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Terminal Velocity ?

  1. Jan 6, 2013 #1
    I want to know about terminal velocity.

    As my point of view, when air resistance and force(=mg) become same then the object

    moving downward has constant velocity.

    This is right?

    If right then v = (initial velocity)v + g*t; becomes v = g*t;

    then velocity increases by time !
    Last edited: Jan 6, 2013
  2. jcsd
  3. Jan 6, 2013 #2
    Velocity increases in time when you have an acceleration. Your equation has a "g" in there as though its accelerating in a vacuum. If you want to consider terminal velocity then you have air resistance providing an opposite force to gravity (not the same). When this happens you dont have accereleration at "g", you have 0 acceleration. This would make your equation v = (initial velocity)v + 0*t; becomes v = (initial velocity)v.
  4. Jan 6, 2013 #3
    I am not understanding what you are saying?

    Example : If you are drop down a ball from 1000 m height of a building, Now As you are saying

    that v = initial velocity(v)?
  5. Jan 6, 2013 #4
    yes. When the drag [or force] of air resistance equals the force of gravity, mg, the object
    falls at some terminal velocity....often taken to be some fixed velocity.

    Of course as air resistance increases as altitude decreases the ACTUAL velocity of a falling object might increase slightly, but this is often ignored.
  6. Jan 6, 2013 #5
    I want to know how terminal velocity is being constant?
  7. Jan 6, 2013 #6
    Are you asking WHY it is constant?? because when air drag equals mg, there is no net force to accelerate an object.

    so no acceleration......

    Just like a car cruising down the highway at say, a steady 60 mph,...rolling friction and air drag and mechanical losses equals the force provided by the engine......
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