theEric
  • theEric
(Classical Mechanics) Would somebody be able to help, maybe with hints, on this projectile motion problem? Thanks! It involves cannons! Scenario is below.
Physics
jamiebookeater
  • jamiebookeater
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

theEric
  • theEric
|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.
theEric
  • theEric
|dw:1383512248696:dw|
theEric
  • theEric
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 \(0\).

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

theEric
  • theEric
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.
anonymous
  • anonymous
What is the question asking for?
theEric
  • theEric
It is asking for \(L\), the maximum length the ground cannon can be from the tower and still hit the top of the tower.
anonymous
  • anonymous
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: \[V_x = \frac{L}{t_{hang}}\]
theEric
  • theEric
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?
theEric
  • theEric
I can't* prove that. I just thought it made sense at the time. But then also I thought it didn't make sense...
anonymous
  • anonymous
That is a constraint we are applying because you want the ball to reach the top of the cliff correct?
anonymous
  • anonymous
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
theEric
  • theEric
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...
anonymous
  • anonymous
Yes but you need to maximize L which means you are looking for the case we are talking about above
theEric
  • theEric
Alright, I'll give that a shot! Thanks!
theEric
  • theEric
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|
theEric
  • theEric
|dw:1383528051969:dw|
theEric
  • theEric
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.
anonymous
  • anonymous
@theEric I think i got an idea, but maybe it's too simple for what you are working on
theEric
  • theEric
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
anonymous
  • anonymous
is the answer suppose to have any specific variables? or can it be two equations or just needs to be 1?
theEric
  • theEric
There is no air resistance, and gravity is constant.
theEric
  • theEric
The answer will be something like \(L=\dots\) And it will have the variable \(v_0\), and maybe \(\theta\)...
theEric
  • theEric
I think...
anonymous
  • anonymous
oh ok, i'll put up what I have then, maybe it helps, then i'm gonna go watch tv ^_^
theEric
  • theEric
Haha, okay, thanks! And have fun watching tv! :)
anonymous
  • anonymous
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.
theEric
  • theEric
Alright, thank you! Take care!
anonymous
  • anonymous
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.
theEric
  • theEric
thank you very much! You didn't have to, but thank you! I'm going to check it out. Thanks!
theEric
  • theEric
I think eq 2 assumes \(v_{y\text{, final}}=0\); but I don't think that is the case, unfortunately.
theEric
  • theEric
same with eq 3. But eq 2 = \(v_y^2\) and eq 3 = \(v_y\), so maybe I'll end up using that!
anonymous
  • anonymous
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.
theEric
  • theEric
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 :)
anonymous
  • anonymous
awwwh, there was an equation from a prior problem that needed to be used, I can wrap my head around that!
anonymous
  • anonymous
good question post, a real brain teaser
theEric
  • theEric
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.
theEric
  • theEric
Thanks for being a part of it @DemolisionWolf and @PhysicsGuru !

Looking for something else?

Not the answer you are looking for? Search for more explanations.