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Guide 4-1. Solving Projectile Problems

Solving kinematics problems in 2 dimensions is similar to solving such problems in 1 dimension. Basically, you solve two 1-dimensional 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 2-dimensional dvat solution.

Attributes of a 2-dimensional dvat solution
  1. 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 ay.

  2. State the goal.

  3. 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.

  4. Write dvat equations--both horizontal and vertical--used to solve for the unknown. Always use as starting points the equations from table 4-1 in the text. An equation such as 4-12 is not an acceptable starting point. You won't find such an equations on the AP equation sheet.

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

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

  7. Apply checks.

Three important points about notation

  • For 2-dimensional motion, the same dvats are used as for 1-dimensional motion; however, additional subscripts are required. The equations for horizontal and vertical motions must be differentiated with x- and y-subscripts. See Table 4-1 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 ay represents the vertical component of the acceleration of an object. When you use ay 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 2-dimensional dvat method. (The comments in blue are for explanation and would not be required in a student's solution.)

Problem:  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 (non-existent in this case)
  • The horizontal distance to the rock collector is shown. This is denoted xc 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.
  • Given:

    Horizontal Vertical
    xo = 0.0 m
    vox = vocosθo
    ax = 0 m/s²
    yo = 0.0 m
    y = -50.0 m
    voy = vosinθo
    ay = -1.62 m/s²
    vo = 10.0 m/s
    θo = 30.0°
    xc = 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.

    Goal: Find | x - xc | when y = -50.0 m.


    Starting with the general dvat formula for the horizontal motion, substitute 0 for xo and ax.
    Starting with the general dvat formula for the vertical motion, substitute 0 for yo.
    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 vox and voy are both known, because vo and θo 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.
    vox = vocosθ and voy = vosinθ are substituted. When simplifying, note that voy/vox = sinθo/cosθo = tanθo.
    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/vox and then substitute that value back in to the dvat formulas for x and y.

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