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Q13E

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Fundamentals Of Differential Equations And Boundary Value Problems

Found in: Page 116

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

When the velocity v of an object is very large, the magnitude of the force due to air resistance is proportional to v² with the force acting in opposition to the motion of the object. A shell of mass 3 kg is shot upward from the ground with an initial velocity of 500 m/sec. If the magnitude of the force due to air resistance is 0.1v², when will the shell reach its maximum height above the ground? What is the maximum height?

The shell reaches at the maximum height in 2.69 sec
The maximum height is 101.9248 m.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Find the thermal speed of the object

Apply the formula for thermal speed

$v_{t} = \sqrt{\frac{mg}{b}} v_{t} = \sqrt{\frac{3 (9 . 81)}{0.1}} = 17.15$ (m = 3, g = 9.81, b = 0.1)

Step 2: Find the value of velocity

$m \frac{dv}{dt} = - mg - {bv}^{2} - \frac{m}{b} \frac{dv}{dt} = \frac{mg}{b} + v^{2} - \frac{m}{b} \frac{dv}{dt} = {v^{2}}_{t} + v^{2} \int \frac{1}{{v^{2}}_{t} + v^{2}} dv = \int \frac{- b}{m} dt \frac{\tan^{- 1} \frac{v}{v_{t}}}{v_{t}} = \frac{- bt}{m} + c$

When v = 500 m/sec and t = 0 then

$c = \frac{\tan^{- 1} \frac{500}{v_{t}}}{v_{t}}$

$\frac{\tan^{- 1} \frac{v}{v_{t}}}{v_{t}} = \frac{- bt}{m} + \frac{\tan^{- 1} \frac{500}{v_{t}}}{v_{t}} \tan^{- 1} \frac{v}{v_{t}} = \frac{- {btv}_{t}}{m} + \tan^{- 1} \frac{500}{v_{t}} \frac{v}{v_{t}} = \tan (\frac{- {btv}_{t}}{m} + \tan^{- 1} \frac{500}{v_{t}}) v = v_{t} . \tan (\frac{- {btv}_{t}}{m} + \tan^{- 1} \frac{500}{v_{t}})$

The shell reaches at maximum height when v=0 and =17.15,

Then

$0 = 17 . 15 \tan (\frac{- 0 . 1 (17 . 15) t}{3} + \tan^{- 1} \frac{500}{17.15}) \frac{- 0 . 1 (17 . 15) t}{3} = \tan^{- 1} \frac{500}{17.15} t = \frac{\tan^{- 1} \frac{500}{17.15}}{0.57} t = 2.69 \sec$

Hence, the shell reaches at the maximum height in 2.69 sec

Step 3: Find the maximum height

$v = v_{t} . \tan (\frac{- {btv}_{t}}{m} + \tan^{- 1} \frac{500}{v_{t}}) v = 17.15 \tan (- 0 . 57 t + \tan^{- 1} \frac{500}{17.15}) (b = 0.1, m = 3,t = 2.69) \frac{dx}{dt} = 17.15 \tan (- 0 . 57 t + 1 . 537) dx = 17.15 \tan (- 0 . 57 t + 1 . 537) dt x = 17 . 15 (\frac{100 \ln (\cos \frac{570 t - 1537}{1000})}{57}) + c$

When x=0 the value of c=-101.925.

$x (t) = 17 . 15 (\frac{100 \ln (\cos \frac{570 t - 1537}{1000})}{57}) + 101 . 925$

Put t=2.69 then

$x (t) = 17 . 15 (\frac{100 \ln (\cos \frac{570 (2 . 69) - 1537}{1000})}{57}) + 101 . 925 - 0.0002 + 101.925 x (t) = 101.9248 m$

Hence, the maximum height is 101.9248 m.

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