Why is the constant velocity model used in a projectile motion derivation? I was re-studying university physics last week, I'm now in the chapter about kinematics in 2 dimensions and specifically the one treating projectile motion. In page 86 of his book (Serway - Physics for scientists and engineers) he derives the equation of the range of the projectile motion to be: $$R=\frac{{v_i}^2\sin2\theta_i}{g}$$
But I don't know why he used one of his assumptions

$\color{red}{\bf Question1:}$ Why $v_{xi}=x_{x\rlap\bigcirc B}$? Where $\rlap\bigcirc {\,\sf B}$ is the time when the projectile stops.
$\color{darkorange}{\bf Question2:}$ Why did he use the particle under constant velocity model to derive that formula, whereas here we deal with a projectile under constant acceleration?
Any responses are welcome, I'm disappointed a lot about those matters!
 A: The force of gravity is in the y direction only.  
There is no force on the particle in the x direction.  
Therefore, the x-component of velocity is constant.
A: I don't understand question 1: where does he equate a speed to a position?
As far as question 2 is concerned, it is basically what DavePhD said, but maybe I can extend it a bit more saying something about the conservation of linear momentum:
Along the x-direction, there is no external force (because gravity points downwards only, assuming a flat surface) so the linear momentum of the projectile is conserved.
Since $p_x = mv_x$, $v_x$ is constant.
A: "Why did he use the particle under constant velocity model to derive that formula, whereas here we deal with a projectile under constant acceleration?"  You appear to have the classic confusion between the vertical component of the motion and the horizontal component.  Vertically you have to include the acceleration due to gravity in your thoughts.  Horizontally, strictly speaking, you should include a decelaration arising from atmospheric friction (at least where there is an atmosphere of some sort to consider). However it's much easier, and simplifies the problem enormously, to assume that you can ignore that deceleration.  Then, all you have to deal with in the horizontal direction is speed = distance / time.  I read however, that in the very practical application of these ideas to gunnery, the air friction is a significant modification to the simple theory.  A shell will have a much shorter range in reality compared to its range computed using 'simple' projectile theory to predict where it will fall.
A: If we assume there is no wind and air, the only force acting on a projectile in the air is the force of gravity. The force of gravity acts in vertical direction, thereby affecting the vertical component or Y component of velocity only.
Newton's 1st law says that a particle will continue to be in its state of rest or motion if no external force is impressed on it. Can you think of any force acting on the projectile in horizontal direction? There is none if we assume no wind/ air resistance etc. Thus there is no reason for velocity of the projectile to change in X direction. Hope this will make the above derivation simpler for you to understand.
You can also watch this video I made, in case it is still not clear
Analysing Projectile Motion in X and Y direction
