How Far Does the Second Fragment Travel from Point A After Explosion?

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Homework Statement



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?

Homework Equations



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


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


ehild

The y components are zero. I am assuming that the projectile was fired from the ground.
 
ehild said:
What are the y components of velocity of the parts, after the explosion? Was the projectile fired from the ground?


ehild

Still a bit confused.
 
ehild said:
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

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?
 
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