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Projectile motion angle problem

  1. Dec 15, 2012 #1
    1. The problem statement, all variables and given/known data

    A projectile is thrown from a point P at an angle θ above the horizontal . It moves in such a way
    that its distance from P is always increasing from its launch until its fall back to the ground. Find all the possible values of θ with which the projectile could have
    been thrown. You can ignore air resistance.




    3. The attempt at a solution

    y=Vosinθ-0.5gt^2

    x=Vocosθ

    Taking the magnitude of a distance vector

    √(x^2 + y^2) = √((Vocosθ)^2+(Vosinθ-0.5gt^2)^2)=√(((vsinθ)-0.5gt^2)^2 +v^2cos^2θ)

    I really have no idea what to do afterwards

    Edit: I checked several threads about this and i found that 70.5 is maximum value, and 45 is also possible, but i don't know how to find ALL the possible values
     
  2. jcsd
  3. Dec 15, 2012 #2

    BruceW

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    your equations aren't quite right. For x and y, your equations say that the position of the object would be constant if gravity was zero. Does this sound right?

    Apart from that, you have the right idea to calculate the magnitude of the distance vector. The next step after that is to think if we require the magnitude of the distance vector to always be increasing with time, then what does that tell us about our function for the magnitude of the distance vector? In other words, we are interested in the rate of change of the magnitude of distance vector, so what should you do to the magnitude of distance vector? And what do you require of this quantity.

    Or, depending on the way you like to think about these problems, think about the graph of magnitude of distance versus time. What do we want it to look like? And what does this mean for the graph of the derivative against time?

    p.s. 70.5 is the maximum value (i.e. if the angle is less than 70.5 then the distance of the projectile will never decrease), you are on the right track to calculate this.
     
  4. Dec 15, 2012 #3
    Typo -_- sorry forgot to multiply by t

    y=Vosinθ*t -0.5gt^2

    x=Vocosθ*t

    √(x^2 + y^2) = √((Vocosθ*t)^2+(Vosinθ*t-0.5gt^2)^2)

    We have to take the derivative of the magnitude to time ?

    dD/dt √((Vocosθ*t)^2+(Vosinθ*t-0.5gt^2)^2) =


    (480.2t^3 -147t^2*vsin(θ)+10tv^2)/(√(t^2(2401.t^2-980.tvsin(θ)+100v^2)))
     
  5. Dec 15, 2012 #4

    BruceW

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    I think you should keep the constants as just symbols, rather then put the numbers in, or it could get pretty messy. And yes, you have the right idea to take the derivative of the magnitude with respect to time. So from here, what do you want dD/dt to be at all times?
     
  6. Dec 15, 2012 #5
    I want dD/dt > 0 so it is always positive

    ∴ after derivation i get

    t(g^2 * t^2 -3Vo*g*sin(θ)t +2Vo^2)>0 ???

    Edit: Not sure if the derivation is correct, if it's wrong correct me
     
    Last edited: Dec 15, 2012
  7. Dec 15, 2012 #6

    BruceW

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    yep!

    Yep, that's right. You've almost finished now.
     
  8. Dec 15, 2012 #7
    Awesome,

    Let the discriminant be ∇

    Now this polynomial here is positive when t>0 if ∇<0

    ∇=9Vo^2 * g^2 *sin(θ)^2 -8Vo^2 *g^2

    ∇<0 ∴ sin(θ)<√(8/9)

    Solving for θ

    2(pi)n - (pi) -sin^-1 ((2√2)/3) < x < 2(pi)n + sin^-1 ((2√2)/3) where n ∈ ℤ

    (ℤ is the set of integers)


    Ps: I solved it as an inequality but i have to find all the possible values of θ, is this solution correct?
     
  9. Dec 15, 2012 #8

    BruceW

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    Nice. That is the answer. The angle theta is restricted between zero and (pi)/2 since it is defined as the angle above the horizontal. So you can just put this into your calculator to get the answer in degrees. (Or just leave it in this form).

    So you have shown that if the angle is less than this special value, then the distance will always be increasing with time. And the questions was "Find all the possible values of θ with which the projectile could have been thrown." So you now know that all values of theta which are less than the special value are possible values of theta.

