Motion of a particle subject to a variable resisitive force

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SUMMARY

The discussion focuses on the motion of a particle subject to a variable resistive force, specifically analyzing the equation for distance, x = (1/2k)ln((g - k(V^2)(tan a )^2)/(g - kv^2). The user expresses confusion regarding the potential energy lost during the motion, particularly when the velocity, v, is set to zero. The conversation highlights the relationship between kinetic energy loss and potential energy gain, emphasizing the importance of understanding the dynamics of vertical motion and terminal velocity during ascent versus descent.

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  • Understanding of Newton's laws of motion
  • Familiarity with concepts of kinetic and potential energy
  • Knowledge of variable resistive forces in physics
  • Basic calculus for analyzing motion equations
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  • Learn about the derivation and implications of terminal velocity
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This discussion is beneficial for physics students, educators, and anyone interested in the dynamics of motion under variable forces, particularly in understanding energy transformations during projectile motion.

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http://img338.imageshack.us/img338/193/33379676.th.jpg

See the image above. I am able to get to (where x = distance):

x = (1/2k)ln((g - k(V^2)(tan a )^2)/(g - kv^2))

If you let v = 0, and then multiply by mg, you find potential energy gained presumably - I don't understand why the question states the potential energy lost.

Obviously the 0.5(MV^2)*(tan a)^2 part of the expression is due to the lost of the kinetic energy, however for the other part of the expression I am struggling.

Thanks for any help in advance.
 
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If the stone is thrown vertically upwards how can the terminal velocity exist?
 
ah right, so why does it say in the ascent, as opposed to descent?

Thanks
 

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