I only dont put the answer that need be proved (I write it in short).

Perhaps my translatio to english in the question is wrong...

Here is again:

calculate the shortes (the littlest) angle that the final velocity of a fragment will do with the horizontal. Solution [tex] tan^{-1}\left[\left(2gh\right)^{\frac{1}{2}} / v \right] [/tex].

If a piece leaves the explosion traveling with its velocity straight DOWN,
what is the angle that its final velocity makes with the ground?

If a different piece leaves the explosion with initial velocity straight UP,
what is the angle that its final velocity makes with the ground?

If some piece leaves the explosion with its initial velocity HORIZONTAL,
what is the angle that its final velocity makes with the ground?

Which final velocity angle is smallest?
Can you find the final velocity angle as a function of initial velocity angle?
If so, you can show that horizontal initial velocity should give the smallest final angle, if it is not obvious already (you do want the largest horizontal component, right? and the smallest vertical component...) set [tex]\frac{d\theta_{final}}{d\theta_{init}} = 0[/tex]