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.

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

answer:

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