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theEric Group Title

(Classical Mechanics) Would somebody be able to help, maybe with hints, on this projectile motion problem? Thanks! It involves cannons! Scenario is below.

  • one year ago
  • one year ago

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  1. theEric Group Title
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    |dw:1383511843844:dw| The question is... How far away can we put the ground cannon so that it shoots a shell at \(v_0\) and still hits the tower cannon.

    • one year ago
  2. theEric Group Title
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    |dw:1383512248696:dw|

    • one year ago
  3. theEric Group Title
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    I can use calc.. \(\ddot{\overrightarrow r}=\overrightarrow a=\hat j g\) where \(g<0\), \(g\approx -9.81\), not that numbers will be used. \(\dot{\overrightarrow r} =\overrightarrow v= \hat i v_0\cos\theta+\hat j \left(v_0\sin\theta+gt\right)\) \(\overrightarrow r = \hat i v_ot\cos\theta + \hat j\left(v_0t\sin\theta+\dfrac{1}{2}gt^2\right)\) since the \(x_0\) and \(y_0\) are both \(0\).

    • one year ago
  4. theEric Group Title
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    I tried to start out with these two assumptions: 1. The vertical velocity will be \(0\) \(\sf at\) or \(\sf after\) a height of \(h\). 2. A range of zero (maximum vertical velocity: \(v_0\)) can reach the top of the tower, or higher.

    • one year ago
  5. PhysicsGuru Group Title
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    What is the question asking for?

    • one year ago
  6. theEric Group Title
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    It is asking for \(L\), the maximum length the ground cannon can be from the tower and still hit the top of the tower.

    • one year ago
  7. PhysicsGuru Group Title
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    You want to find the components of velocity with the angle (Vx and Vy). You want to find the hangtime so that the final vertical velocity is just zero when it reaches the cliff \[V_{yfinal} =V_{yinital} -gt_{hang}\] This would be the maximium time allowed in the air.... then use this time with the equation for average velocity in the horizontal direction: \[V_x = \frac{L}{t_{hang}}\]

    • one year ago
  8. theEric Group Title
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    I was thinking about saying \(v_y=0\) at \(y=L\), but I can prove that, and I don't know if it's true... But it is true?

    • one year ago
  9. theEric Group Title
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    I can't* prove that. I just thought it made sense at the time. But then also I thought it didn't make sense...

    • one year ago
  10. PhysicsGuru Group Title
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    That is a constraint we are applying because you want the ball to reach the top of the cliff correct?

    • one year ago
  11. PhysicsGuru Group Title
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    You dont need to prove that point ... all math problems have constraints to solve for a quantity otherwise they would be just general equations and you could not solve them for a specific case

    • one year ago
  12. theEric Group Title
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    Right, thanks! :) My only issue was, what if have the vertical velocity not be zero when it reaches the cliff? It might go farther that way, is what I thought afterwards.|dw:1383516481049:dw|Greater \(v_y\) means more time in the air, and we have to balance that with \(v_x\) and \(v_y\) being such that it hits the tower...

    • one year ago
  13. PhysicsGuru Group Title
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    Yes but you need to maximize L which means you are looking for the case we are talking about above

    • one year ago
  14. theEric Group Title
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    Alright, I'll give that a shot! Thanks!

    • one year ago
  15. theEric Group Title
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    I've rejected the theory that \(v_y=0\) when \(x=L\). I was working on a proof, but I think it will take too much time away from my homework assignment. However, there is a picture here: http://upload.wikimedia.org/wikipedia/commons/6/61/Ideal_projectile_motion_for_different_angles.svg Now, if \(v_y=0\) at the top of the tower, that is our max height. And for the max height to be the greatest range, we look to the \(\theta=45^\circ\) situation. By saying this is the solution, you are saying that it gives you the longest range. That creates a contradiction, because it is \(\sf\color{red}{not}\) the longest range, as there are longer ranges with larger angles.|dw:1383527791809:dw|For more mathematical evidence, look at the link introduced previously. Now, the range from one end on the ground to the other end on the ground is not as long. But the range from the ground to the top of the tower is larger.|dw:1383527973180:dw|

    • one year ago
  16. theEric Group Title
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    |dw:1383528051969:dw|

    • one year ago
  17. theEric Group Title
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    So, I'm not done working on it, but I think I found a strong hint. The angle at which the shell hits the tower is \(\alpha\), given as \(\csc^2\alpha=2\left(1+\dfrac{gh}{v_0^2}\right)\). It was found in an earlier problem, and I'll try to apply it.

