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± acceleration due to gravity?

  1. Jul 16, 2008 #1
    A ball is thrown from the top of one building toward a tall building 50 m away. The initial velocity of the ball is 20 m/s, 40° ABOVE HORIZONTAL. How far above or below its original level will the ball strike the opposite wall?





    [tex]v_{ix} = 20 cos 40°[/tex] = 15.3 m/s
    [tex]v_{iy} = 20 sin 40°[/tex] = 12.9 m/s

    Using [tex]x = v_{x}t[/tex] to find time;
    50 = 15.3 t => t = 3.2 seconds.

    We use [tex]y = v_{iy}t + \frac{1}{2}a_{y}t^2[/tex] to find How far above or below its original level will the ball strike the opposite wall;

    Now I don't know wether I should take acceleration due to gravity, a = 9.8 as positive or negative for this. (What about viy = 12.0?)


    [tex]12.9 \times 3.27 + \frac{1}{2} \pm 9.8 \times 3.27[/tex] ?

    Thanks.

     
  2. jcsd
  3. Jul 16, 2008 #2

    tiny-tim

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    … be consistent …

    Hi roam! :smile:

    Either … but you must be consistent!

    Decide whether you're going to measure y upward or downward.

    If downward, then of course g is +9.8, but viy is -12.

    If upward, then g is -9.8, but viy is +12. :smile:
     
  4. Jul 16, 2008 #3
    Yes, thank you, Tiny tim. I had to make sure since I was skeptic about this.

    I understand how it goes now, thanks. :smile:
     
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