Explanation Given information: The first ball is thrown from the ground with an initial speed of 25 m / s , and the second ball is dropped at the same instant from a 15 m height. For the condition of object moving in upward direction, the acceleration due to gravity is considered as negative. Similarly, the acceleration due to gravity is considered positive for object moving in downward direction. Formula to calculate the final position of the first ball is, y f1 − y i1 = v i1 t f + 1 2 g t f 2 y f1 is the final height of the first ball. y i1 is the initial height of the first ball. v i1 is the initial speed of the first ball. t f is the final time instant. g is the acceleration due to gravity. Substitute 0 m for y i1 , 25 m / s for v i1 , and − 9.8 m / s 2 for g in the above equation to find y f1 . y f1 − ( 0 m ) = ( 25 m / s ) t f + 1 2 ( − 9.8 m / s 2 ) t f 2 y f1 = ( 25 m / s ) t f − ( 4.9 m / s 2 ) t f 2 Formula to calculate the final position of the second ball is, y f2 − y i2 = v i2 t f + 1 2 g t f 2 y f2 is the final height of the second ball. y i2 is the initial height of the second ball. v i2 is the initial speed of the second ball. Substitute 15 m for y i2 , 0 m / s for v i2 , and − 9

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University

9th Edition

ISBN: 9781305372337

Author: Raymond A. Serway | John W. Jewett

Publisher: Cengage Learning

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Chapter 2, Problem 52P

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The time instant at which both the ball are at same height above the ground.

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Chapter 2, Problem 51P

Chapter 2, Problem 53P

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Chapter 2 Solutions

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University

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The time instant at which both the ball are at same height above the ground.

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Chapter 2 Solutions