Solving kinematics problems in 2 dimensions is similar to solving such problems in 1 dimension. Basically, you solve two 1dimensional problems simultaneously. In setting up the solutions, you must choose unambiguous symbols for positions, displacements, and accelerations in both horizontal and vertical directions.
For handy reference, we provide the following encapsulation of the
components of a 2dimensional dvat solution.

List the givens using standard symbols
subscripted with x or y to distinguish between horizontal and vertical.
List givens in two columns, one for horizontal and one for vertical. If values are known to be 0, say so. Express the vertical acceleration as a_{y}.

State the goal.

Draw a diagram which shows the relevant quantities, position of
the origin, the directions of +x and +y, and the directions of velocity and acceleration.

Write dvat equationsboth horizontal and verticalused to solve for the unknown.
Always use as starting points the equations from table 41 in the text. An
equation such as 412 is not an acceptable starting point. You won't find
such an equations on the AP equation sheet.

Solve the dvat equations
algebraically to obtain a reduced equation in symbolic form for the unknown.

Substitute values and units and solve for the
numerical value of the unknown.

Apply checks.


For 2dimensional motion, the same dvats are used as for 1dimensional motion; however, additional subscripts are required. The equations for horizontal and vertical motions must be differentiated with x and ysubscripts. See Table 41 on p. 82. The only symbols that you don't subscript are the horizontal and vertical positions, x and y, and the time, t, which is the same for horizontal and vertical motions.

The symbol g represents the magnitude of the vertical acceleration that an object
would have in free fall, that is, subject to the force of gravity only. Thus g =
9.8 m/s² near the surface of the Earth. g is a number and not a vector. Since g represents a magnitude, g is always positive by the definition of magnitude.

The symbol a_{y} represents the vertical
component of the acceleration of an object. When you use a_{y} for that
purpose, the symbol represents a vector. As such, it has both magnitude and
direction. It can have a positive or negative value depending on your choice of
the direction of +y.

Below is an example of a problem solved using the 2dimensional dvat method. (The comments in blue are for explanation and would not be required in a student's solution.)
A Moon rock is fired at 10.0 m/s from a Moon dune 50.0 m above the surface of the Moon at an angle of 30.0° above the horizontal. A rock collecting vehicle is stalled on the surface a horizontal distance of 75.0 m from the launch point of the rock. Will the rock land in the collector? If not, by what horizontal distance will the rock miss the collector? The acceleration due to gravity on the Moon is 1.62 m/s².


Here are some things to note in the diagram:
Both the horizontal and vertical axes are shown.
The origin of coordinates is shown. (In this case, the origin is selected to be the initial position of the rock.)
The vertical acceleration is subscripted with y to distinguish it from any horizontal acceleration (nonexistent in this case)
The horizontal distance to the rock collector is shown. This is denoted x_{c} to distinguish it from the symbol x, which will be used to indicate the
horizontal position of the rock when it’s 50.0 m below the launch position.
The signs of the known quantities are assigned based on the choice of coordinate axes.

Horizontal 
Vertical 
x_{o} = 0.0 m
v_{ox} = v_{o}cosθ_{o}
a_{x} = 0 m/s² 
y_{o} = 0.0 m
y = 50.0 m
v_{oy} = v_{o}sinθ_{o}
a_{y} = 1.62 m/s² 
Common 
v_{o} = 10.0 m/s
θ_{o} = 30.0°
x_{c} = 75.0 m 

The givens are organized by
horizontal and vertical components as well as values that are common to both dimensions.
Components are distinguished by subscripts.
The
horizontal component of acceleration is stated. 
Find  x  x_{c}  when y = 50.0 m.

Starting with the general dvat formula for the horizontal motion, substitute 0 for x_{o} and a_{x}. 

Starting with the general dvat formula for the vertical motion, substitute 0 for y_{o}. 

In order to eliminate the unknown time, t, from the x and y equations, begin by solving for t in the linear equation. 

Substitute the result for t into the y equation, leaving x as the only unknown. Note that v_{ox} and v_{oy} are both known, because v_{o} and θ_{o} are known. 

Rewrite in the math form Ax² + Bx + C = 0 for use with the quadratic formula. 

The coefficients of the quadratic formula are expressed in terms of the problem’s givens. 

The equation Ax² + Bx + C = 0 is solved for x. 

The known expressions for A, B, and C are substituted into the solution. 

v_{ox} = v_{o}cosθ and v_{oy} = v_{o}sinθ are substituted. When simplifying, note that v_{oy}/v_{ox} = sinθ_{o}/cosθ_{o} = tanθ_{o}. 

Substitute given values and reduce. 

Use the positive value for x. Solve for the unknown. The rock misses the collector by 24.9 m. 
The positive root for x was selected. The units reduce correctly to meters. The value for x seems reasonable. However, a convincing check is to solve for the time using t = x/v_{ox} and then substitute that value back in to the
dvat formulas for x and y.
