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The height of a building

  1. Oct 22, 2014 #1
    1. The problem statement, all variables and given/known data

    A rocket is being launched along the line y=x from the base of a building, and another rocket is being launched from the top of the building parallel to the bottom rocket. Another rocket is being launched directly upwards away from the building. Find:

    The height of the building, how far away the third rocket is from the building and find the area the first two rockets cover before crossing the path of the third rocket given that shapes perimeter is 10m.

    2. Relevant equations
    The shape of the area is a parallelogram (A=bh). Based on the perimeter, and assuming the height of the building is the one diagonal side of the shape, its height can be represented as height = 5 - base.

    3. The attempt at a solution

    Since these are rockets, I assume they are a trajectory, forming a parabola. I think that I can find the total height of building by finding the difference between the two vertices. However, I am a little stumped coming up with the formulas for these. The linear equation for the second rockets launching is y=x+height (y=x+5-base).

    What am I missing? I feel like I am suppose to be assuming something but I don't see it. I have this question for one student of mine that I tutor math in and I just can't see it....







    Thanks!
     
  2. jcsd
  3. Oct 22, 2014 #2

    phinds

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    I don't think the geometry of this works out the way you seem to think it does, and based on the exact wording you have, the problem makes no sense. Please draw a picture of what you think is happening.
     
  4. Oct 22, 2014 #3
    My apologies. The green area is the area I am trying to find along with the height of the blue rectangle (building). I think my reasoning is not too far off that one of the diagonals of the area is the height of the building. I believe my equation for rocket 2, based on the perimeter of the parallelogram being 10m.

    upload_2014-10-22_11-22-47.png
     

    Attached Files:

  5. Oct 22, 2014 #4

    phinds

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    Ah. I misinterpreted "y=x" in your original post. You were clear but my brain told me you had said, effectively, "y=K". Sorry. My bad.

    Now, technically, your drawing is (as you recognize in your post) incorrect because it shows straight lines and thus does not account for gravity but it is possible that you are supposed to neglect this. It's not clear, since you do realize that the trajectories are parabolic, whether that's supposed to be taken into account or ignored. Do you know?
     
  6. Oct 22, 2014 #5
    I am assuming that we have to take into account the trajectories since my student is doing the quadratics unit in class right now. However, as my student tells me, this is the diagram her teacher gave her.

    My idea was to find the vertices of the two trajectories, and that would give us the height of the building by finding the difference between them. I just can't see how to answer this question fully although I think my approach and assumptions are not entirely off the wall.
     
  7. Oct 22, 2014 #6

    phinds

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    Ah ... another simplification by a teacher, who failed to mention the ymplification.

    There are two MORE simplifications here (both reasonable ones I think) which is that all the rockets have the same power AND we are supposed to ignore the fact that even so they won't have QUITE the same equations because the higher one will have a tiny difference in the starting force of gravity because it's higher.

    Given all the assumptions they yeah, I think your approach is reasonable but you'll only be able to come up with an equation based on the the height of the building and the distance of the vertical rocket from the building, not a numerical answer.

    Ignoring for a moment the fact that the trajectories are parabolic, this is exactly like being given a parallelagram of perimeter 10m and being asked to find its area.


    OH ... OOPS. Now I think I see what's going on. You ARE supposed to ignore the parabolic nature of the two sides. Forget rockets entirely. You've been given a parallelogram and its perimeter and the point of the problem is most likely COME UP WITH a quadratic equation that includes the area and either the height or the width and thus can be resolved to express the area as a function of either the height (the distance of the vertical rocket from the building) or the width (the height of the building). That's easy.
     
  8. Oct 22, 2014 #7
    Ah, well then thank you! It is after all an easy question. Thank you for working through the logic with me :)
     
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