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Question about Air Resistance

  1. Nov 14, 2009 #1
    [NEED HELP+DIAGRAM INCLUDED]Question about Air Resistance

    http://img410.imageshack.us/img410/3732/airr.jpg [Broken]

    Red line- No air resistance
    Blue line-Air resistance present

    Ok, so I thought on the way UP, resistive forces (oppose motion) acts in same way as g-force => Deceleration is very large. Time taken is very short.

    On the way down, air resistance acts in opposite to g-force, resultant force and acceleration are smaller. Time is much longer.

    I thought it was B since the a is smaller. (Longer time)

    But the answer suggested that the first is correct.

    Reasoning given by answer (Im still confused):
    Air resistance opposes motion and decreases both vertical and horizontal velocity=>Lower vertical height and shorter horizontal range [OK, I get this]
    The path of A after maximum height is progressively steeper due to the diminishing magnitude of horizontal velocity.

    It makes sense also but I wonder why am I wrong? Can anyone help clarify this?

    Thanks.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Nov 15, 2009 #2
    Bump....Anyone?
     
  4. Nov 15, 2009 #3
    Bumpy Bumpy?
     
  5. Nov 15, 2009 #4

    PhanthomJay

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    Is the the trajectory of a projectile with air resistance symmetrical or non-symmetrical? When it starts to fall, it is a steep dive or a shallow one?
     
  6. Nov 15, 2009 #5
    ^^Without air resistance, I guess symmetrical.

    With air resistance, both are not. But where is the peak?
    I thought on the way down, resultant force/acceleration is less resulting in a gentler slope, but apparently the answer decides otherwise.
     
  7. Nov 16, 2009 #6

    ideasrule

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    I don't think either you or the answer sheet gave a correct explanation. You were doing fine until you said "I thought it was B since the a is smaller. (Longer time)". What's a? Is it acceleration?

    The answer sheet falls flat when it claims "The path of A after maximum height is progressively steeper due to the diminishing magnitude of horizontal velocity." Sure, but the path of B also becomes progressively steeper. Besides, an increase in vertical velocity would also cause the path to become steeper.

    I thought about the problem this way. The time the object takes to reach the vertex is short, but the object's horizontal speed is also high. Once it passes the vertex, the object takes a long time to fall back to the ground, but its horizontal speed is now low. Since short time * high speed might be equal to long time * low speed, this line of reasoning doesn't prove that the graph isn't symmetrical.

    I can't think of any way, aside from using calculations involving calculus, of proving that A is the right answer. Are you sure this is something you're supposed to calculate, and not something you're supposed to memorize?
     
  8. Nov 16, 2009 #7
    I dont think Im involved with physics calculus at my level. Yes, I do mean that acceleration is less.

    My line of reasoning: Same distance vertically->Smaller acceleration (due to air resistance opposing the gravitation [compare with upwards])->Longer time in air (looking at vertical distance)->Longer range?
     
  9. Nov 16, 2009 #8

    PhanthomJay

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    Figure A clearly shows that the BLUE trajectory with air resistance is not symmetrical. Figure B clearly shows that the BLUE trajectory with air resistance is symmetrical. Are we looking at the same figures?
    As shown
    For the air resistance case, the acceleration in the x direction slows down the projectile's speed in the x direction, resulting in a steeper slope downward in the projectiles fall from peak height. Imagine that the projectile came to a near stop in its x direction speed, it'd fall near straight down, wouldn't it, regardless of its vertical slow downward acceleration? Figure A BLUE clearly shows a steeper slope after max height is reached than Figure B BLUE shows. The answer and explanation is correct.
     
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