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Verticle Projectile Problem

  1. Oct 22, 2007 #1
    1. The problem statement, all variables and given/known data

    Note: This is supposed to be simplistic so we're only concerned with the vertical movement. All horizontal factors are completely ignored and should not be included in the solutions or the procedures to get to the solutions.

    Phil wants to test his new potato launcher at the edge of a cliff overlooking the ocean. He launches his potato from the edge of the cliff and observes that it rises to a maximum height of 56.5 meters and hits the water in a total time of 12.31 seconds. Ignoring the effects of air resistance, answer the following:

    What is the initial velocity v(0y) of the potato at the launch point.

    2. Relevant equations

    [tex]g = \frac {v_y - v_0_y} {t}[/tex]
    [tex]v_y = v_0_y + g t[/tex]
    [tex]y = y_0 + v_0_y t + \frac{1}{2} g t^2[/tex]
    [tex]v_y^2 - v_0_y^2 = 2 g \Delta y[/tex]
    [tex]\Delta y = (\frac {v_y + v_0_y} {2}) t[/tex]
    [tex]t = \frac {-b \pm \sqrt {b^2 - 4ac}}{2a} [/tex]

    (g= 9.80665 OR g=-9.80665, I'm not sure which applies when.)

    3. The attempt at a solution

    I tried to use multiple formulas but none of them seem to work right. I only have two pieces of information here and I don't know how to use them in place of any variable. The time given is total. The distance given is only the displacement up to the highest point.

    I'm completely stumped. Is there just not enough information? Or maybe I'm over thinking the whole thing and the solution is right there? Either way, I just can't even begin to understand this. I'm not particularly looking for an answer, I just need to know how I'm supposed to do this. There are more questions after this one, but if I can do this, I'll probably be able to do the rest too.
    I copied down the problem here exactly as it on my paper. And I reread the problem and this post multiple times to ensure there are no errors. Also, here is a link to a scan of the original problem and diagram as it is on the paper.

    EDIT: As I wrote this, I realized something I didn't notice before and have started to work on the problem again. But I would still appreciate your help. Thanks in advance.

    http://s100.photobucket.com/albums/m32/smzee27/?action=view&current=Untitled.jpg

    I would really appreciate any help on this. I've done these kinds of problems before, but this particular one has me stumped. Although everyone else in the class seems to have gotten through this one, albeit with some difficulty.
     
    Last edited: Oct 22, 2007
  2. jcsd
  3. Oct 22, 2007 #2
    Energy conservation approach is better here, just equate KE at initial stage to change in PE at final stage (decide what to take as final state).

    Alternatively
    Newton's way : use your 4th equation.
     
    Last edited: Oct 22, 2007
  4. Oct 22, 2007 #3

    Integral

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    You can do this problem just using the 2nd and 3rd of your listed equations.

    Use the 2nd to elminate V0y from the 3rd. Then solve for the time to the top of flight, using the given height. Once you have the time to the top, you know the time to return to the starting point. Now use that time and your knowledge of the velocity to solve the 2nd equation for the initial velocity.
     
  5. Oct 22, 2007 #4
    The class is not at the point yet where we can use something like that; your method seems too advanced right now. Thanks for your reply though.

    Assuming I did that first part right, I get:

    [tex] y = y_0 + v_y - gt + \frac {1}{2} gt^2 [/tex]

    Okay, so [tex] y = 56.5 m [/tex], and [tex]y_0 = 0[/tex]? Also, how would I get [tex]v_y[/tex]? If [tex]v_y[/tex] is supposed to be the average velocity for the whole thing, I only know the time for the whole motion, not the distance.
     
  6. Oct 22, 2007 #5
    Velocity at the peak of the projection is 0. You have the height of 56.5m.

    EDIT:Read post below! no need to find time first.
     
    Last edited: Oct 22, 2007
  7. Oct 24, 2007 #6
    Thanks for your help everyone.

    By the way, I don't see a "Mark solved" option under Thread tools.
     
    Last edited: Oct 24, 2007
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