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Horizontal projectile

  1. Sep 5, 2006 #1
    The problem states; A projectile is fired horizontally with an initial speed of 50m/s. neglect air resistance. What is the magnitude of the displacement of the projectile 3 seconds after it is fired? What is the speed of the projectil 3 seconds after it is fired? I used this equation to get my X component: X=Vox(t)+(1/2)(-9.8m/s^2)(t^2) and I got 105.9m. I also did this for the Y component I used 50m/s for initial velocity and I think that is what my mistake is, because when I use the pathagorean theorm I don't get the correct answer. Can anyone help?
     
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  3. Sep 5, 2006 #2

    tony873004

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    It's fired horizontally, so what is the initial x velocity? What is the original y velocity?
     
  4. Sep 5, 2006 #3
    I got the answer finally the initial velocity for Y is 0m/s. I knew that is where I was going wrong but I still don't really understand why it is 0.
     
  5. Sep 5, 2006 #4

    tony873004

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    Because that's what "horizontally" means. If it had any velocity at all in the y direction, either up or down, then it would not be moving horizontal.

    It's similar to your other question where it said "60 degrees above the horizontal". "Horizontal" means "0 degrees above the horizontal". You can even do trig on it if you want:

    Your other problem:
    initial velocity = 30 m/s
    x-component = cos(60)*30 = 15 m/s
    y-component = sin(60)*30 = 25.98 m/s

    This problem
    initial velocity = 50 m/s
    x-component = cos(0)*50 = 50 m/s, since cos(0)=1
    y-component = sin(0)*50 = 0 m/s, since sin(1)=0.

    Did you get the right answer? This one was trickier than your last problem.
     
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