Guide to 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, it’s important to choose unambiguous symbols for positions, displacements, and accelerations in both horizontal and vertical directions.

We present as an example a problem of greater algebraic complexity than in the textbook. However, the method of setting up the problem is the same as for any projectile problem.  Practice using the method in solving projectile problems over the next several days, and you'll become adept at it.

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:

    vo = 10.0 m/s
    qo = 30.0°
    xc = 75.0 m

    Horizontal

    Vertical

    Comment
    xo = 0.0 m

    x = ?

    vox = vocosqo = (10.0 m/s)cos30.0°

    ax = 0 m/s²

    		 yo = 0.0 m

    y = -50.0 m

    voy = vosinqo = (10.0 m/s)sin30.0°

    ay = -1.62 m/s²

    The givens are organized by horizontal and vertical components and are distinguished by subscripts.

     

     

    The horizontal component of acceleration is stated.

    Goal: Find the difference between the horizontal position of the collector, xc, and that of the rock, x, when y = -50.0 m. In other words, find .

    Solution:

    Starting with the general dvat formula for the horizontal motion, we substitute 0 for xo and ax.
    Starting with the general dvat formula for the vertical motion, we substitute 0 for yo.
    We need 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. The only unknown left is x. Note that vox and voy are both known, because vo and qo are known.
    Rewrite in the math form ax² + bx + c = 0 for use with the quadratic formula.
    Solve for x using the quadratic formula.
    When simplifying, note that voy/vox = sinqo/cosqo = tanqo.
    Substitute given values and reduce.
    Use the positive value for x. Solve for the unknown. The rock misses the collector by 24.9 m.

    Checks:

    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.