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Throwing a ball: Parabolic motion

  1. Feb 3, 2012 #1
    Hopefully I learned a few things from the previous thread and got some of this right.

    1. The problem statement, all variables and given/known data
    Refer to the diagram. The ball is thrown with the parameters given on the diagram. a) Compute the total height the ball reaches and the height the ball will reach above the building. b) Compute the x-displacement the ball reaches from the base of the building. c) With what velocity does the ball hit the ground?
    Height of building = 45.0m
    Initial velocity = 20.0 m/s
    Angle thrown = 30.0 degrees



    2. Relevant equations
    Ymax = Vot + (1/2)at^2
    V = sqrt Vx^2 + Vy^2
    Quadratic
    Delta(y) = Voy(time) - (1/2)gt^2


    3. The attempt at a solution

    a) v = Vo + at
    0 = 20 + (-9.8)t
    -20 = -9.8t
    t = 2.04s

    Ymax = Vot + 1/2 at^2
    20(2.04) + 1/2 (9.8)(4.16)
    40.8+20.38 = 61.18m

    61.18 - 45 = 16.18m

    b) Vox = Vo Cos(Theta) = (20m/s)(cos30) = 17.3 m/s
    Voy = Vo Sin(Theta) = (20m/s)(sin30) = 10 m/s

    Delta(y) = Voy t - 1/2gt^2
    -45 = (10m/s)t + (-4.9)t^2
    -4.9t^2 + 10t + 45 = 0
    4.9t^2 - 10t - 45 = 0
    x = 10 + sqrt 100-4(4.9(-45)/9.8
    x = 10 + sqrt 100+882/9.8
    x = 10 + 31.34/9.8
    x = 4.22 seconds

    Vo Cos(Theta)t = (20m/s)(cos30)(4.22)
    = (20)(.866)(4.22)
    = 73.09m

    c)
    Vy = Vo Sin(Theta) - gt
    = (20 m/s)(1/2) - (9.8)(4.22)
    = 10 m/s - 41.36 = -31.4 m/s

    V = sqrt Vx^2 + Vy^2
    = sqrt (17.3)^2 + (-31.4)^2
    = sqrt 299.30 + 985.96
    = sqrt 1285.26
    = 35.85 m/s
     
  2. jcsd
  3. Feb 3, 2012 #2
    You seem to have forgotten your diagram, I assume the guy is throwing the ball from the top of the building?
     
  4. Feb 3, 2012 #3
    Yes, sorry, I'm not aware of how I can attach a diagram so I added the important values after the directions.

    To answer your question, yes, he is throwing it straight off the side of the building at 20 m/s at a 30 degree angle.
     
  5. Feb 3, 2012 #4
    it looks like you have the right idea about splitting the velocity into x and y components, except in part a the total time you've figured out is the time it would take for the ball to reach its peak if he threw the ball straight up at 20m/s.

    The acceleration due to gravity points straight down, and doesn't point at all into the x direction. This means that only velocities in the y direction will be affected, while velocities in the x direction will remain constant.
     
  6. Feb 3, 2012 #5
    Bummer. >< Any tips as to what else I could try?
     
  7. Feb 3, 2012 #6
    well, if you consider that only the y velocity is changing, and it will be decreasing, then at some point your ball will reach a peak and then start heading back down. We don't need to worry about the x velocity if we want to find the time. It's similar to your last problem, find the time it takes for the ball to reach zero velocity in the y direction, but make sure you consider only the initial velocity in the y direction.

    actually you don't even need to worry about time, if you just want to find the height that it reaches when its y velocity equals zero

    you might find this equation helpful:

    v2 = v02 + 2a(x-x0)

    but remember to consider only the y velocities
     
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