ELEMENTRY DIFFERENTIAL EQUATIONS 2.3

ELEMENTRY DIFFERENTIAL EQUATIONS 2.3 - c) What is the...

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ELEMENTRY DIFFERENTIAL EQUATIONS CHAPTER 2 2.3 PROBLEMS 1-
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23- Given: Weight of the diver = 180lb, g 32ft/s (i)--- mg = 180, m = 180/32 (ii)--- x(0) = 5000ft (iii)--- air resistance = 0.75|v| when parachute is closed (iv)--- air resistance = 12|v| when parachute is open (v)--- parachute opens after 10s a) Find the speed of the sky diver when the parachute opens. v(10) must be found and an I.V.P. must be set up for when the sky diver is free falling. The I.V.P. is: we know g, and m, so
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and v(0) = 0, so and and after 10 seconds when the parachute opens. b) Find the distance fallen before the parachute opens. We know that so we must integrate our equation for v to find x.
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distance travelled is then
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Unformatted text preview: c) What is the limiting velocity v L after the parachute opens? Here we must find the limit as t approaches infinity of the velocity after the parachute opens, so we must set up a new I.V.P. that is: using the integrating factor we solve we also know that when the parachute opens, so so the limiting veocity v L =15ft/s d) Determine how long the sky diver is in the air after the parachute opens. Here we want x(t) again, so we have to integrate our v(t) and x(0) is 0, since we are finding the distance from the point where the parachute opens. and we also know the parachute travels 5000 - 1074.5 = 3925.5 ft so we set x = 3925.5 ft and solve for t and we find t = 256.6s 25-27-29-31-...
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This note was uploaded on 09/25/2011 for the course MATHS 12 taught by Professor Ramarao during the Spring '10 term at Punjab Engineering College.

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ELEMENTRY DIFFERENTIAL EQUATIONS 2.3 - c) What is the...

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