1. The problem statement, all variables and given/known data A shell is shot with an initial velocity http://edugen.wileyplus.com/edugen/shared/assignment/test/session.quest2529559entrance1_N10030.mml?size=14&ver=1436312588112 [Broken] of 17 m/s, at an angle of θ0 = 63° with the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass (see the figure). One fragment, whose speed immediately after the explosion is zero, falls vertically. How far from the gun does the other fragment land, assuming that the terrain is level and that air drag is negligible? http://edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/c09/fig09_46new.gif 2. Relevant equations m2/M d = xcom 3. The attempt at a solution I have found the correct answer for this one its 35.77m which i solve by inspection and symmetry. My problem is how do i find the distance of the other particle if it was not symmetric? I tried using, xcom = m1x1 + m2x2 / Mwhere M = sum of all masses. And solving this eq for x2 But i keep getting 23.86m which is the distance of xcom.