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31 May, 09:37

A baseball player catching a ball can soften the impact by pulling his hand back. Assuming that a 5-oz ball reaches his glove at 90 mi/h and that the player pulls his hand back during the impact at an average speed of 30ft/s over a distance of 6in. bring the ball to a stop, determine the average impulsive force exerted on the player's hand.

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  1. 31 May, 10:29
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    F_avg = 76.9 lb

    Explanation:

    Given:

    - The speed of the ball v_i = 90 mi/h = 40.2336 m/s

    - The force of the ball F = 5-oz = 0.141748 kg

    - The ball is stopped in gloves

    - The velocity with which the hand is moved back v_h = 30 ft/s

    - The distance pulled back s = 6 in = 0.5 ft

    Find:

    Determine the average impulsive force exerted on the player's hand.

    Solution:

    - The conservation of momentum is applied before and after impact:

    P_i = m_b*v_i

    P_i = 0.141748*40.2336 = 5.70303 kg-m/s

    P_f = 0

    - The total time with which the hand is moved back:

    t = s / v_avg

    t = 0.5 / 30

    t = 1 / 60 s

    - The impulse is defined as:

    Impulse = F_avg*t = change in momentum

    F_avg*t = P_f - P_i

    F_avg = 5.70303*60 = 342.18194 N

    - Conversion from SI to Imperial:

    F_avg = 342.18194/4.45 = 76.9 lb
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