Projectile motion velocity If a projectile breaks into two pieces at the highest point of its trajectory and one of them retraces the same path, why will that piece have the same magnitude in velocity as the whole projectile?
 A: At every point along the trajectory you have two velocity components $\left\{v_x(t),v_y(t)\right\}$. You have asked what it means for a projectile to retrace its path. That the original projectile broke up and/or the second part now has a smaller mass is not relevant here since this is a kinematics problem.
This is simply $\left\{-v_x(t),-v_y(t)\right\}$. Reversing the sign of the velocity components retraces the path and also obeys Newton's laws for a free falling object. Moreover, this is the only solution that retraces the path and is consistent with a body in free-fall. 
Now, the magnitude of the velocity is simply $|v_\text{forward}(t)|=\sqrt{v_x^2(t)+v_y^2(t)}$ in the first case. In the second case we have $|v_\text{return}(t)|=\sqrt{\{-v_x(t)\}^2+\{-v_y(t)\}^2}$. 
Thus $|v_\text{forward}(t)|=|v_\text{return}(t)|$
A: Because for the x-component you have:
$$ma_x=0\to a_x=0, $$
and for the y-component
$$ ma_y=-mg \to a_y=-g. $$
In both cases the equations of motions are independent of mass. If you have a particle which seperates into smaller pieces at one point and one of those pieces follows the same trajectory as the initial part, the smaller piece follows the same solution of the equations of motion. This means it will also have the same magnitude of velocity as the initial part.
