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Angle of inclination of the projectile: θ

  1. Mar 28, 2008 #1
    d = at + ½ a t^2

    Horizontally: 7.12m = a cos θ t

    at = 7.12m/ cos θ

    Vertically: 10m = a sin θ t + ½ g t^2

    10m = (7.12m/ cos θ ) *sin θ + ½ g t^2 = 7.12m tan θ + ½ g t^2

    10m = 7.12m tan θ + ½ g t^2

    10m = 7.12m tan θ + ½ g 1.1^2g

    Solve for θ

    Angle of inclination of the projectile: θ = 29.8o

    This answer (29.8) is wrong...that much I know; what did I do/where did I go wrong in my calcs??
     
  2. jcsd
  3. Mar 28, 2008 #2

    cristo

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    Well, what's the question? What is "a"-- acceleration? Your first equation is incorrect, if so: d=ut+(1/2)at^2
     
  4. Mar 28, 2008 #3
    that's interesting; my teacher says to use a for acceleration... what is 'u'
     
  5. Mar 28, 2008 #4

    cristo

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    u is initial speed, sometimes denoted v_i. One does normally use a for acceleration, but that equation is not correct. See here for the kinematic equations.
     
  6. Mar 28, 2008 #5
    'u' is the variable for what??

    if 'a' is acceleration and I'm not supposed to use 'u', then shouldn't the equation be d=at+(1/2)at^2 ??
     
  7. Mar 28, 2008 #6
    ok, thanks!
     
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