    • one year ago
  18. DemolisionWolf Group Title
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    @theEric I think i got an idea, but maybe it's too simple for what you are working on

    • one year ago
  19. theEric Group Title
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    Never hurts to give it a shot! There has to be something I'm missing... I would think this can be solved algebraically. Calculus just gives equations :P

    • one year ago
  20. DemolisionWolf Group Title
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    is the answer suppose to have any specific variables? or can it be two equations or just needs to be 1?

    • one year ago
  21. theEric Group Title
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    There is no air resistance, and gravity is constant.

    • one year ago
  22. theEric Group Title
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    The answer will be something like \(L=\dots\) And it will have the variable \(v_0\), and maybe \(\theta\)...

    • one year ago
  23. theEric Group Title
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    I think...

    • one year ago
  24. DemolisionWolf Group Title
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    oh ok, i'll put up what I have then, maybe it helps, then i'm gonna go watch tv ^_^

    • one year ago
  25. theEric Group Title
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    Haha, okay, thanks! And have fun watching tv! :)

    • one year ago
  26. DemolisionWolf Group Title
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    I used the kinematic equation: \[s = s +vt + \frac{ 1 }{ 2 } (a)t^2\]now in the x direction it becomes: \[L = 0 + v_0Cos(\theta)(t) + \frac{ 1 }{ 2 }(0)t^2\]and for the y-direction it becomes: \[h = 0 + v_0\sin(\theta)(t) - \frac{ 1 }{ 2 }(g)t^2\]these are my two equations. now, if it needs to become 1 equation. I'd take the y-dir equation and set it equal to v_0 and substitute it into x-dir eq, but then you are left with an equation that doesn't depend on v_0 but on t, which I suppose can make sense. I haven't looked at the other kinematic equations to reduce the 2eqs down into 1 eq, but those are my two cents. ^_^ oh, and t must be t_final, not anything else.

    • one year ago
  27. theEric Group Title
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    Alright, thank you! Take care!

    • one year ago
  28. DemolisionWolf Group Title
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    came back, I worked it out more becuase I wasn't satisfied myself with it and I ended up with \[L=\cot(\theta)\sqrt{2gh}\frac{ \sin(\theta)v_0 }{ g}\]becomes \[L=\cos(\theta)\sqrt{2gh}\frac{ v_0 }{ g}\] (if i did my fractions and trig right) to get here I used the following 3 equations L = cos(theta)vt [eq 1] 0=(sin(theta)v)^2 -2gh (y-direction) [eq 2] 0 = sin(theta)v-gt (y-direction) [eq 3] I took eq 2, set it equal to v, plugged it into eq 1. then took eq 3, set it equal to t and plugged it into eq 1.

    • one year ago
  29. theEric Group Title
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    thank you very much! You didn't have to, but thank you! I'm going to check it out. Thanks!

    • one year ago
  30. theEric Group Title
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    I think eq 2 assumes \(v_{y\text{, final}}=0\); but I don't think that is the case, unfortunately.

    • one year ago
  31. theEric Group Title
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    same with eq 3. But eq 2 = \(v_y^2\) and eq 3 = \(v_y\), so maybe I'll end up using that!

    • one year ago
  32. DemolisionWolf Group Title
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    when i went to bed last night, i realized i had made that mistake, v final in the y-direction can't be zero. I then back checked my final equaiton with a scenario i made up, and the final equation didn't hold true.

    • one year ago
  33. theEric Group Title
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    That's alright. I have the answer. If anyone wants to see it, let me know and I'll put it up! It was complex, made even more complex by using an equation defined in an earlier problem as a result that gives the maximum distance along a slope for any projectile with a given slope (\(\phi\)) and initial velocity \(v_0\). I think that was it.. But I can find the answer and post it if it's desired! Thanks to all who helped :)

    • one year ago
  34. DemolisionWolf Group Title
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    awwwh, there was an equation from a prior problem that needed to be used, I can wrap my head around that!

    • one year ago
  35. DemolisionWolf Group Title
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    good question post, a real brain teaser

    • one year ago
  36. theEric Group Title
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    Yeah! It was definitely interesting. So, the formula and thinking behind the formula that was in the previous problem was just as much work. To solve it on the spot would be difficult.

    • one year ago
  37. theEric Group Title
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    Thanks for being a part of it @DemolisionWolf and @PhysicsGuru !

    • one year ago
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