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.

Tue Sep 16 16:33:10 EDT 1997