1. The problem statement, all variables and given/known data An instrument carrying rocket accidentally explodes at the top of its trajectory. The horizontal distance between the launch point and the point of explosion is L. The rocket breaks into two pieces that fly apart horizontally. The larger piece has three times the mass of the smaller piece. To the surprise of the scientist in charge, the smaller piece returns to Earth at the launching station. How far away does the larger piece land? Neglect air resistance and effects due to the Earth's curvature. 2. Relevant equations Momentum conserved Kinematics equations 3. The attempt at a solution Let M4 = initial mass Let 3m = mass that flies away from the launch site. Let m = mass that flies towards the launch site. 4m * V(M4) = - m * V(m) + 3m * V(3m) Since velocity at the top is 0: M4(0) = - m * V(m) + 3m * V(3m) m * V(m) = 3m * V(3m) Therefore V(3m) = V(m)/ 3 Since m travels L horizontally, L = V(m) * t therefore V(m) = L/ t If we plug in V(m) = V(3m) * 3 we get V(3m) = L / (t*3) (since velocities are all horizontal, both take the same amount of time to fall down) Plugging this into the Kinematics Eqn for distance, we get D = V(3m) * t - .5 * 9.8 * t^2 D = (L/(3*t)) * t - .5 *9.8 * (L/(3*(v(3m)))^2 I dont know how to get rid of the v(3m). thanks in advance.