Following the prescription discussed previously we derive the trajectory for
projectile motion by eliminating t from the equations for the
horizontal and vertical component of the projectile motion:
We get the following formula:
Here is the launch angle, ie
As we had mentioned previously the trajectory for projectile motion is parabolic. There are a few useful expressions which can be derived from this equation. The maximum range on level ground is the distance between y=0 solutions:
We see that this expression is maximal for where the range is
Note that the smaller g the larger range, so we ca throw almost 10 times further on the moon than on earth! Also note that the range decreases for steeper as well as for shallower angleswhich means that for any range less than max-range there are two elevations which will mach. Highest elevation always keeps the projectile in air longer so if we want to get there fast we use the lower elevation.
Another interesting characteristic of projectile motion is the max height.
To determine it we need only consider the motion in the y-direction.
The following equation holds for this motion with constant acceleration:
At maximum height we have . We insert this value in the equation and solve for h to get
Remember that these formulae only hold for projectile motion on flat ground. For more complicated cases you will have to work the formulae for and the trajectory yourself so make sure you understand what went on in these derivations.