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Please check this answer for me! (Projectile Motion Question)

  1. Sep 28, 2005 #1
    Two points A and B are located on the ground a certain distance d apart. Two rocks are launched simultaneously from points A and B with equal speeds but at different angles. Each rock lands at the launch point of the other. Knowing that one of the rocks is launched at an angle [tex]\vartheta [/tex]> 45 degrees with the horizontal, what is the minimum distance between the rocks during the flight? Express your answer in terms of d and [tex]\vartheta[/tex]

    Hm.. this is a very hard problem for me so I hope anyone could help me check this over, as there are lots of rooms for mistakes. Thanks in advance.

    4 base equations that I used for solving:
    [tex](1) x = V_{0}cos\vartheta t[/tex]
    [tex](2)V_{2x} = V_{0}cos\vartheta[/tex]
    [tex](3)y = V_{0}sin\vartheta t + \frac{1}{2}(-g)(t)^2[/tex]
    [tex](4)V_{2y} = V_{0}sin\vartheta + (-g)(t)[/tex]

    [tex]D^2 = (x_{2} - x_{1})^2 + (y_{2} - {y_{1})^2[/tex]

    Rock 1:
    [tex]x_{1} = V_{0}cos \vartheta t[/tex]
    [tex]y_{1} = V_{0}sin \vartheta t - \frac{1}{2}gt^2[/tex]

    Rock 2:
    [tex]x_{2} = d - V_{0}cos \phi t[/tex]
    [tex]y_{2} = V_{0}cos \phi t - \frac{1}{2}gt^2[/tex]

    [tex] cos \phi = cos(90 - \vartheta) = sin \vartheta[/tex]
    [tex] sin \phi = sin (90 - \vartheta) = cos \vartheta[/tex]

    [tex]x_{2} - x_{1} = d - V_{0}(cos \vartheta + sin \vartheta)t[/tex]
    [tex]y_{2} - y_{1} = V_{0}(cos \vartheta - sin \vartheta)t[/tex]

    [tex]Let A = d, B = -V_{0}(cos \vartheta + sin \vartheta) and c = V_{0}(cos \vartheta + sin \vartheta)[/tex]

    [tex]D^2 = (A + Bt)^2 + (Ct)^2[/tex]
    [tex]= A^2 + 2ABt + B^2t^2 + C^2t^2[/tex]
    [tex]= A^2 + 2ABt + (B^2 + C^2)t[/tex]

    Recall quadratic equation = ax^2 + bx + c
    [tex]Let a = B^2 + C^2, b = 2AB and c = A^2[/tex]

    Minimum = vertex
    [tex]x = \frac{-b}{2a}[/tex]
    [tex]... y_{min} = a(\frac{-b}{2a})^2 + b(\frac{-b}{2a}) + c[/tex]
    [tex] = \frac{ab^2}{4a^2} - \frac{b^2}{2a} + c[/tex]

    [tex]... d_{min} = \sqrt\frac{4A^2B^2 - 2(4A^2B^2) + A^2}{4(B+C)}[/tex]
    [tex] = \sqrt\frac{-A^2B^2 + A^2(B^2 + C^2)}{B^2 + C^2}[/tex]
    [tex] = \sqrt\frac{A^2C^2}{B^2 + C^2}[/tex]
     
  2. jcsd
  3. Sep 28, 2005 #2
    Substitute for A, B and C

    [tex] d_{min} = \sqrt\frac{d^2V_{0}^2(cos \vartheta + sin \vartheta)^2}{[-V_{0}(cos \vartheta - sin \vartheta)]^2 + [V_{0}(cos \vartheta - sin \vartheta)]^2}[/tex]

    [tex] = \frac{dV_{0}(cos \vartheta + sin \vartheta)}\sqrt { V_{0}^2(1- 2cos \vartheta sin \vartheta) + V_{0}^2(1- 2cos \vartheta sin \vartheta)}[/tex]
     
  4. Sep 28, 2005 #3
    sorry for the separate messages.. but the preview keeps showing me my previous equations -_-

    [tex] = \frac{dV_{0}(cos \vartheta + sin \vartheta)}{V_{0} \sqrt 2 ( 1 - 2 cos
    \vartheta sin \vartheta)}[/tex]

    [tex] = \frac{d(cos \vartheta + sin \vartheta)}{\sqrt 2 ( 1 - 2 cos
    \vartheta sin \vartheta)}[/tex]
     
  5. Sep 29, 2005 #4

    Päällikkö

    User Avatar
    Homework Helper

    With d = 1 m, and [itex]\vartheta[/itex] = 40 degrees, from your equations I get the minimum distance as 66 m, which does seem a bit too big.

    In the quadratic equation you've assumed the distance hits zero. This is not the case:
    [tex]A^2 + 2ABt + (B^2 + C^2)t^2 = D^2 \ne 0[/tex]

    So, your c should be: c = A2 - D2

    EDIT:
    I got [tex]|\mathbf{r}|_{min} = \frac{d}{2}\sqrt{(\sin(2 \theta)-1)^2+(2 \sin ^2 \theta - 1)^2}[/tex]
    But I'm not too convinced :)
     
    Last edited: Sep 29, 2005
  6. Sep 29, 2005 #5
    oh yes 66m does seem unreasonably big.
     
  7. Sep 29, 2005 #6

    Päällikkö

    User Avatar
    Homework Helper

    Oh, I suppose there's no use trying to solve t from the quadratic equation (you'd just end up with the same variables).
    You should rather figure out a way to compute t when distance is at its minimum.

    Hint: There's a mathematical method for this.
     
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