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anonymous

  • 4 years ago

This question is confusing the hell out of me: A plane is climbing 150m/s at 37 degrees. It must clear a 450 meter tall obstruction that is 1440 meters away. Will the plane make it over?

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  1. anonymous
    • 4 years ago
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    |dw:1327963156017:dw|

  2. anonymous
    • 4 years ago
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    |dw:1327963259674:dw|

  3. anonymous
    • 4 years ago
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    I figured, if I found out the difference between time = 1 and time = 2, then I would get the rate of climb and distance traveled over one second. So: Found the angle for c which is: sqrt of 450^2 + 1440^2 = 1509 meters. Divide that number by 150m/s = 10.1 seconds for the plane to travel the full 1440 meters.

  4. anonymous
    • 4 years ago
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    ok yeah let's see now

  5. anonymous
    • 4 years ago
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    This is algebra, right? not Calculus?

  6. anonymous
    • 4 years ago
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    Then I subtract 150 from 1508 to get 1359m and x (base of triangle) and y (height of triangle) plus 37 degrees from angle theta. So cos37 = x/1359 for 1085 meters remaining. 1440 minus 1085 = 355 meters traveled horizontally in one second. This cannot be, the plane is only moving 150m/s, how can it possibly move 355m horizontally in the time it traveled 150m's at 37 degrees. What did I mess up on?

  7. anonymous
    • 4 years ago
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    |dw:1327963508479:dw| just writing my work out here...hold on

  8. anonymous
    • 4 years ago
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    I am pretty sure it is algebra/vector math

  9. anonymous
    • 4 years ago
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    yeah, I think the plane makes it over...let me explain it in, haha, less messy terms

  10. anonymous
    • 4 years ago
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    okay so you really have two triangles. one is the one you drew, the other is the one I just drew. They are similar, in that you have to figure out what distance in the x direction the plane travels every second

  11. anonymous
    • 4 years ago
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    so if you set up the drawing that I just did a few posts up, you solve for x using cos and you get x = 129.904 meters. so every second, the plane travels 150 m/s up, but the x-component of that is 129.904. You could think of the plane traveling in two dimensions: it travels a certain distance horizontally (x) and a certain distance vertically (y) to get the 150 m/s speed

  12. anonymous
    • 4 years ago
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    Now, take the hypotenuse from the first triangle (1508.67) and divide that number by 150. You get 10.0578 seconds. So it takes 10.0578 seconds for the plane to travel the hypotenuse of the bigger triangle and clear the obstruction.

  13. anonymous
    • 4 years ago
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    Now take that 10.0578 seconds and multiply by the value of x we found in the 2nd triangle, 129.904, and you get 1306.55. So at the plane's current speed, that's the horizontal distance it'll travel to clear that 450 meter obstruction. That's smaller than how far horizontally the plane is from the obstruction (1440), so no plane crash this time. Did that make sense?

  14. anonymous
    • 4 years ago
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    This makes sense, but since the angle is 37 degrees, shouldn't x = 119.80?

  15. anonymous
    • 4 years ago
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    Poop. Haha.

  16. anonymous
    • 4 years ago
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    Yeah it should. No idea why I typed 30 in

  17. anonymous
    • 4 years ago
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    So let's see now...okay, plane crashes.

  18. anonymous
    • 4 years ago
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    wait

  19. anonymous
    • 4 years ago
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    Nope, haha. Wrong again. So the time that the plane takes is still the same, so the calculation is actually (cos(37)*150)*10.0578 = 12.04

  20. anonymous
    • 4 years ago
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    argh 1204.88

  21. anonymous
    • 4 years ago
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    not 12.04. Okay, so I think that's the final answer. No, plane doesn't crash, because it can clear the obstruction with ~ 200 meters to spare

  22. anonymous
    • 4 years ago
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    horizontally. Heh, sorry!

  23. anonymous
    • 4 years ago
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    I see what you are saying, but I am having difficult visualizing it. Thanks though, just need to process it in my head.

  24. anonymous
    • 4 years ago
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    The problem I see is with knowing the height once it has traveled 1204.88 meters. It is based on the time, just, not visualizing it yet

  25. anonymous
    • 4 years ago
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    Well, at that point, you know the horizontal distance, right? 1204.88. And you know the hypotenuse...let's draw this bit out...

  26. anonymous
    • 4 years ago
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    |dw:1327964487654:dw| this gives you the height at that very moment

  27. anonymous
    • 4 years ago
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    put a little exponent of 2 on the 1204.88 there, I missed that bit

  28. anonymous
    • 4 years ago
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    when I do that calculation I get x = 907.937 meters. So the height of the plane at that moment is much higher than the obstruction.

  29. anonymous
    • 4 years ago
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    Now, I could also do y = 150*tan37 and then multiply that result by 10.0578 to get the height traveled in 10 seconds?

  30. anonymous
    • 4 years ago
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    no, it'd have to be sin. tan is opposite over adjacent. You could do tan, but you'd have to do this: y = 129.904*tan(27)

  31. anonymous
    • 4 years ago
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    But good initiative, yes, you can find the height traveled in one second.

  32. anonymous
    • 4 years ago
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    Alright dude (or gal w/e), I need to head off to class. I hope I was of some help

  33. anonymous
    • 4 years ago
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    you were, thanks

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