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Magnitude and Direction of Velocity (projectile motion)

  1. Apr 4, 2009 #1
    1. The problem statement, all variables and given/known data
    A baseball is thrown. The ball leaves your hand with a velocity of 15 m/s at 60 degrees above the horizontal. acceleration due to gravity is 9.8 m/s^2. Determine the velocity (magnitude and direction) of ball at the maximum height.


    2. Relevant equations
    im not sure, but i think im supposed to find the x and y components of the velocity first:
    V_x = Voxcos60 = 15 cos 60 = 7.5 m/s
    V_y = Voysin60 = 0 sin 60 = 0
    does this mean that the velocity at max height is just the horizontal velocity component? i know with a horizontal projectile motinon problem, you have to find the components and then plug them into this equation: V^2 = Vx^2 + Vy^2 and then use tan theta to find theta for the direction, but im not sure if this applies to a projectile motion problem which involves an angle.

    3. The attempt at a solution
    see above
     
  2. jcsd
  3. Apr 4, 2009 #2

    Hootenanny

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    Are you sure about these equations? In general,

    [tex]v_x = \left| \vec{v}\right|\cos\theta[/tex]

    [tex]v_y = \left| \vec{v}\right|\sin\theta[/tex]

    where [itex]\left| \vec{v}\right|[/itex] is the magnitude of velocity, or speed.
     
  4. Apr 4, 2009 #3
    so the velocity at max. height for this problem is just 15 m/s in the horizontal direction? Why for a problem that doesnt involve an angle, do you have to use the V^2 = Vx^2 + Vy^2 equation to get the magnitude and direction of velocity when an object lands, but for this problem the magnitude and direction is just the initial velocity to begin with?
     
  5. Apr 4, 2009 #4

    Doc Al

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    Good. I assume you meant that the initial V_x = Vo cos60.
    How did you get this for the initial vertical speed? Vo sin60 ≠ 0.
    I don't know what you mean by "does this mean", since I don't understand what you tried to do. But it's certainly true that at the highest point, the vertical component of velocity is zero. (Otherwise it would keep rising.) And that's all you need to answer the question. :wink:
     
  6. Apr 4, 2009 #5
    "Good. I assume you meant that the initial V_x = Vo cos60." yes-thats what i meant.
    "How did you get this for the initial vertical speed? Vo sin60 ≠ 0." i assumed that there is no inital vertical speed (Y direction). you just said that the vertical component of velocity is zero, isnt that generally the case for a projectile? but if you say that Voysin60 does not equal zero, then why?? because when is the Voy ever not zero?? 0timessin60 = 0
     
  7. Apr 4, 2009 #6
    ooooohhhh! i just realized something- the Vo for these equations is both the inital horiztonal speed: Vocos60 and Vosin60 not Voxcos60, Voysin60!!! oops now i understand
     
  8. Apr 4, 2009 #7
    that is, 15 m/s cos 60 = 7.5 m/s and 15 m/s sin 60 = 12.9 m/s. so these are just the components of the velocity
     
  9. Apr 4, 2009 #8
    i have a question though-why dont i have to use the V^2 = Vx^2 + Vy^2 equation to get the magnitude and direction of velocity at max height? is this only used to find magnit/direct. for velocity when it lands?
     
  10. Apr 4, 2009 #9

    Doc Al

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    I think you've figured it out, but just in case:
    Not true, the initial vertical speed is Vo sin60.
    The vertical component of velocity at the highest point is zero, not at any point.
    Don't confuse Vo_y (the vertical component of the initial velocity, which equals Vo sin60) with V_y (the vertical component at any point, which may equal zero but does not equal Vo sin60).
     
  11. Apr 4, 2009 #10

    Doc Al

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    That equation, V^2 = Vx^2 + Vy^2, is always true. You just don't need it here, since the "trick" is to realize that at the highest point Vy = 0, so:

    V^2 = Vx^2 + Vy^2 = Vx^2 + 0^2 = Vx^2

    Thus V = Vx.
     
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