Explanation Given: The object is of mass 160 lbs was dropped 500 feet down into a 75 foot tall tank of liquid. Formula used: The differential equation for calculating the time and impact velocity is given below. v = m k g − m k g e − k m t where k is coefficient of air resistance. Calculation: When an object falls through the air its velocity can be formulated by the differential equation: m d v d t = m g − k v The above equation can be also written as, d v d t + k m v = g … … ( 1 ) Solve the above differential equation by first order differential equation. Here, Integrating factor = e ∫ k m d t = e k m t Multiply the equation (1) with the integrating factor and integrate. Therefore, e k m t d v d t + e k m t k m v = g e k m t d d t ( v e k m t ) = g e k m t v = m k g + c e − k m t … … ( 2 ) The velocity of the object, as it leaves the helicopter is zero. So, use the initial condition v ( 0 ) = 0 to determine the constant c . v = m k g + c e − k m t 0 = m k g + c e − k m .0 c = − m k g Substitute c = − m k g for c in the equation (2). Then, the equation ( 2 ) becomes, v = m k g − m k g e − k m t . Substitute 160 lbs for m , 32 ft/s 2 for g and 0.25 for k in the above equation to calculate the velocity of the object. v = 20480 − 20480 e − 1 640 t d s d t = 20480 − 20480 e − 1 640 t Integrate the above equation to find the position equation. ∫ d s = ∫ ( 20480 − 20480 e − 1 640 t ) d t s = 20480 t + 13107200 e − 1 640 t + d The position of the object at the level of the helicopter is zero. Therefore, use the initial condition s ( 0 ) = 0 in the above equation to get the constant term d s = 20 , 480 t + 13 , 107 , 200 e − 1 640 t + d 0 = 20 , 480 × 0 + 13 , 107 , 200 e − 1 640 .0 + d d = − 13 , 107 , 200 Therefore the position equation of the object is given by, s = 20 , 480 t + 13 , 107 , 200 e − 1 640 t − 13 , 107 , 200 The object enters the tank of liquid when it covered a distance of 500 feet . Then, substitute 500 feet for s in position equation, s = 20480 t + 13107200 e − 1 640 t − 13107200 500 = 20480 t + 13107200 e − 1 640 t − 13107200 Solve the above equation for t

6th Edition

ISBN: 9781284127003

Author: ZILL

Publisher: JONES+BARTLETT LEARNING,LLC (CC)

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Differential Equations

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Chapter 2.8, Problem 30E

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The time and the impact velocity when the object hits the bottom of the tank.

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Chapter 2.8, Problem 29E

Chapter 2.9, Problem 1E

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The time and the impact velocity when the object hits the bottom of the tank.

Solution Summary: The author explains the differential equation for calculating the time and impact velocity when the object hits the bottom of the tank.

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The time and the impact velocity when the object hits the bottom of the tank.

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Chapter 2 Solutions