    (Take a moment to think about what the question has asked, and what you have found so far). So the final part of the problem: What about the values of theta greater than the special value? Will they definitely be excluded from being possible values of theta?
     
  10. Dec 15, 2012 #9
    The question has asked to find the possible values of θ I found the maximum value





    Yeah I think they'll be exluded since the maximum value of θ = 70.53 , so any value higher than this value won't be a possible angle.
     
  11. Dec 15, 2012 #10

    BruceW

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    Well, you are right, but you have not shown it yet. You have only shown that θ less than 70.53 are possible values, you have not shown that θ greater than 70.53 must not be possible values.

    Edit: to be honest, what you have done already will probably get you a good grade, but you might get a better grade if you also show that any θ greater than 70.53 is not a possible value.
     
  12. Dec 15, 2012 #11
    I'm stuck -_- how do i show that no values bigger than 70.53 are possible
    (hint?)
     
  13. Dec 16, 2012 #12

    BruceW

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    have a look at your polynomial again. (The thing which needs to be positive so that distance is increasing). This must be positive for all positive values of t, right? So as long as you can show that it is negative for some value of t when the angle is greater than 70.53 then you will have a counter-example to the claim that there exists some possible angle which is greater than 70.53.

    Edit: did I explain that alright? I'm sure that you can logically go through it yourself, but maybe you just need a push in the right direction. For any angle greater than 70.53 then what can you say about your discriminant? And so for every angle greater than 70.53 can you provide a time (t) which disagrees with the claim that the distance must always increase?
     
    Last edited: Dec 16, 2012
  14. Dec 19, 2012 #13
    Sorry

    I am VERY sorry for the late reply i had to get in contact with my teacher, he didn't respond first but now he did.

    He said i didn't have to prove that no values bigger than 70.53 are possible values, but he did say that i have to "list" all the values.

    i got the maximum value, and i know that the possible values are between 0 and 70.53. now what should i do ?

    PS : I want to thank you for helping me Bruce, only a little more left.
     
  15. Dec 19, 2012 #14

    mfb

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    [0,70.53°] is the interval of all possible angles. I don't see how you can make a list in a meaningful way.
     
  16. Dec 19, 2012 #15
    Yes, 70.53 is the maximum angle but i have to list them (find all the possible values of θ)

    Edit : So i can only say that the values of theta are between 0 and 70.53, but can't list them ?
     
  17. Dec 19, 2012 #16

    mfb

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    $$[0,arcsin(\sqrt{\frac{8}{9}})] = \{\theta|0\leq \theta \leq arcsin(\sqrt{\frac{8}{9}})\}$$
    That is the best approximation to a list you can get, as the number of possible values is uncountable.
     
  18. Dec 19, 2012 #17
    Thank you mfb, so the full solution should go like this:

    y=Vosinθ*t -0.5gt^2

    x=Vocosθ*t

    √(x^2 + y^2) = √((Vocosθ*t)^2+(Vosinθ*t-0.5gt^2)^2)

    dD/dt √((Vocosθ*t)^2+(Vosinθ*t-0.5gt^2)^2)

    I want dD/dt > 0 so it is always positive

    after derivation i get:

    t(g^2 * t^2 -3Vo*g*sin(θ)t +2Vo^2)>0

    Let the discriminant be ∇

    Now this polynomial here is positive when t>0 if ∇<0

    ∇=9Vo^2 * g^2 *sin(θ)^2 -8Vo^2 *g^2

    ∇<0 ∴ sin(θ)<√(8/9)

    [0,arcsin(√(8/9))]={θ|0≤θ≤arcsin√(8/9))}
     
  19. Dec 19, 2012 #18

    BruceW

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    Re: Sorry

    Yes. What your professor is saying you need to do is what I referred to as a 'good' grade. (which is what you have already done). In other words, you know that all angles between zero and 70.53 are possible angles. To give a full answer, you should really then show that any angle greater than 70.53 is not a possible angle. But your professor is saying you don't need to do this to get full marks.
     
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