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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.
ok yeah let's see now
This is algebra, right? not Calculus?
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?
|dw:1327963508479:dw| just writing my work out here...hold on
I am pretty sure it is algebra/vector math
yeah, I think the plane makes it over...let me explain it in, haha, less messy terms
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
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
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.
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?
This makes sense, but since the angle is 37 degrees, shouldn't x = 119.80?
Yeah it should. No idea why I typed 30 in
So let's see now...okay, plane crashes.
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
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
horizontally. Heh, sorry!
I see what you are saying, but I am having difficult visualizing it. Thanks though, just need to process it in my head.
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
Well, at that point, you know the horizontal distance, right? 1204.88. And you know the hypotenuse...let's draw this bit out...
|dw:1327964487654:dw| this gives you the height at that very moment
put a little exponent of 2 on the 1204.88 there, I missed that bit
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.
Now, I could also do y = 150*tan37 and then multiply that result by 10.0578 to get the height traveled in 10 seconds?
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)
But good initiative, yes, you can find the height traveled in one second.
Alright dude (or gal w/e), I need to head off to class. I hope I was of some help
you were, thanks