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Two-Dimensional Motion help with a Flatcar

  1. Sep 27, 2009 #1
    1. The problem statement, all variables and given/known data

    A man is riding on a flatcar traveling at a constant speed of 8.85 m/s. He wishes to throw a ball through a stationary hoop 3.60 m above the height of his hands in such a manner that the ball will move horizontally as it passes through the hoop. He throws the ball with a speed of 12.0 m/s with respect to himself.

    (a) What must the vertical component of the initial velocity of the ball be?
    (b) How many seconds after he releases the ball will it pass through the hoop?
    (c) At what horizontal distance in front of the hoop must he release the ball?
    (d) When the ball leaves the man's hands, what is the direction of its velocity relative to the frame of reference of the flatcar?
    Relative to the frame of reference of an observer standing on the ground?



    2. Relevant equations

    I am not sure at all what equations I need :(
    I was thinking:

    x-xo = vot + .5at^2
    vh = vcos(theta)
    vv= v sin(theta)

    I don't know what else I could possible need.

    3. The attempt at a solution
    I attempted part a first assuming that when he tossed it up at 12 m/s that meant the vertical velocity was 12 m/s (wrong). I then redid it drawing a triangle and doing Pythagorean theorem (with 12 being the hypotenuse and 8.55 m/s being the horizontal component) to get 8.104 m/s, but that was wrong too.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 27, 2009 #2

    Q_Goest

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    Hi wishingwell. If the hoop is 3.6 meters above his hands when released, and since the ball stops its upward motion at that point and begins dropping again (ie: all velocity is horizontal and not up/down as it passes through the hoop) then the velocity upwards when it leaves the man's hands is exactly the same as if it were dropped from a height of 3.6 meters into his hands.
    (a) So can you find the upward velocity of the ball first?
    (b) The time it takes to reach a height of 3.6 meters is the same time it takes for it to fall that far.
    (c) The ball has a verticle and horizontal component of velocity when it leaves the man's hands. You should have found the verticle component per (a) above. You also are given the velocity with respect to the man which has a verticle and horizontal component. Can you now find the horizontal component? Once you have this horizontal component and the time from (b) along with the velocity of the flat car, you should then be able to determine the horizontal distance.
    (d) This answer follows from the (c) above. Find the angle it is thrown at with respect to the man throwing the ball and with respect to the ground.
     
  4. Sep 29, 2009 #3
    Your thoughts for (a) are what first come to mind when you look at the problem; but, once you break the problem down, you will realize that it is easier than it first appears.

    For the first step, you need to notice that the problem states "the ball will move horizontally as it passes through the hoop." Without this, you will be in a world of pain. Interpreting this, you now know that the y-velocity is zero when it passes through the hoop.

    This simplifies the problem; now, you are basically just looking for the solution to this:

    [tex]y = y_{0} + v_{0_{y}}t + \frac{1}{2}a_{y}t^2[/tex]

    We know the following pieces of information, from the problem:

    [tex]y = 3.60\text{m}[/tex]

    [tex]y_{0} = 0\text{m}[/tex]

    [tex]a_{y} = -g = -9.80\frac{\text{m}}{\text{s}^2}[/tex]

    This leaves us with two free variables for our previous equation (v and t), meaning we need another equation.

    [tex]v_{y} = v_{0_{y}} + a_{y}t[/tex]

    With algebraic manipulation (and knowing that v with respect to the y-axis will be zero at the hoop), you can now solve the second equation for t and substitute the result for t in the first equation.

    [tex]0 = v_{0_{y}} + a_{y}t[/tex]

    [tex]t = \frac{-v_{0_{y}}}{a_{y}}[/tex]

    [tex]t = \frac{-v_{0_{y}}}{-g}[/tex]

    The rest of the work from here on out is just algebra. (I am too tired to type the rest of it out in LaTeX; and I'm sure you will figure it out from this point onward.)

    To give you some reference (to make sure you're getting the right answers): if you search Google for "flatcar hoop," you will find a similar http://www.google.com/url?sa=t&source=web&ct=res&cd=29&url=http%3A%2F%2Fwww.luc.edu%2Ffaculty%2Fdtribbl%2Fhw111%2F111hw04.doc&ei=TMTCSsbRJZ6Jtgfiu4HnBA&usg=AFQjCNHUil7h5AK6Q9ZPSkjfzqtR_RDRZg&sig2=V2tkaBbQZr-0N3o-UeB35Q" [Broken] in *.doc format.

    Good luck!
     
    Last edited by a moderator: May 4, 2017
  5. Jan 17, 2010 #4
    I am doing this problem as well, and I have gotten the answers to A and B, but C is totally stumping me. How do I get the horizontal component of the velocity with respect to the man?
     
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