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Dart trajectory, trig proof.

  1. Sep 18, 2008 #1
    A target is suspended on a platform. A dart launcher is placed at ground level and aimed directly at the target along the line of sight (the distance between dart and target can vary infinitely). Assume a bottomless pit below the target. The dart is launched, and regardless of speed, it will hit the target (although the height at which it hits the target varies). Prove that this will always happen.

    Half of the assignment is to solve it with given numbers (10m horizontal distance, initial velocity at 24 m/s, 35 deg above horizontal). I had no trouble solving and verifying that part. However, when I try to repeat the same equations without numbers, I end up at -1 = 1.

    I scanned the work I've done so far. I'd appreciate it if someone could point out where I went wrong.

    My scanned work is available at:
    http://webserver.smphoto.com/physics12/ [Broken]
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Sep 19, 2008 #2


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    Homework Helper

    Hi LANS,

    I believe you have two closely related errors in your work. The first occurs when you are deriving a form for [itex]\Delta h[/itex]. You first write the general formula:

    \Delta h=v_i\ \Delta t+\frac{1}{2}a\ \Delta t^2
    and below that you have:

    \Delta h=\sin\theta\ z\ \left(\frac{\Delta d}{\cos\theta\ z}\right)+\frac{1}{2}a \left(\frac{\Delta d}{\cos\theta\ z }\right)^2
    but this is not right. You have chosen downwards to be positive, but the term with the initial velocity is providing an upwards displacement, and so must be negative. (The point is since the two terms on the right are in different directions they must have different signs.) So this equation should be:

    \Delta h=\ -\sin\theta\ z\ \left(\frac{\Delta d}{\cos\theta\ z}\right)+\frac{1}{2}a \left(\frac{\Delta d}{\cos\theta\ z }\right)^2

    The next problem occurs right after that, where you say:

    \tan\theta\ \Delta y -\Delta y = \Delta h
    which is not correct based on the way you have defined [itex]\Delta y[/itex] and [itex]\Delta h[/itex]. If you look back at the diagram at the top of the page, what this equation is saying is

    y=\Delta y +\Delta h
    Now y is a length, so it is a positive number. [itex]\Delta y[/itex] is the displacement of the target, and since it is moving downwards, that would be a positive number. But [itex]\Delta h[/itex] is the vertical displacement of the dart, and since that is upwards, that would be a negative number.

    So if [itex]y[/itex] is some number like 8m, and [itex]\Delta y[/itex] is some number like 2m, then [itex]\Delta h[/itex] would be -6m. And so your equation above should be:

    y&=\Delta y -\Delta h\nonumber\\
    \tan\theta\ \Delta y -\Delta y &= \ -\Delta h\nonumber

    (More mathematically we would say the lengths of these three add together, so that what we would actually use is absolute values:

    |y|=|\Delta y| +|\Delta h|
    and then since [itex]\Delta h[/itex] is negative, [itex]|\Delta h|=\ -\Delta h[/itex].)

    Once you make these changes, I think you'll see that all the terms cancel out in the line immediate below where you have a circled 1 in your work.
    Last edited by a moderator: May 3, 2017
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