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Help with horizontal velocity

  1. Apr 23, 2014 #1
    Help plzzz with horizontal velocity

    1. The problem statement, all variables and given/known data
    Ms. Rob launches a rocket from the school roof at an angle of 45 degrees and observes it's horizontal range. What other angle could she launch the rocket so that it lands in the same spot, ignoring air resistance?


    2. Relevant equations
    V=d/t
    D=v/t
    D=1/2gt^2
    V=gt
    A^2+b^2+c^2


    3. The attempt at a solution

    I just can't get it at all I'm getting really frustrated and it's due Friday! :cry::frown:
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 23, 2014 #2

    Simon Bridge

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    What is special about the angle of 45deg when it comes to the range?

    You can find the horizontal and vertical components of velocity from trigonometry.
    recall: sin(45)=cos(45)=1/√2

    You can work out the equations of motion by sketching velocity-time graphs for the components.

    Please make more of an attempt at problems before posting - there is some very strict moderation going on here at the mo and you have to show some little effort beyond saying "I have no idea". You do have some ideas - you must have spent at least a year in physics class learning how to solve problems in general: do try to apply that knowledge.
     
    Last edited: Apr 23, 2014
  4. Apr 23, 2014 #3
    Thanks

    Thanks for the help I'm kinda still in Conceptual physics 1 and I don't know about trigonometry yet. Is their a simpler way? I just started science this year...thanks AGIAN
     
  5. Apr 23, 2014 #4

    Simon Bridge

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    You can always draw the triangle and measure - but since this is conceptual physics, you have had a result in your notes about what to do. There's something special about the range at 45degs.

    Note: If you know trigonometry anyway - do use it.
     
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