- #31
romiomustdie
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i didnt get the equation ... how did you get the above equation ?
How Do You Derive the Distance Traveled by a Projectile in a Viscous Liquid?
- Thread starter romiomustdie
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SUMMARY
The discussion centers on deriving the distance traveled by a projectile in a viscous liquid, where the retarding force is proportional to the square of the velocity, expressed as a = -kV². The correct expressions derived are D = 0.693/K for distance and t = 1/(kVo) for time to reduce the velocity to Vo/2. Participants emphasize the importance of solving the differential equation using separation of variables and integrating the velocity to find the distance, highlighting common mistakes in the process.
PREREQUISITES- Understanding of differential equations, specifically separation of variables.
- Familiarity with projectile motion concepts in physics.
- Knowledge of calculus, particularly integration techniques.
- Basic principles of forces and acceleration in fluid dynamics.
- Study the method of separation of variables in differential equations.
- Learn how to integrate velocity functions to derive distance traveled.
- Explore the application of retarding forces in fluid dynamics.
- Review calculus techniques for solving integrals involving exponential functions.
Students in physics or engineering courses, particularly those studying fluid dynamics, differential equations, and calculus. This discussion is beneficial for anyone preparing for exams involving projectile motion in viscous mediums.
- #32
romiomustdie
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ok got it
- #33
romiomustdie
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but how will i get a numerical value like 0.693 in numerator by splving this . i am not getting what to do , ?
- #34
romiomustdie
- 29
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but how will i get a numerical value like 0.693 in numerator by solving this by integration to obtain S .
i am not getting what to do , ? Totally puzzed , please do the rest please
i am not getting what to do , ? Totally puzzed , please do the rest please
- #35
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Start by using the velocity equation to find the time at which the velocity is half its initial value.
Chet
Chet
- #36
romiomustdie
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- #37
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Yes. Now integrate to get the distance. Let's see your work.romiomustdie said:
Chet
- #38
romiomustdie
- 29
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Failure failure failure
Unable to find out the distance please show
Unable to find out the distance please show
- #39
romiomustdie
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Log and all coming
- #40
romiomustdie
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Please show
- #41
vela
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We don't do the problems for you. You need to figure them out.
- #42
romiomustdie
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Give some hint then
- #43
romiomustdie
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I am struggling with this problem for a long time. One question I solved but unable to find Distance
- #44
vela
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Supposedly you got the velocity, and then all you've written about getting the position is "failure failure failure."
Here's a hint: Try again.
The relationship between acceleration and velocity is analogous to that between velocity and position. If you understood how to get v from a, it's exactly the same mathematical process to get x from v.
Here's a hint: Try again.
The relationship between acceleration and velocity is analogous to that between velocity and position. If you understood how to get v from a, it's exactly the same mathematical process to get x from v.
- #45
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dx=Vdtromiomustdie said:
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