a. Explanation Given data: Mass of the pitched baseball ( m ) is 145 g, which is written as 145 × 10 − 3 kg . Diameter of the ball ( d ) is 7.4 cm. Initial speed of the ball ( v i ) is 40.2 m s . Formula used: Refer to the equation (5.12) in the textbook, and modify the expression for drag force on the on the baseball as follows. D = 1 2 C D ρ A ( v i ) 2 (1) Here, C D is the drag coefficient for the pitched baseball, ρ is the density of the air, which is 1.2 kg m 3 at sea level, A is the cross-sectional area of the baseball, and v i is the initial speed of the baseball. Write the expression for force on the baseball as follows. F = m a (2) Here. a is the acceleration of the baseball. Write the expression for cross-sectional area of the baseball as follows. A = π 4 d 2 (3) Here, d is the diameter of the ball. Explanation: Calculation of cross-sectional area of the baseball ( A ) : Substitute 7.4 cm for d in equation (3), A = π 4 ( 7.4 cm ) 2 = π 4 ( 7.4 × 10 − 2 m ) 2 = 43.0084 × 10 − 4 m 2 The drag force on the baseball is equal in magnitude and opposite in direction with the force on the baseball. Therefore, write the expression for the drag force on the baseball as follows. D = − F Substitute m a for F , D = − m a (4) Calculation of acceleration of the baseball due to the drag force ( a ) : Refer to the TABLE 5.4 in the textbook for Drag coefficients. From the table, the drag coefficient for the pitched baseball is 0.35. C D = 0.35 From equation (4), substitute ( − m a ) for D in equation (1), − m a = 1 2 C D ρ A ( v i ) 2 Rearrange the expression to find the acceleration of the baseball as follows b. To determine To find: The final speed of the ball.

4th Edition

ISBN: 9780137574728

Author: Knight

Publisher: PEARSON

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Question

Chapter 5, Problem 69GP

To determine

To find: The magnitude of the ball’s acceleration due to the drag force.

To determine

To find: The final speed of the ball.

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Chapter 5, Problem 68GP

Chapter 5, Problem 70GP

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