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Trajectory of a Base Ball

  1. Sep 25, 2009 #1
    Now I'm seriously just tired and can't get these, but I dunno why this is wrong at all. This would help if I had a study partner in my physics class (to bounce ideas off of), but no one else in my class is overly strong in the subject.

    1. The problem statement, all variables and given/known data
    A batter hits a pitched ball when the center of the ball is 1.22m above the ground. The ball leaves the bat at an angle of 45 degress with the ground. With that launch, the ball should have a horizontal range (returning to the launch level) of 107m

    (a) Does the ball clear a 7.32m high fence that is 97,5 n horizontally from the launch point? (b) At the fence, what is the distance between the fence top and the ball center?


    2. Relevant equations
    y = (tan theta)x - (g(x^2))/(2(vicos45)^2)

    R = (2Vi^2/g)(Sin 2 Theta)



    3. The attempt at a solution

    107 = (2Vo^2/g) * Sin 2(45)
    107 * g = 2Vo^2 * 1
    1048.6 / 2 = Vo^2
    524.3 = Vo^2
    22.89 = vo^2

    y = (tan 45)(97.5) - (9.8(97.5)^2) / ((2)(-524.3)(cos45)^2)

    y = 97.5 - 177.58

    y = -80

    The ball clears the fence, so why am I getting -80 for the y coordinate?

    Okay, how about this... Lets try using basic kinematics

    vix = d/t

    t = d/vix
    t = 97.5/(22.89cos45)
    t = 6.023

    d = Viy(t) + .5(g)t^2
    d = 22.89sin45(6.023) + .5(-9.8)(6.023)^2
    d = 97 - 177.7
    d = -80

    Why am I getting -80 as my displacement? It's supposed to clear the fence. I had to obviously have done something wrong when calculating the Vi in the range.
     
  2. jcsd
  3. Sep 25, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Your range formula is incorrect.
     
  4. Sep 25, 2009 #3
    Yeah, after I went to bed I realized that there was a random 2 that I forgot to get rid of in my Sin 2 Theta identity... Thanks for confirming it.
     
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