A shell is shot with an initial velocity 0 of 29 m/s, at an angle of θ0 = 56° with the horizontal. At the top of the trajectory, the shell explodes 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 that air drag is negligible?
Equation 1: R=((v0^2)/g)*sin(2θ0)
The Attempt at a Solution
R gives the horizontal range of the projectile, however, due to the change in mass and velocity of the shell, I divided the R by 1/2 to get the distance traveled by the shell before its explosion. After that, I'm not sure what to do.