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1 mdv dt 10 dv dt = mg-bv; v(0) = v = (10) (9.81)-50 v

1 mdv dt 10 dv dt = mg-bv; v(0) = v = (10) (9.81)-50 v

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Can someone please tell me what I have done wrong or what I am missing in my work?  I keep getting the wrong answer on this question, and I've attempted this same problem over 23 times.  I DO NOT understand what I am doing wrong. Please tell me where I am messing up.

17 m dv = mg-bv; v(0) = v dt •10dv = (10) (9.81)-50 v dt • dv = 9.81 - 5vdt . dv = dt 9,81-5v dv 9.81-5v S f= du = S dt -s(u) In 19.81-5₁ = + + c = Sat -S (9.81-5v) = Cet 75-22 = 15-18'6 x = 9.81 X = 9.81 S 7 -Sv=Cest - 9.81 -St v = Ce S -e dx = 9,81 (¹-1) dt S Sdx = 9.81 -1)dt flest - 241 (-6²² - +)* S ·st -e t) S ^5-186=1 75 du = -sdv dv = du -S S ce = 0 م -t + 9.81 25 ↑ v 1 (025- s 0=(-9,81 |C=9.81 -St 9.81 ; √(0)=0 √= 9.81 st - 9.81 S - 9.81 S ~0=(07x = 2 + S = 9.81 (e²³² - 1) S + (0) - - - > -) 15³6 = 0 < S 0=9,81 S 0= -9.81 3¹ (-/-) + +C as C = +9.81 as +( +(
Transcribed Image Text:17 m dv = mg-bv; v(0) = v dt •10dv = (10) (9.81)-50 v dt • dv = 9.81 - 5vdt . dv = dt 9,81-5v dv 9.81-5v S f= du = S dt -s(u) In 19.81-5₁ = + + c = Sat -S (9.81-5v) = Cet 75-22 = 15-18'6 x = 9.81 X = 9.81 S 7 -Sv=Cest - 9.81 -St v = Ce S -e dx = 9,81 (¹-1) dt S Sdx = 9.81 -1)dt flest - 241 (-6²² - +)* S ·st -e t) S ^5-186=1 75 du = -sdv dv = du -S S ce = 0 م -t + 9.81 25 ↑ v 1 (025- s 0=(-9,81 |C=9.81 -St 9.81 ; √(0)=0 √= 9.81 st - 9.81 S - 9.81 S ~0=(07x = 2 + S = 9.81 (e²³² - 1) S + (0) - - - > -) 15³6 = 0 < S 0=9,81 S 0= -9.81 3¹ (-/-) + +C as C = +9.81 as +( +(
An object of mass 10 kg is released from rest 3000 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? Assume that the acceleration due to gravity is 9.81 m/sec² and let x(t) represent the distance the object has fallen in t seconds. First write m = mg - bv, v(0) = v in terms of the given information. dv dt 10 = 98.1 - 50v; v(0) = 0 dv dt Determine the equation of motion of the object. x(t) = 1.962t - 0.3924 (1 - e¯5t) When will the object hit the ground? The object will hit the ground after seconds. (Round to two decimal places as needed.)
Transcribed Image Text:An object of mass 10 kg is released from rest 3000 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? Assume that the acceleration due to gravity is 9.81 m/sec² and let x(t) represent the distance the object has fallen in t seconds. First write m = mg - bv, v(0) = v in terms of the given information. dv dt 10 = 98.1 - 50v; v(0) = 0 dv dt Determine the equation of motion of the object. x(t) = 1.962t - 0.3924 (1 - e¯5t) When will the object hit the ground? The object will hit the ground after seconds. (Round to two decimal places as needed.)
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