Explanation Given information: The weight of the ball ( m ) is 5 oz or 5 16 lb . The average speed ( v a v ) is 30 ft / s . The velocity of the ball ( v ) is 90 mi / h . The distance where he pulls his hand back ( d ) is 6 in . or 0.5 ft . The acceleration due to gravity ( g ) is 32.2 ft / s 2 . Calculation: Show the free body diagram of the ball as in Figure (1). Calculate the mass of the ball ( m ) using the relation: W = m g m = W g Substitute 5 16 lb for W and 32.2 ft / s 2 for g . m = 5 16 lb 32.2 = 0.009705 lb ⋅ ft / s 2 Convert the unit of velocity of the ball ( v ) from mph to ft / s : v = ( v ) mph × ( 5280 ft mi ) × ( 1 3600 hr sec ) Here, ( v ) mph is the velocity of the ball in mph . Substitute 90 mi / h for ( v ) mph . v = 90 × ( 5280 ft mi ) × ( 1 3600 hr sec ) = 132 ft / sec Calculate the time ( t ) to reach his hand using the relation: t = d v a v Substitute 0.5 ft for d and 30 ft / s for v a v . t = 0

Vector Mechanics for Engineers: Statics and Dynamics

12th Edition

ISBN: 9781259638091

Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self

Publisher: McGraw-Hill Education

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Chapter 13.3, Problem 13.139P

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Find the average impulsive (Fav) force exerted on the player’s hand.

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Chapter 13.3, Problem 13.138P

Chapter 13.3, Problem 13.140P

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Find the average impulsive ( F a v ) force exerted on the player’s hand.

Solution Summary: The author shows the free body diagram of the ball as in Figure (1).

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