a. Explanation Given: The position of the object without air is governed by the differential equation s ' ' ( t ) = − g , where g = 9.8 m / s 2 is the acceleration due to gravity. Calculation: Consider the object is fired at time t = 0 and that s ( t ) is the position of the object at time t ≥ 0 . Integrating both sides of the differential equation s ″ ( t ) = − 9.8 gives the velocity v ( t ) . ∫ s ″ ( t ) d t = − ∫ 9.8 d t s ′ ( t ) = − 9.8 t + C 1 Therefore, the velocity is v ( t ) = − 9.8 t + C 1 (1) Use the initial condition v ( 0 ) = 49 to evaluate the constant C 1 . v ( 0 ) = − 9.8 × 0 + C 1 49 = 0 + C 1 49 = C 1 Therefore, the values is C 1 = 49 . Substitute 49 for C 1 in equation (1), v ( t ) = − 9.8 t + 49 . Therefore, the velocity function is v ( t ) = − 9.8 t + 49 . Again integrate both sides of the velocity equation v ( t ) gives the position function s ( t ) . ∫ v ( t ) d t = ∫ ( − 9 b. To determine To find: The time and the height of the object when reached at the point of trajectory.

Student Solutions Manual, Single Variable for Calculus: Early Transcendentals

2nd Edition

ISBN: 9780321954329

Author: William L. Briggs, Lyle Cochran, Bernard Gillett

Publisher: PEARSON

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Chapter D1.1, Problem 30E

To determine

To find: The position and velocity functions for the object which is above the ground.

To determine

To find: The time and the height of the object when reached at the point of trajectory.

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Chapter D1.1, Problem 29E

Chapter D1.1, Problem 31E

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Chapter D1 Solutions

Student Solutions Manual, Single Variable for Calculus: Early Transcendentals

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The position and velocity functions for the object which is above the ground.

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