the hook?

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  1. 9 September, 11:21
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    15 m/s or 1500 cm/s

    Explanation:

    Given that

    Speed of the shoulder, v (h) = 75 cm/s = 0.75 m/s

    Distance moved during the hook, d (h) = 5 cm = 0.05 m

    Distance moved by the fist, d (f) = 100 cm = 1 m

    Average speed of the fist during the hook, v (f) = ? cm/s = m/s

    This can be solved by a very simple relation.

    d (f) / d (h) = v (f) / v (h)

    v (f) = [d (f) * v (h) ] / d (h)

    v (f) = (1 * 0.75) / 0.05

    v (f) = 0.75 / 0.05

    v (f) = 15 m/s

    Therefore, the average speed of the fist during the hook is 15 m/s or 1500 cm/s
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