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2D Projectile Motion

  1. Sep 9, 2012 #1
    A ball is tossed from an upper story window of a building. It has an initial velocity of 8.00 m/s at an angle of 20 degrees below the horizontal. How long does it take the ball to reach a point 10.0m below the level of launching.

    I tried using the equation
    yf=vi(sin(theta))-.5gt^2
    Which I simplified to
    (yf-vi(sin(theta)))/.5g = t^2
    Then
    (10-8(sin(20)))/(4.91) =t^2
    Finally t=1.23, however my book says I should be getting 1.18s, what am I doing incorrectly?
    Thanks in advance
     
  2. jcsd
  3. Sep 9, 2012 #2
    Your equation is missing a "t" in the term containing the initial velocity.
    The general form should be
    [tex]y=y_0+v_{y0}t+1/2gt^2[/tex]
     
  4. Sep 9, 2012 #3
    Would that mean that I need to solve the equation using the quadratic formula?
     
  5. Sep 10, 2012 #4

    CAF123

    User Avatar
    Gold Member

    You are missing a 't' after the [itex] v_o\sin\theta [/itex] term. After you have this, rearrange your equation into the form [itex] at^2 + bt +c = 0 [/itex] and use methods of solving quadratics.
     
  6. Sep 10, 2012 #5
    Yes.
     
  7. Sep 10, 2012 #6

    rl.bhat

    User Avatar
    Homework Helper

    Check the sign convention. If g is negative in the downward direction then yf -yo and vi*sin(theta) should be negative in the downward direction.
     
  8. Sep 10, 2012 #7
    Got it, thanks guys
     
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