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Which is Faster, Going Up or Coming Down?

  1. Apr 14, 2014 #1
    1. The problem statement, all variables and given/known data

    Suppose you throw a 1kg ball into the air. Determine whether the ball takes
    longer to reach its maximum height or fall back to Earth from its maximum
    height?

    Assume the forces acting on the ball are the force of gravity and a retarding
    force of air resistance with direction opposite to the direction of motion.
    (-mkv) (assume k = 0.1)


    Would it not just me the same? How would I show this

    PS. This is not being used for an assignment
     
  2. jcsd
  3. Apr 14, 2014 #2

    LCKurtz

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    My intuition says they would be the same too, but that might be wrong. To prove it you need to set up the DE and solve it.
     
  4. Apr 14, 2014 #3

    Dick

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    I don't think so. There's an intuitive way to deal with this. Suppose you are at a height h above where you started. Do you have more or less kinetic energy coming up than going down? What does that mean about velocity going up versus coming down?
     
  5. Apr 14, 2014 #4
    I'm not sure....that there is less kinetic upwards that coming down? How would one algebraically show this?
     
  6. Apr 14, 2014 #5

    Dick

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    You don't have to. The air resistance is always taking energy away because it's opposite to the direction of motion. Do you know about kinetic and potential energy?
     
  7. Apr 14, 2014 #6
    Not really, I know an object has gravitational potential energy?
     
  8. Apr 14, 2014 #7

    Ray Vickson

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    I don't think you can rely on conservation of energy in this problem. The ball's total energy is not necessarily conserved, because frictional forces may be dissipative, and some energy may be transferred to the atmosphere in the form of heat, etc.
     
  9. Apr 14, 2014 #8

    LCKurtz

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    Agreed. I wasn't thinking in those terms. Solving the DE would be the hard way.
     
  10. Apr 15, 2014 #9
    Can't you just use conservation of energy? (in case this is for physics and not math)
    Show that the potential energy when the ball is at max h = the kinetic energy right before the ball hits the ground.
     
  11. Apr 15, 2014 #10

    LCKurtz

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    Dick's point is that energy is lost. That gives some energy inequalities.
     
  12. Apr 15, 2014 #11
    Hm fair enough. I don't know then. It's beyond my capabilities.
     
  13. Apr 15, 2014 #12
    Does anyone know how I could algebraically prove this?
     
  14. Apr 15, 2014 #13

    LCKurtz

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    Sure. Like I said in post #2, set up the DE and solve it. Once you have the solution you can figure out how long it takes to hit the max height and how much additional time it takes to hit the ground. That will leave the intuition out of it. In the DE give it an initial height of ##0## and an initial velocity of ##v_0##. Or take ##v_0=1##, it shouldn't matter.
     
  15. Apr 15, 2014 #14

    HallsofIvy

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    But the whole point is that the total kinetic energy plus potential energy is NOT conserved since you lose some energy to air resistance. At any given height, the potential energy is the same so coming down, the kinetic energy, and so the speed, must be less than it was going up.
     
  16. Apr 15, 2014 #15

    D H

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    That was Dick's point in post #3.
     
  17. Apr 16, 2014 #16

    Ray Vickson

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    It is impossible to tell which message you are now responding to, and what you mean by 'this' (as in 'prove this algebraically'). To avoid this in future please respond by using the 'quote' button.

    Anyway, on the upward part both forces act against the direction of motion; that is, forces add. On the way down, one force acts in the direction of motion and the other acts in the opposite direction, so there is a partial cancellation of forces on the downward trip. This suggests that the upward trip decelerates faster than the downward trip accelerates, so the up-trip should take less time than the down-trip. You can verify this explicitly in numerical examples, by solving the differential equations and solving some transcendental equations that arise from the DE solutions.
     
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