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Batted Baseball

  1. Feb 3, 2009 #1
    1. The problem statement, all variables and given/known data

    A batted baseball leaves the bat at an angle of 33.0 above the horizontal and is caught by an outfielder 400 ft from home plate at the same height from which it left the bat

    What was the initial speed of the ball?
    How high does the ball rise above the point where it struck the bat?


    2. Relevant equations

    the 3 acceleration equations

    v = vo + at
    v^2 = vo^2 + 2a(deltax)
    x = xo + vot + 1/2at^2


    3. The attempt at a solution

    I am at a lost to find velocity. Cannot seem to even start the problem.
     
  2. jcsd
  3. Feb 3, 2009 #2

    Kurdt

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    You will need to split the problem up into horizontal and vertical components of motion. Once you have done that it should be relatively easy to find the initial speed of the baseball. All you have to do is find an equation with variables you know and one that you want.
     
  4. Feb 3, 2009 #3
    I believe I will be using components of initial velocity. Still alittle lost on how to proceed.

    U0y = U0sin(33)
    U0x = U0cos(33)
     
  5. Feb 3, 2009 #4

    Kurdt

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    Ok so you have the components of the initial velocity there. Now you're told that the ball is caught 400ft away at the same height it left the bat. You know there is no acceleration in the horizontal component so which equation can you use to solve for the initial velocity?
     
  6. Feb 3, 2009 #5
    Vx=x/t then 0=Vyit = 1/2gt^2

    Vcos33*t = ?
    Vsin33 - (1/2)gt = 0

    I think im in the right track, still alittle lost.
     
  7. Feb 3, 2009 #6

    Kurdt

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    How far does it travel in the horizontal direction? It tells you in the question. Now you need to eliminate the t variable by rearranging the vertical equation and plug it into the horizontal equation and solve for V.
     
  8. Feb 3, 2009 #7
    Goes 400 ft in the horizontal direction.

    t=cos33/v

    Vsin33 - (.5) -9.8 * cos33/v = 0

    Am I getting warmer?
     
  9. Feb 3, 2009 #8

    Kurdt

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    Nearly. [itex]v\cos(33) t = ?[/itex], what distance does the horizontal equation equal? Plus you need to be careful with units. The distance is given in feet and the gravitational constant you're using is in metres per second squared. Decide which units you're going to use.
     
  10. Feb 3, 2009 #9
    would it vcos(33)t= 121.92?


    400ft = 121.92 m
     
  11. Feb 3, 2009 #10

    Kurdt

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    Yes :smile:
     
  12. Feb 3, 2009 #11
    ok haha, what is the 121.92 m? besides the horizontal direction?
     
  13. Feb 3, 2009 #12

    Kurdt

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    Nothing else, thats it.
     
  14. Feb 14, 2009 #13
    Physics question!!!

    AHHHHHH I am in serious need of help with my physics problems. I have this assignment due by 11pm tonight, and I am absolutely lost. This stuff will not "click". I was wondering if anyone would be able to help me.
     
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