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Finding Velocity given angle and distance.

  1. Mar 10, 2009 #1
    1. The problem statement, all variables and given/known data
    The questions is, that if you throw something at exactly 45 degrees above the horizontal and it travels 60 m before hitting the ground whats it velocity?

    2. Relevant equations
    Physic's equation, see below

    3. The attempt at a solution

    Horizontal Component
    Vh = cos 45
    = 0.707

    now t = 60/0.707

    Vertical Component
    Vv = 0.707 - gt

    What do I do now? Am I right to this point as well?
  2. jcsd
  3. Mar 10, 2009 #2


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    Homework Helper

    Welcome to PF.

    The first thing to do is start over.

    Thrown at 45 degrees means that vertical and horizontal speeds are the same.

    So how long is it in the air? Vy/g is the time to go up ... so total time (up and down) is 2*Vy/g

    And how fast is it going to go 60 m?

    a = 60 = Vx*t = Vx*2*Vy/g

    Since Vx = Vy then just solve.
  4. Mar 10, 2009 #3

    Ok cool, so is Vy cos 45 = 0.707? so then Vy = Vx = 0.707?
  5. Mar 10, 2009 #4


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    Homework Helper


    .707 is merely the value of cos45 and sin45.

    Your mission is to figure out what the initial velocity is.
  6. Mar 10, 2009 #5
    ok so is this right,

    Vo = [tex]\sqrt{(g*d)/(2 sin Theta*cos Theta)}[/tex]

    Vo = [tex]\sqrt{(9.8*60) / (2 sin 45*cos 45)}[/tex]

    Vo = 24.24
  7. Mar 10, 2009 #6


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    Homework Helper

    That would be correct.
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