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2D Kinematics On Planet Zorg

  1. Oct 17, 2014 #1
    Hello everybody. Although this is just a kinematics problem, im having a hard time figuring out what to do. The question is:

    An archer on the planet zorg spots an octomorph that is resting in a tree above water. The octomorph's position in the tree is 100m away horizontally and 100m vertically away from the bow of the archer. The archer shoots a quantum arrow at the creature with an initial speed of 100m/s, however, at the instant the arrow is fired, the octomorph drops straight out of the tree into the water below. If the acceleration of gravity on planet zorg is 98.1m/s^2 , in what direction should the archer aim his arrow if it is to hit the octomorph in stride on the way down?

    2. Relevant equations
    y= -1/2g(t^2)+Voy(t) + yo
    x= Vxt

    3. The attempt at a solution
    I have figured out the position of the octomorph as a function of time : y=(-49.05m/s^2)t^2 + 100m

    I have no idea how to make the arrow and octomorph reach the same position in the same time. We cannot simply set their equations equal to eachother as the arrow has an x component to take into consideration. Any help or ideas will be greatly appreciated!
  2. jcsd
  3. Oct 17, 2014 #2
    You are going to have two equations in two unknowns (the time and the angle). The horizontal location of the arrow has to match the horizontal location of the octomorph, and the vertical location of the arrow has to match the vertical location of the octomorph. Let ##\theta## be the angle (measured from the horizontal) that the arrow is launched. In terms of ##\theta##, what are the x and y components of the arrow velocity? In terms of ##\theta## and t, what are the horizontal and vertical locations of the arrow at time t (relative to the archer)?

  4. Oct 17, 2014 #3
    The horizontal component of velocity is (Vo)cosθ which is (100m/s)cosθ

    And the vertical component of velocity will be (100m/s)sinθ

    The position equation in the horizontal direction will be

    X = (100m/s)cosθ(t)

    And for the y direction it will be

    Y = (-49.05m/s^2)t^2 + (100m/s)sinθ(t)
  5. Oct 17, 2014 #4
    Im thinking now we might have to combine and somehow manipulate these equations in order to solve for the angle and time. Not sure exactly how since there are two unknowns. Thats what im thinking atleast.. but i will wait for your response
  6. Oct 17, 2014 #5
    Also i believe that the value for x will be 100m regardless of what occurs in the y direction. Meaning that the full equation for the horizontal direction will be

    100m = (100m/s)cosθ(t)
    cosθ = 1s(second)/t
    θ = arcos( s/t )

    is that right?
  7. Oct 17, 2014 #6
    How are the x and y coordinates of the arrow related to the x and y coordinates of the octorph when the arrow hits the octomprph?

  8. Oct 17, 2014 #7
    The will both have the same x and y coordinates.

    Both x coordinates will be at X = 100m

    I think also means that the y equations will be equal to each other.

    (-49.05m/s^2)t^2 +(100m/s)sinθ(t) = (-4.905m/s^2)t^2 +100m
  9. Oct 17, 2014 #8
    So, you can cancel the acceleration terms from both sides of the y equation. What does that leave you?

  10. Oct 17, 2014 #9


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    Gold Member

    I don't think any equations or calculations are needed to answer this question.
    Last edited: Oct 17, 2014
  11. Oct 17, 2014 #10
    I don't think so either, but that is only because we both have lots of experience. I don't think I would have known this as a new initiate like the OP. (Probably you would have).

  12. Oct 17, 2014 #11
    Yes they can be canceled.

    This will leave the equation as

    (100m/s)sinθ(t) = 100m

    Im sorry.. but still cant see where this is going. Sorry for the late reply. i had to attend some classes
  13. Oct 17, 2014 #12
    So you have:

    100 sinΘ t = 100


    100 cosΘ t =100

    What do you get if you divide the first equation by the second?

  14. Oct 17, 2014 #13
    You divide them so
    θ=arctan(1)=45 degrees

    Is that right
  15. Oct 17, 2014 #14
  16. Oct 17, 2014 #15
    Thank you so much i appreciate it! Thumbs up!!
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