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theEric

  • 2 years ago

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

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  1. theEric
    • 2 years ago
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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.

  2. theEric
    • 2 years ago
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    |dw:1383512248696:dw|

  3. theEric
    • 2 years ago
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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 \(0\).

  4. theEric
    • 2 years ago
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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.

  5. anonymous
    • 2 years ago
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    What is the question asking for?

  6. theEric
    • 2 years ago
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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.

  7. anonymous
    • 2 years ago
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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: \[V_x = \frac{L}{t_{hang}}\]

  8. theEric
    • 2 years ago
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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?

  9. theEric
    • 2 years ago
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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...

  10. anonymous
    • 2 years ago
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    That is a constraint we are applying because you want the ball to reach the top of the cliff correct?

  11. anonymous
    • 2 years ago
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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

  12. theEric
    • 2 years ago
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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...

  13. anonymous
    • 2 years ago
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    Yes but you need to maximize L which means you are looking for the case we are talking about above

  14. theEric
    • 2 years ago
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    Alright, I'll give that a shot! Thanks!

  15. theEric
    • 2 years ago
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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|

  16. theEric
    • 2 years ago
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    |dw:1383528051969:dw|

  17. theEric
    • 2 years ago
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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.

  18. anonymous
    • 2 years ago
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    @theEric I think i got an idea, but maybe it's too simple for what you are working on

  19. theEric
    • 2 years ago
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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

  20. anonymous
    • 2 years ago
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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?

  21. theEric
    • 2 years ago
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    There is no air resistance, and gravity is constant.

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

  23. theEric
    • 2 years ago
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    I think...

  24. anonymous
    • 2 years ago
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    oh ok, i'll put up what I have then, maybe it helps, then i'm gonna go watch tv ^_^

  25. theEric
    • 2 years ago
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    Haha, okay, thanks! And have fun watching tv! :)

  26. anonymous
    • 2 years ago
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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.

  27. theEric
    • 2 years ago
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    Alright, thank you! Take care!

  28. anonymous
    • 2 years ago
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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.

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

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

  31. theEric
    • 2 years ago
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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!

  32. anonymous
    • 2 years ago
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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.

  33. theEric
    • 2 years ago
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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 :)

  34. anonymous
    • 2 years ago
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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!

  35. anonymous
    • 2 years ago
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    good question post, a real brain teaser

  36. theEric
    • 2 years ago
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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.

  37. theEric
    • 2 years ago
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    Thanks for being a part of it @DemolisionWolf and @PhysicsGuru !

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