I have no idea how i keep getting the wrong answer. Here's what i did At the top of the trajectory the horizontal velocity is the same which is cos(35)(125). By using conservation of momentum i can find the velocity of fragment 2 MVix = MVfx (28)(cos(35)(125)) = (14)(0) + (14)(Vf2x) So i just find Vf2 which is 204.788011 m/s Well to calculate range i need to know the max height, but we also need the time it reaches max height. So to find time that reaches max height: Vfy = Voy + ayt 0 = sin(35)(125) + (-9.81)t t = 7.3085s Max Height: dy = Voyt + 0.5ayt dy = (sin(35)(125))x(7.3085) + 0.5(-9.81)(7.3085) dy = 262m We now know the max height, horizontal velocity of fragment 2, isn't this like the simple projectile motion question where: (Horizontal velocity of fragment 2) / (time it takes for the object to fall 262m) = the range of fragment 2 then you add that up to the range it will take the projectile to get to max height. Isn't that the answer? If it's not clear enough please tell me so i can rephrase it or something.