Explanation Given: Mass of the object is 500 kg . Height above the ground is 1000 m . Proportionality constant b is 50 N-sec / m . Calculation: Consider the equation for the net force acting on the object. m d v d t = m g − b v Substitute 500 kg for m , 500 for m g and − 50 v for b v in the above equation. 500 d v d t = 500 g − 50 v d v d t = g − 0.1 v d v g − 0.1 v = d t − 10 ln | g − 0.1 v | = t + C Solve the equation further, ln | g − 0.1 v | = − t 100 + C 1 …… ( 1 ) Here, C 1 = − C 10 Take exponent on the both the sides of equation ( 1 ) . | g − 0.1 v | = C 2 e − t 10 Here, C 2 = e C 1 − g + 0.1 v = − C 2 e − t 10 v = g − C 2 e − t 10 0.1 = 10 g − C 3 e − t 10 …… ( 2 ) Substitute 0 for v and 0 for t in the above equation. 0 = 10 g − C 3 C 3 = 10 g Substitute 10 g for C 3 in equation ( 2 ) . v = 10 g − 10 g e − t 10 Substitute d x d t for v in the above equation. d x d t = 10 g ( 1 − e − t 10 ) d x 10 g = ( 1 − e − t 10 ) d t x 10 g = t + 10 e − t 10 + C 4 x = 10 g t + 100 g e − t 10 + C 5 …… ( 3 ) Here, C 5 = − 100 g Substitute 0 for x in the above equation

Fundamentals of Differential Equations and Boundary Value Problems

7th Edition

ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

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Chapter 3.4, Problem 3E

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The time when the object strikes the ground.

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The time when the object strikes the ground.

Solution Summary: The author explains that the time taken by the object to strike the ground is 18.6mathrmsec.

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To find:

The time when the object strikes the ground.