Displacement:

The overall displacement in the x-direction is the range R. The overall displacement in the y-direction is zero. Remember that displacement is a vector that is drawn from the initial point to the final point.

The displacement shown as s has two components, s_x and s_y. This displacement is to the projectile at the peak of its motion. The next displacement vector shown (unlabeled) has a larger x-component and a smaller y-component. The displacement vector shown as R is entirely in the x-direction; it has a zero y-component.

Distance:

The distance traveled for each of the three positions of the projectile (shown above) is not the magnitude of the displacement (vector). The three distances are along the arc of the parabola!

Velocity:

The diagram below shows that the initial velocity (as well as all other velocities) has two components, which are:

The components at other points in the path vary. At the top of the motion, for example, there is no y-component of the velocity.

Force and acceleration:

Gravity acts on the projectile, and so there is an acceleration of 9.8 m/s² down (I won't call this negative since the sign depends on which way we take as positive).

There are forces (such as friction) that act on the projectile in the x-direction. But for short distances, such forces are much smaller than the gravitational force (weight of the projectile) and we will ignore them. Since Newton's 2^nd Law states that a net force produces an acceleration, there is no acceleration in the x-direction. So a_y = 9.8 m/s² (positive or negative, depending on the direction we choose as positive), and a_x = 0.

Kinematics:

Velocity...Let's take the usual directions as positive (up and to the right). In addition, let's agree to use the initial firing point as the origin (this means that x_o and y_o are zero). Because velocity is a vector, it can be written as two components instead of the overall v:

The x-direction equation merely says that the x-component of the velocity is constant (does not depend on time) and equals the magnitude of the overall initial velocity times the cosine of the firing angle.

The y-direction equation could be used to find the time to the maximum height, since the y-component of the velocity is zero there. And since the projectile's motion is symmetric, we could double that time to get the total time (don't try this for a projectile that impacts at a different height from the firing point!).

Displacement...Since x_o and y_o are zero, I can write

These equations describe x and y for the projectile at any point in its motion. Consider the impact point, where x = R and y = 0:

The second equation has two solutions, t = 0 and . Substituting this into the x-direction equation,

There's a trig identity ("identity" means something that is always true) that states:

To apply it to the formula I just derived, I need to multiply the top and bottom by 2:

Because I substituted "g" for 9.8, I can use this for any value of g anywhere (for example, on the moon). Here's an example that illustrates how useful equations that are written in general terms can be.