Short Answer

An object of mass 5 kg is released from rest 1000 m above the ground and allowed to fall under the influence of gravity. Assuming the force due to air resistance is proportional to the velocity of the object with proportionality constant b = 50 N-sec/m, determine the equation of motion of the object. When will the object strike the ground?

The equation of motion of the object is $x (t) = 0 . 981 t - 0.0 981 (1 - e^{- 10 t}) m$

The time taken by the object is 1019 sec.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Find the equation of motion of an object

The given values are ${m = 5, g = 9.81, v}_{0} = 0, b = 50,$

The equation of motion is role="math" localid="1664170601861" $x (t) = \frac{mgt}{b} + \frac{m}{b} (v - \frac{mg}{b}) (1 - e^{\frac{- bt}{m}})$

Put all the given values

$\frac{5 (9 . 81) t}{50} + \frac{5}{50} (0 - \frac{5 (9 . 81)}{50}) (1 - e^{\frac{- 50 t}{5}}) \frac{(9 . 81) t}{10} - \frac{(9 . 81)}{100}) (1 - e^{- 10 t}) x (t) = \frac{(9 . 81) t}{10} - \frac{(9 . 81)}{100}) (1 - e^{- 10 t}) x (t) = 0 . 981 t - 0.0 981 (1 - e^{- 10 t}) m$

Hence the equation of motion is role="math" localid="1664170624317" $x (t) = 0 . 981 t - 0.0 981 (1 - e^{- 10 t}) m$

Step 2: Finding when will the object strike the ground

$1000 = 0.981 t - 0.0 981 (1 - e^{- 10 t}) 1000 = 0.981 t - 0.0981 (B e c a u s e v a l u e o f i s s o s m a l l s o n e g l e c t i n g) 1000 + 0.0981 = 0.981 t 1000.0981 = 0.981 t t = 1000.0981 / 0.981 t = 1019 \sec$

Hence, the object will strike the ground in 1019 sec.

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Fundamentals Of Differential Equations And Boundary Value Problems

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