Explanation Given: Initial velocity, v ∘ = 0 , mass of the object, m = 400 − lb , proportionality constant of air resistance, b = 10 lb-sec/foot , acceleration due to gravity, g = 32 ft / sec 2 and x ( t ) = 30 ft . Approach: The free body diagram of the falling object is shown below Forces on falling object. The force due to gravity is F 1 = mg , where g is the acceleration due to gravity. The force due to air resistance is F 2 = − bv , where b ( > 0 ) is the proportionality constant and v is the velocity of the object. Total force on the falling object, F = F 1 + F 2 = mg − bv ( t ) ... ( 1 ) According to Newton’s second law F = m d v d t ... ( 2 ) ⇒ m d v d t = mg − bv ( t ) ⇒ m d v d t +bv ( t ) = mg ⇒ d v d t + b m v ( t ) = g Integrating factor, I . F . = e ∫ b m d t = e b m t ⇒ v × I .F . = g ∫ I .F . × d t ⇒ v = m g b + C e − b m t ... ( 3 ) Substituting v ( 0 ) = 0 in (3), C = − mg b ⇒ v = m g b − mg b e − b m t

Pearson eText Fundamentals of Differential Equations with Boundary Value Problems -- Instant Access (Pearson+)

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Author: R. Nagle, Edward Saff

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Chapter 3.4, Problem 4E

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The time when the falling object will hit the ground.

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Chapter 3.4, Problem 3E

Chapter 3.4, Problem 5E

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The time when the falling object will hit the ground.

Solution Summary: The author presents the free body diagram of the falling object. The force due to gravity is ft/mathrmsec

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To find: The time when the falling object will hit the ground.

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To find:

The time when the falling object will hit the ground.