One really good way to solve problems of constant acceleration

One way to visualize two-dimensional projectile motion:
Projectile motion problems can sometimes seem overwhelming because there seems to be so many
equations to keep track of, and it can be difficult to visualize just what is going on. This is one suggestion as
to how you can plow through any projectile motion problem. But this is not a “recipe”; rather it’s a way of
looking at the problem that makes it easier to set up your strategy. (And, indeed, setting up is usually the
hardest part of a problem anyway.)
First, make sure you understand some very key things about any projectile motion situation:
1. The projectile is changing position depending on time t: [x(t), y(t)]. (The notation, x(t) represents a math
expression where the independent variable is t and the dependent variable is x.)
2. The projectile’s velocity v is also changing with time t : both magnitude v (=speed) and direction
(=angle from +x axis).
3. Acceleration a is always 9.81 m/s2 downward. If you define vertical to be the y axis, then this means a
has only a y-component, = -g; the x component of a is zero. Therefore: only the y-component vy of the
velocity v is changing, while the x-component vx is the same number at all times during the motion.
Second, try to imagine this: the position [x(t), y(t)] of your object is indicated by two markers: one on the xaxis and the other on the y-axis. As the motion takes place, the x-marker slides along the x axis, and the ymarker slides along the y-axis. Now, focus on the behavior of these two markers. Notice that they move
independently of each other – while following the same time t. The end effect is two-dimensional motion: as
the two markers move, as the point (x, y) moves at velocity v, it traces out a trajectory on the x-y graph.
Here’s the punch line:
The motion of the x-marker is one-dimensional constant-velocity motion along the x- axis, and
The motion of the y-marker is one-dimensional free-fall motion along the y-axis.
Details: the velocity of the x-marker is the x-component of the object’s velocity vector, and the velocity of
the y-marker is the y-component of the object’s velocity vector. Remember that the projectile’s velocity is v
= vx x^ + vy y^. Now, vx(t) is only along the horizontal, and vy(t) is only along the vertical. Therefore you can
think of vx as the velocity of the x-marker, and you can think of vy as the velocity of the y-marker.
Since vx does not change with time, then the x-marker acts like an object traveling with constant velocity (=
vox ) along the x-axis. vy looks like the vertical velocity of an object initially thrown up, or thrown down, or
released from rest; therefore the y-marker acts like an object in free fall along the y-axis.
You know the procedure for solving a one-dim constant-velocity problem along the x axis. You know the
procedure for solving a one-dim free-fall problem along the y-axis. Therefore you can treat the motion of
the x-marker and the motion of the y-marker as two separate problems – while keeping in mind that they
share the same time t (they do NOT have different times).
NOTE: Remember that the speed v is equal to the square root of (vx2 + vy2). So even if the two components
are independent of each other, the speed (in two dimensions) does depend on both of these.
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