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Projectile Fired from Point A

  1. Feb 26, 2013 #1
    1. The problem statement, all variables and given/known data

    A projectile is fired from point A at an angle above the horizontal. At its highest point, after having travled a horizontal distance D from its launch point, it suddenly explodes into 2 identical fragments that travel horizonatally with equal but opposite velocitites as measured relative to the projectile just before it exploded. If one fragment lands back at point A, how far from A (in terms of D) does the other fragment land?

    2. Relevant equations

    V1/e-x=V1/p-x + Vp/e-x


    3. The attempt at a solution
    1 is going to the right(+)
    V1/e=V1/p + Vp/e

    V1/e = velocity of 1 relative to earth.
    V1/p= velocvity of 1 relative to projectile original
    Vp/e= velocity of projectile relative to earth

    2 is going to the left(-)
    V2/e-x=V2/p-x + Vp/e-x

    V2/e = velocity of 2 relative to earth.
    V2/p= velocvity of 2 relative to projectile original
    Vp/e= velocity of projectile relative to earth


    This is what I have for one

    X=Xo +vox*t

    X=D+(V1/p + Vp/e)*t

    Ok so at first I thought that the time it took for the 2nd to fall down to origin(A) was the same time it took the first to come to the ground. Then I saw it different. I realized that the 1st fragment(+) actually has a higher velocity after explosion relative to the earth. So that must mean that it could have taken longer for it to fall to the ground. I am stuck here though and I was wondering if I could get some hints.
     
    Last edited: Feb 26, 2013
  2. jcsd
  3. Feb 26, 2013 #2

    ehild

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    What are the y components of velocity of the parts, after the explosion? Was the projectile fired from the ground?


    ehild
     
  4. Feb 26, 2013 #3
    The y components are zero. I am assuming that the projectile was fired from the ground.
     
  5. Feb 26, 2013 #4
    Still a bit confused.
     
  6. Feb 26, 2013 #5

    ehild

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    Both parts starts to fall from the same height, with zero vertical velocity. Any reason why they do not reach the ground at the same time?

    ehild
     
  7. Feb 26, 2013 #6
    Yeah that makes sense. That is what I originally thought. I would think that the answer would be 2D.

    But then this came up.

    V1/e-x = V1/p-x+ Vp/e-x and this V2/e-x=V2/p-x + Vp/e-x

    V1/p and V2/p are both equal in magnitude but opposite in direction right?
    Let us say that it is 5 and vp/e is 10.

    It turns out to be this. V1/e= 15m/s and V2/e= 5m/s. They would have different velocities relative to the earth. Doesn't that change the time for them to come down to the ground?
     
    Last edited: Feb 26, 2013
  8. Feb 26, 2013 #7

    ehild

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    The x component of the velocities are different, but the y components are the same. The parts descend together, but cover different horizontal distances.

    ehild
     
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