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Help with Projectiles

  1. Sep 15, 2007 #1
    1. The problem statement, all variables and given/known data

    A shot putter releases the shot some distance above the level ground with a velocity of 12.0 m/s, 51.0 degree above the horizontal. The shot hits the ground 2.08s later. You can ignore air resistance.


    What is the x-component of the shot's velocity at the end of its trajectory?
    What is the y-component of the shot's velocity at the end of its trajectory?
    2. Relevant equations

    x = v initial cos theta time

    3. The attempt at a solution

    I managed to get the x and y component of acceleration.

    but for this I would suppose its going to be 12cos51 2.08 = 15.71

    I am lost...as this is wrong.
     
    Last edited: Sep 15, 2007
  2. jcsd
  3. Sep 15, 2007 #2

    learningphysics

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    The 2.08 should be s, not m/s right?

    What is the initial x and y component?

    What is the acceleration of the shot put?
     
  4. Sep 15, 2007 #3
    x-component of the shot's velocity at the beginning of its trajectory

    7.55m/s

    y-component of the shot's velocity at the beginning of its trajectory
    9.32

    y-component of the shot's acceleration while in flight
    -9.80
    x-component of the shot's acceleration while in flight
    0

    from what I got thusfar.

    now I need to find the x and y component of the shot's velocity at the end of its trajectory
     
    Last edited: Sep 15, 2007
  5. Sep 15, 2007 #4

    learningphysics

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    cool. you have everything you need... you have initial velocity... acceleration... time... you should be able to get final velocity.

    EDIT: why do you have -9.32 instead of 9.32?
     
  6. Sep 15, 2007 #5
    so the equation would be vx = v initial +gt

    vx = 11.99 + (9.8)(2.08)

    = 32.37

    but now how do I get x and y, (I don't have degree for the final velocity)
     
  7. Sep 15, 2007 #6

    learningphysics

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    No... vertical acceleration affects vertical velocity... horizontal acceleration affects horizontal velocity...
     
  8. Sep 15, 2007 #7
    sorry , but i don't see where you are going with this...
     
  9. Sep 15, 2007 #8

    learningphysics

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    The x-component of acceleration is 0... so does the x-component of velocity change?
     
  10. Sep 15, 2007 #9
    no...same for y also....
     
  11. Sep 15, 2007 #10

    learningphysics

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    Ah... but there is a y-component to acceleration... calculate the final y-velocity using initial y velocity and the y-acceleration.
     
  12. Sep 15, 2007 #11

    learningphysics

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    vx = v0x + ax*t

    vy = v0y + ay*t
     
  13. Sep 15, 2007 #12
    vx = v0x + ax*t

    vy = v0y + ay*t

    vx = 7.55 + (0)(2.08)

    vy = 9.32 + (-9.8)(2.08)

    vx = 7.55

    vy = -11.064

    Which is correct, thank you learning physics.
     
    Last edited: Sep 15, 2007
  14. Sep 15, 2007 #13
    also,

    How far did she throw the shot horizontally?

    this is what I did

    I decided to use the following equation: x-x initial = t/2 (vinitial + vx)

    I know the vf and vi for x and y.

    so I decided to do a^2 + b^2 = C^2

    so the velocity for initial would be 11.99
    for final it would be 13.39

    so now its

    x-0 = 2.08 /2 (11.99+13.39)

    x= 26.39

    so that would be distance right? but I am wrong :(
     
  15. Sep 15, 2007 #14

    learningphysics

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    you're making it too complicated... horizontal displacement, you only need horizontal velocity... you know the horizontal velocity, you have the time... you know acceleration is 0...
     
  16. Sep 15, 2007 #15
    I tend to do that.

    x-0 = 7.55(2.08) + .5........

    final being 15.70 :)
     
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