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Kinematics In 2D problems

  1. Oct 5, 2005 #1
    Hello there : )
    I'm new to this forum and I will probably be a regular because I alwasy get stuck on one or two questions from hw ^^;

    Well, I have two questions for now.

    In both cases, I don't know where to start from. It seems some thing is missing (which is highly unlikely).

    Help will be greatly appreciated!
     
  2. jcsd
  3. Oct 6, 2005 #2

    Fermat

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    Homework Helper

    Q1 is, to all intents and purposes , identical with this problem.

    In Q1, you have distance, angle and take-off speed, with angle as the unknown.

    In Q2, you have distance, angle and take-off speed, with take-off speed as the unknown.
    Use the exact same eqns as you used for Q1, but solve them for a different unknown.
     
  4. Oct 6, 2005 #3
    First of all make a draw.
    Than ankyze each dimension seperatly. Do it in a table with x and y in the header.
    Here is an example of the beginging of the table.

    Code (Text):

    _X______|_____Y__
    V=const.| [tex]V=V_0+at[/tex]
    a=0     |a=-g
    .       |.
    .       |.
    .       |.
     
    After you do so, find in the the equations that will help you slove the problem
     
  5. Oct 6, 2005 #4
    Thank you both of you for such a quick replies.
    However, I'm not quite sure if I'm following your instruction.

    For Q1.
    I checked out the similar problem, but can't seem to get any furher than this:
    0 = sin(theta)35 - 9.8*t which gives me the half of the traveling time.
    Well, as for the xy table, I'm don't think I understnad how it works +_+
    Sorry...^^; I tried!
    Wel, thank you very much for your help and I will check this forum out in a few hours again.
    Good day.
     
  6. Oct 6, 2005 #5
    Step by step drawing
    Draw the Y axis and the X axis
    find the point (27, 0)
    Draw a line between (0,0) and (27,0)
    Draw half a circle with this line as diameter.
    Draw the speed:
    The speed is a vector tangent to the circle at (0,0).
    The angle [tex]\alpha[/tex] between the speed and the X axis is the angle we are looking for.
    Break the speed apart, to find the "real speed at each axis"
    The speed at the X axis is V*cos alpha [tex]V_x=V\cos \alpha[/tex]

    The speed at the Y axis is V* sin alpha [tex]V_y_0=V\sin \alpha[/tex]

    At this stage make a table with the following fields for each axis:
    [tex]\sum F[/tex]
    a
    V
    X
    t (equal for both axis)

    Fill the table with equatations and you will be able to see the answer .
    Or you can use this quotation:

    [tex]S=\frac{V^2\sin 2\alpha}{g}[/tex]
    (S is the distance)
     
    Last edited: Oct 6, 2005
  7. Oct 6, 2005 #6

    Fermat

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    Well, you've done half the work.
    You got an eqn for movement in the vertical direction.
    Now do the same and work out an eqn for movement in the horizontal direction.
    When you compare these two eqns with each other, will you be able to work out the answer ?
     
  8. Jun 13, 2008 #7
    i have a similar problem, tried and tried and just cant get it

    A cannon shoots at 400m/s towards a target 400m high and 15000m away from it, what angles must the cannon be at?

    this is a kinematics question i have for homework and just dont know where to turn to. i tried with the equations but seem to always have something missing... i keep getting one equation with T and Sin(theta) or one with T and Cos(theta),, 2 equations 3 missing numbers, dont know what to do??
     
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