Question

well we'll fired with a velocity U at an angle of theta with the horizontal at highest point of a trajectory split up into three segments mm and 2 m first part Falls vertically downward with the zero initial velocity and second part returns via same path to the point of projection the velocity of third past of mass 2m that after explosion will be

Answer 1

Projectile at:  speed = u , angle = Ф,  mass = m+m+2m = 4m

At the highest point:  vertical speed = 0.   Horizontal speed = u CosФ.

The part retracing the path will have horizontal speed = u CosФ.

Apply the momentum conservation at the highest point.
           m * 0 + m * (- u CosФ) + 2 m * v = 4m * u CosФ
           v = 5/2 * u CosФ