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Projectile Motion: two positive times?

  1. Sep 10, 2010 #1
    I explained to my class that in projectile motion one always chooses the positive time answer with the negative time answer being dropped. Then a student asked if two positive time answers were possible. I just looked at it assuming that t+/- are both positive and got the result that for t+ to be positive, then

    yf > yi.​

    But for t- to be positive (for the same problem)

    yi > yf.​

    I believe this is a contradiction and so the answer to the student's question is 'no' - two positive time answers, in the same problem, are not possible. I need confirmation of this before I give the answer to the class.

    Thanks in advance for any help you may give me.

  2. jcsd
  3. Sep 10, 2010 #2

    Doc Al

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    I don't understand the context. Why can't you have a projectile motion question with two answers, both at positive times? Throw a ball in the air. At what time(s) will it be 10 ft above the ground?
  4. Sep 10, 2010 #3
    If the initial and final positions are fixed, then solving for the time of flight should yield only one positive time.

    yf = yi + voy*t - 4.9*t2

    The solution of this is the quadratic formula:

    [tex]t = \frac{-v_{oy} \pm \sqrt{v_{oy}^{2} + 19.6*\Delta y}}{-g}[/tex]​

    There are two answers: t+ and t-.

    For both to be positive, one sets up the inequality for each and deduces the result above, namely:

    For t+ > 0 this implies Dy > 0; and
    for t- > 0 this implies Dy < 0.

    Thus the contradiction.
  5. Sep 11, 2010 #4

    Doc Al

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    I don't understand how you are drawing this conclusion. If Yf > Yi, then both times will be positive if Vo is positive; If Yf < Yi, then one of the times will be negative.
  6. Sep 12, 2010 #5
    Using the quadratic solution for the time (link is to the above formula):

    https://www.physicsforums.com/latex_images/28/2874555-0.png [Broken]

    I set t+ > 0 and t_ > 0 and look for the necessary condition to ensure this.

    For t+ > 0, yf > yi, and

    for t_ > 0, yi > yf.

    But this was my original question, was my deduction correct?
    Last edited by a moderator: May 4, 2017
  7. Sep 12, 2010 #6

    Doc Al

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    How can your deduction be correct? You claim that you cannot have a case where you have two positive time solutions and I gave a trivial example that shows you can! Here's one with numbers: Throw a ball upward with initial velocity = +10 m/s. At what time(s) will it be 5 m above the starting position?

    Show the details of how you made your deductions--I suspect either a mathematical error or that I am misinterpreting the problem you are trying to solve.
  8. Sep 12, 2010 #7
    Ok, here goes...

    [tex]t_{+} = \frac{v_{oy}}{g} + \frac{\sqrt{v_{oy}^2 + 19.6\Delta y}}{g} > 0[/tex]

    Canceling the common denominator, g, and bringing the voy term to the other side then I get:

    [tex] \sqrt{v_{oy}^2 + 19.6\Delta y} > -v_{oy} [/tex]

    Squaring both sides I get:

    [tex] v_{oy}^2 + 19.6\Delta y > v_{oy}^2[/tex]

    Now the initial y-velocity component can be canceled, which yields:

    [tex] 19.6\Delta y > 0[/tex]

    Dividing out the constant factor of 19.6,

    [tex] \Delta y > 0[/tex]

    which yields the result for t+, namely:

    yf > yi.


    For t- the result is:

    [tex]t_{-} = \frac{v_{oy}}{g} - \frac{\sqrt{v_{oy}^2 + 19.6\Delta y}}{g} > 0[/tex]

    Canceling the common denominator, g, and bringing the square-root to the other side then I get:

    [tex] v_{oy} > - \sqrt{v_{oy}^2 + 19.6\Delta y}[/tex]

    Squaring both sides, and canceling the voy term leaves:

    [tex] 0 > 19.6\Delta y[/tex]

    Dividing out the constant factor of 19.6,

    [tex] 0 > \Delta y[/tex]

    which yields the result for t-, namely:

    yi > yf.


    I see something now I didn't see before - that in one attempt I moved the voy term, and in the other I moved the square-root term. Thus I obtained two different results. However, this isn't wrong, and your previous, intuitive, answer is correct.

    So what's going on here???

    Thanks for your help Doc Al.

  9. Sep 12, 2010 #8

    Doc Al

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    I see the problem. You made a sign error with respect to Δy:

    yf = yi + voy*t - 4.9*t2

    0 = -Δy + voy*t - 4.9*t2

    Now apply the quadratic formula.
  10. Sep 12, 2010 #9
    That minus sign is common to both t+ and t-. The only thing that changes is that the results are switched. Thanks for that. But I still think that something is weird about the "square both sides of the inequality" step.

    Thanks for all your help.

  11. Sep 12, 2010 #10

    Andy Resnick

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    It's definitely possible to have two positive times- for example, launch two projectiles at two angles, each equidistant from 45 degrees. Both projectiles will hit the target, but take 2 different paths (and 2 different times).

    Does that help?
  12. Sep 12, 2010 #11

    Doc Al

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    Good point. Putting the sign of Δy aside for the moment: Yes, the problem is where you square both sides. Not valid!


    2 > -3

    Square both sides:

    4 > 9 :yuck:

    Sorry about not spotting that earlier! :uhh:

    Starting with this:
    Clearly the inequality will be satisfied for any value of Δy that keeps the discriminant positive. (If it's not positive, there's no real solution.)
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