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Q5E

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Fundamentals Of Differential Equations And Boundary Value Problems

Found in: Page 115

Book edition 9th

Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages 616 pages

ISBN 9780321977069

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Short Answer

An object of mass 5 kg is given an initial downward velocity of 50 m/sec and then allowed to fall under the influence of gravity. Assume that the force in newtons due to air resistance is -10v, where v is the velocity of the object in m/sec. Determine the equation of motion of the object. If the object is initially 500 m above the ground, determine when the object will strike the ground.

The equation of motion of the object is $x (t) = 4 . 91 t + 22 . 55 (1 - e^{- 2 t})$
The time takes the object hits the ground 97.24 sec.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Find the velocity

$ma = W - bv (t) ma = mg - bv (t)$

Velocity,

$v (t) = \frac{mg}{b} + (v - \frac{mg}{b}) e^{\frac{- bt}{m}} v (t) = \frac{5 (9 . 81)}{10} + (50 - \frac{5 (9 . 81)}{10}) e^{\frac{- 10 t}{5}} v (t) = 4 . 905 + 45.095 e^{- 2 t}$

Step 2: Find the equation of motion

$x (t) = \frac{mgt}{b} + \frac{m}{b} (v - \frac{mg}{b}) (1 - e^{\frac{- bt}{m}}) x (t) = 4 . 905 t + \frac{5}{10} (45 . 095) (1 - e^{- 2 t}) = 4.91 t + 22 . 55 (1 - e^{- 2 t})$

Hence the equation of motion is $x (t) = 4 . 91 t + 22 . 55 (1 - e^{- 2 t})$

Step 3: Find the value of t

Put x=500 and neglecting the exponential part

$500 = 4.91 t + 22.55 t = 97.24 \sec$

Hence, the time takes the object hits the ground 97.24 sec.

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