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Projectiles launched at complementary angles

  1. Apr 19, 2017 #1
    1. The problem statement, all variables and given/known data
    a) Show that for a given velocity [itex]V_0[/itex] a projectile can reach the same range [itex]R[/itex] from two different angles [itex]\theta = 45 + \delta[/itex] and [itex]\theta = 45 - \delta[/itex], as long as [itex]R[/itex] doesn't go over the maximum range [itex]R_{max} = \frac{V_0^2}{g}[/itex]. Calculate [itex]\delta[/itex] in function of [itex]V_0[/itex] and [itex]R[/itex].

    b) Generalize your previous result, showing that a projectile launched from the ground with an initial velocity [itex]V_0[/itex] can reach a point [itex](x, y)[/itex] from two different angles, as long as the point [itex](x, y)[/itex] is within the parabola of safety:[tex]y = \displaystyle{\frac{1}{2}\left(\frac{V_0^2}{g} - \frac{x^2}{\frac{V_0^2}{g}}\right)}[/tex]
    2. Relevant equations
    Will edit if there is need of any.

    3. The attempt at a solution
    So far this was my progress:

    First in regards to a), I went ahead to show that complementar angles resulted in the same range:[tex]\frac{V_0^2sin(2 \theta)}{g} = \frac{V_0^2sin(2 \times 90- \theta)}{g} = \frac{V_0^2sin(180 - \theta)}{g}\Longleftrightarrow sin \theta = sin(180 - \theta)[/tex]From that, I continued:[tex]R = \frac{V_0^2sen(90-2 \delta)}{g} \Rightarrow Rg = V_0^2cos(2 \delta) \Rightarrow cos(2 \delta) = \frac{Rg}{V_0^2} \Rightarrow \delta = \frac{1}{2}cos^{-1}\left(\frac{Rg}{V_0^2}\right)[/tex]My initial question is wether the first part of my demonstration was satisfactory, because for some reason I don't really think it was, considering I didn't include the [itex]\theta = 45 \pm \delta[/itex] on it...

    And now for the b) part, I tried to consider the position equations:[tex]x(t)=(V_0cos \theta)t \\ y(t)=(V_0sin \theta)t - \frac{gt^2}{2}[/tex]By eliminating t, we have:[tex]y = xtan \theta - \frac{g}{2} \frac{x^2}{V_0^2 cos^2 \theta}[/tex]And now I don't know how can I continue the demonstration... x___x

    I'd really appreciate some feedback on both of these problems!
     
  2. jcsd
  3. Apr 19, 2017 #2

    BvU

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    you may have lost something on the way. We can't tell if you don't post your steps :smile: so please guide us...
    It's not totally incorrect but somewhat trivial. It's not what the exercise composer meant :rolleyes: .
     
  4. Apr 19, 2017 #3
    Uhm. I'm confused now. What did I lose? D:
    Doesn't that notation imply they are complementar?

    And yes, I realized it's not really what was asked, but I don't know how should I proceed.
     
  5. Apr 19, 2017 #4
    The range R is given by V02sin(2θ)/g. The sin(2θ) solves part (a).

    For part (b), the parabola of safety is the parabola centered at x = 0, whose height is the maximum height of ½V02/g and maximum range is ±V02/g. To show that any point under this envelope can occur twice, use

    sec2θ = 1 + tan2θ

    to reduce your bottom most equation to

    αtan2θ + βtanθ + δ = 0,

    where the coefficients are in terms of x and y. tanθ ranges as (-∞,∞), so all θ in the range (-π/2,π/2) are viable.
     
  6. Apr 19, 2017 #5

    BvU

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    How do you solve ##\frac{V_0^2\sin(2 \theta)}{g} = X## ?
    $$\frac{V_0^2\sin(2 \theta)}{g} \ne \frac{V_0^2\sin(\pi- \theta)}{g} $$Instead you want to solve
    $$\frac{V_0^2\sin(2 \theta)}{g} = \frac{V_0^2\sin(\pi- 2\theta)}{g} $$
    (note: use \sin instead of sin in ##\LaTeX##)
     
  7. Apr 19, 2017 #6

    PeroK

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    Hmm! If you start from:

    ##R_{\pm} = \frac{V_0^2}{g} \sin(2\theta) = \frac{V_0^2}{g} \sin(90 \pm 2\delta)##

    I suggest you can simply state that these are equal by symmetry of the ##\sin## function. And every value strictly between ##0## and ##R_{max}## is possible by continuity of the ##\sin## function.

    PS Although, I see you have to calculate ##\delta## in any case.
     
    Last edited: Apr 19, 2017
  8. Apr 19, 2017 #7

    PeroK

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    This is not strictly correct, as points vertically above the launch point can only be reached by one angle.
     
  9. Apr 20, 2017 #8
    I'm sorry, I didn't really follow what you mean x_x. But here's what I tried:

    Noting [itex]R_{max} = \frac{V_0^2}{g}[/itex] and [itex]r = \tan \theta[/itex], we find a quadratic equation for [itex]r[/itex]: [tex]y = rx- \left( \frac{1+r^2}{2R_{max}} \right) x^2[/tex]Which we rewrite as[tex]x^2r^2-(2R_{max})r + (2R_{max}y + x^2) = 0[/tex]The equation has two real solutions if[tex](R_{max}x^2)-x^2(2R_{max}y + x^2) \geq 0[/tex]or[tex]2R_{max}x^2 \left[ \frac{1}{2}\left(R_{max}-\frac{x^2}{R_{max}}\right)-y\right] \geq 0[/tex]which means (x, y) is a point within the parabola. What do you guys think?
     
  10. Apr 20, 2017 #9
    Yes, that's true, but still, I need to suggest a solution. D:
     
  11. Apr 21, 2017 #10

    PeroK

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    You could set up a quadratic equation in ##\tan \theta## then use the discriminant to identify in terms of ##x## and ##y## when it has two solutions.
     
  12. Apr 21, 2017 #11
    Yes, I did that, check my above comments.
    Do you think it's good?
     
  13. Apr 21, 2017 #12

    PeroK

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    I think it's right, but setting ##R_{max}## equal to half of what it should be was confusing.
     
  14. Apr 21, 2017 #13
    Hmm... But why half? That's the value the exercise gives.
     
    Last edited: Apr 21, 2017
  15. Apr 21, 2017 #14

    PeroK

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    Yes, you're right.
     
  16. Apr 21, 2017 #15
    Thank you! By the way, what do you think in regards to a)? Is there a way I could make the answer more appropriate or is it fine?
     
  17. Apr 21, 2017 #16

    PeroK

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    Re part a), I would prefer:

    ##R_{\pm} = \frac{V_0^2}{g} \sin(2\theta_{\pm}) = \frac{V_0^2}{g} \sin(90 \pm 2\delta) = R_{max} \cos(2\delta)##
     
  18. Apr 21, 2017 #17
    That seems good. Thanks a lot!
     
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