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Vertical Component of a Velocity Vector

  1. Nov 10, 2007 #1
    1. The problem statement, all variables and given/known data
    [​IMG]

    A particle of mass 0.450 kg is shot from P. The particle has an initial velocity [itex]V_0[/itex] with a horizontal component of 30.0 m/s. The particle rises to a maximum height of h = 21.0 m above P. Assume that h2 = 62.0 m.

    Determine the horizontal and the vertical components of the velocity vector when the particle reaches B.

    2. Relevant equations

    Ay = Asin[tex]\theta[/tex] ?? i think

    3. The attempt at a solution

    Okay well I already know the horizontal component is still 30.0 m/s because no force is acting upon it. But I dont even know if I'm even using the right equation to solve for the vertical component.
     
    Last edited: Nov 11, 2007
  2. jcsd
  3. Nov 10, 2007 #2
    [tex]V_{0V} = V_{0}sin{\theta}[/tex]

    The easy way to remember if it's sin or cos is to know that sin0 is 0 and cos0 is 1.
     
  4. Nov 10, 2007 #3
    im sorry that doesn't really help me..
     
  5. Nov 11, 2007 #4
    Sorry if I'm wrong about this, but shouldn't there be an initial velocity? If that's zero, it isn't moving at all, and definitely won't be going up. However, If I read that wrong, or have forgotten some rule of dynamics, forgive me.

    Anyway, I'd take the vertical component at a known point; the maximum of the arc. As Vi is then known, as is acceleration, you can solve for Vf. I didn't see any value for the distance from P to B, so I'm assuming you have to use variables, and arrive at a general solution?
     
  6. Nov 11, 2007 #5
    The diagram shows an initial velocity, [itex]V_0[/itex].
     
  7. Nov 11, 2007 #6
    ya, mae, look at the problem again.. you MUST have forgotten something or typed wrong.. how can something have an initial velocity of 0, but have an X component of 30m/s at the same time?? anyway.. keep in mind that the X component of the vector doesn't change because there is no acceleration in that direction, and when the particle reaches its peak, the vertical component of the velocity is 0.. but the problem isn't clear..lol
     
  8. Nov 11, 2007 #7
    There's a typo in "The particle has an initial velocity 0 with a horizontal component of 30.0 m/s". Should be [itex]V_0[/itex], not 0.
     
  9. Nov 11, 2007 #8
    oh ok
     
  10. Nov 11, 2007 #9

    D H

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    The "velocity 0" is a typo. I take it that mean meant v0.

    mae, you are correct in that the horizontal velocity doesn't change. All you have to worry about is the vertical velocity.

    =====

    Suppose you release an object with zero initial velocity. What velocity does it attain after falling some distance d? Ignoring the horizontal velocity, the particle does have zero velocity at the point 21 meters above initial point. Treat this as your new initial point, with zero initial velocity. How far does the particle fall? Can you now determine the particle's final vertical velocity?
     
  11. Nov 11, 2007 #10
    im sorry guys, yeah i did make a typo and i changed plus i added the value of h2.

    D H, i know i must have the wrong equation or something because i dont even know [tex]\theta[/tex]. So to solve for how far the particle falls, i should use a kinematic equation instead, am i right? And Vidatu, i think you're right here. I already know the initial velocity and the acceleration, so I should just end up with a value for the final velocity?
     
    Last edited: Nov 11, 2007
  12. Nov 11, 2007 #11
    OK, now we've got that sorted would you like to share any thoughts with us about how to answer the question?
     
  13. Nov 11, 2007 #12

    D H

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    You don't need to know the angle [itex]\theta[/itex]. All you need is the total height the particle falls, and that is [itex]h + h_2[/itex].
     
  14. Nov 11, 2007 #13
    okay i think i might've figured it out.

    i used the kinematic equation: [tex]V_{f}^2 = V_{i}^2 + (2ad)[/tex]
    i assume the initial velocity is 0, so we can ignore it in this case.
    what im solving for should be the final velocity.
    if the acceleration is pointed downwards, then a = -9.81 m/s^2?
    the displacement is also downwards, so it should be negative too... i guess
    so i add both 21 and 62 which is 83, but in this case it would be -83?

    if i plug in what i know i end up with this:

    [tex]V_{f}^2 = V_{i}^2 + (2[-9.8][62+21])[/tex]
    [tex]V_{f} = \sqrt{(2[-9.8][-62+-21])}[/tex]
    [tex]V_{f} = \sqrt{(2[-9.8][-83])}[/tex]
    [tex]V_{f} = \sqrt{(1626.8)}[/tex]
    [tex]V_{f} = 40.33[/tex]
     
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