I need some help with the problem. A shell is shot with an initial velocity v0 of 20 m/s, at an angle of 60 degrees with the horizontal. At the top of the trajectory, the shell explodeds into two fragments of equal mass. 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 the air drag is negligible First, I labled three parts of the trajectory. 1. At the initial position K= (1/2) m * v^2 U = 0 2. At the max height of the trajectory K = (1/2) (m/2) * v^2 ----> Because the shell exploded in half U = (m/2) * g * h 3. At the end of the the trajectory K = 0 U = 0 Then using the conservation of energy from points 1 to 2 I found the velocity at point 2 to be 22.36 m/s. Now I have to finish the problem and find the total distance traveled. Can someone check my work up to now and give me some advice on how to find the total distance.