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3D Cartestian Coordinate Vector Problem

  1. Oct 2, 2014 #1
    EDIT: There was an issue where half of the post was missing, so I apologise but i have redone it

    I was not sure if this was the most appropriate forum for this or not, so feel free to move if needed.

    1. The problem statement, all variables and given/known data

    A football (soccer ball) with a diametre of 0.44m is tracked using a motion tracking camera system and is found to have a trajectory of

    [itex] x=(-2.5+5t)m \\
    y=(2+3t)m \\
    z=(0.5 +2t-5t^2)m[/itex]

    There are a few parts to the question and the part before the one I am stuck on asked to find the time when the centre of the ball touches the ground, so what I did was set z=0.22 and solved for t using quadratic forumla which produced the answer 0.513 s which I am happy with. The next part asks:

    What is the velocity immediately before it hits the ground?

    2. Relevant equations

    N/A



    3. The attempt at a solution

    It is this bit which I am a bit stuck on, mainly because the way the question is set out is different to any similar kind of problems I have done previously. The only thing I can think of is differentiating each of those equations for the x, y and z components to get the x,y and z components of the velocity but that would only leave the z component as a function of time, I suppose that doesnt matter, but its the only thing I can think of, even though we have not covered this in the first two lectures we have had of this module.

    Does this make sense?

    [itex]
    \frac{dx}{dt} = v_x = 5 \\
    \frac{dy}{dt} = v_y = 3 \\
    \frac{dz}{dt} = v_z = 2-10t \\
    [/itex]

    And then @ t=0.513
    [itex]
    v_z = 2 - 10(0.513)=-3.13 \\
    \, \\
    \therefore \vec{v}= (5 \hat{i} + 3 \hat{j} - 3.13 \hat{k} ) m/s \\
    \therefore |\vec{v}|=\sqrt{5^2+3^2+3.13^2} = 6.62 m/s

    [/itex]
    Any help is really appreciated.

    Thanks :)
     
    Last edited: Oct 2, 2014
  2. jcsd
  3. Oct 2, 2014 #2

    Orodruin

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    You know the time it hits. What is the relationship between position and velocity? How does it look at an arbitrary time?
     
  4. Oct 2, 2014 #3
    Yes I do know the time it hits (as long as I worked it out correctly, but for the sake of argument lets say that I did). Velocity is the derivative of postiion with respect to time?? What do you mean by how does it look at an arbitrary time?

    EDIT: I just realised maybe you only say the post when half of it was missing? Something happened when I first posted it and half of the post got deleted, but its back there now. If not and you saw it as you see it now then I apologise.
     
  5. Oct 2, 2014 #4

    Orodruin

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    Yes, I posted before your edit of the OP. Your approach looks reasonable. Without a force acting in the x and y directions, the velocity components in those directions will not change.
     
  6. Oct 2, 2014 #5
    Ah right ok, thanks for taking a look :)
     
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