What is the distance that a projectile lands from its origin?

[Hungry!]

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?
Please explain by using projectile equations absent from air resistance.

[Melanie]

let's call the two smaller masses m
then the mass before explosion was 2m
so
the momentum at the moment of explosion is 2mV
after the explosion one mass comes back to A retracing its original path so it must have -V and its momentum is -mV
momentum is conserved so the total must still be 2mV
2mV = -mV + XmV
X= 3 so the second fragment must have velocity 3V

the highest point is the halfway point of the trajectory
if there were no explosion the mass would have travel led another distance D for a total of 2D
but it is going 3 times as fast so it will go another 3D
this gives a total distance from A of 4D

[federox]

2D, the second fragment lands where the projectile was going to land had it not exploded.

Projectile motion follows a parabolic path. The maximum height is the vertex of the parabola that models the projectile's path. At this point, the velocity is 0 and the projectile has traveled half its total distance. For the fragment that traveled backwards from the vertex to land on A, it must have had the same horizontal velocity that the projectile initially did. This means that it traveled the same path of the projectile except backwards. Since the fragment that went forwards had the same horizontal velocity, it also followed the path the projectile would have taken before exploding and landed in the same place. As I said before, the projectile followed a parabolic curve. A is one root of that curve. D is the distance from A to the vertex, the point of explosion which occurs at the max height and is half the total distance. Let's label the point at which the fragment lands (where the projectile was supposed to land) as B. The horizontal distance from the vertex to B is also D. So the second fragment lands 2D from A.
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Correction for Old Science Guy's comment:

When conserving momentum, 2mV = m(-V) + mV, realize that 2mV = 0 because this point is at the vertex where the velocity is 0. So the second fragment DOES NOT go at 3V  but rather at V. Also, such a thing is absurd considering the problem tells you that both fragments travel at EQUAL but opposite velocities.

[curlz]

Old Science Guy is right.
Note to A. Square:  At the vertex only the vertical component of the velocity is zero.  The horizontal component is a constant throughout the problem if you ignore air resistance which is typically done in an elementary problem such as this