Lesson 2: Projectile
Motion
Horizontally Launched
Projectile Problems
One of the powers of physics is to be able to use
physics principles to make predictions about the final
outcome of a moving object. Such predictions are made
through the application of physical principles and
mathematical formulas to a given set of initial
conditions. In the case of projectiles, a student of
physics can use information about the initial velocity
and position of a projectile to predict such things as
how much time the projectile is in the air and how far
the projectile will go. The physical principles which
must be applied are those discussed
previously in Lesson 2. The mathematical formula
which are used are commonly referred to as kinematic
equations. Combining the two allows one to make
predictions concerning the motion of a projectiles. Such
predictions are often made in response to a problem posed
by a teacher, known as projectile problems.
There are two basic types of projectile
problems which we will discuss in this course. While
the general
principles are the same for each type of problem, the
approach will vary due to the fact the problems differ in
terms of their initial conditions. The two types of
problems are
A
projectile is launched with a horizontal velocity from an
elevated position and follows a parabolic path to the
ground; predictable unknowns include the initial speed of
the projectile, the initial height of the projectile, the
time of flight, and the horizontal distance of the
projectile.
Examples of this type of problem
are
 A pool ball leaves a 0.60meter high
table with an initial horizontal velocity of 2.4
m/s. Predict the time required for the pool ball to
fall to the ground and the horizontal distance
between the table's edge and the ball's landing
location.
 A soccer ball is kicked horizontally off a
22.0meter high hill and lands a distance of 35.0
meters from the edge of the hill. Determine the
initial horizontal velocity of the soccer ball.
A
projectile is launched at an angle to the horizontal and
rises upwards to a peak while moving horizontally; upon
reaching the peak, the projectile falls with a motion
which is symmetrical to its path upwards to the peak;
predictable unknowns include the time of flight, the
horizontal range, and the height of the projectile when
it is at its peak.
Examples of this type of problem
are
 A football is kicked with an initial
velocity of 25 m/s at an angle of 45degrees with
the horizontal. Determine the time of flight, the
horizontal distance, and the peak height of the
football.
 A long jumper leaves the ground with an initial
velocity of 12 m/s at an angle of 28degrees above
the horizontal. Determine the time of flight, the
horizontal distance, and the peak height of the
longjumper.
The latter
type of problem will be the subject of the
next part of Lesson 2. In this part of Lesson 2, we
will focus on the first type of problem  sometimes
referred to as horizontallylaunched projectile problems.
Three common kinematic equations which will be employed
for both type of problems include the following:
The above equations work well for motion in
onedimension, but a projectile is usually moving in two
dimensions  both horizontally and vertically. Since
these two components of motion are independent of each
other, two distinctly separate sets of equations are
needed  one for the projectile's horizontal motion and
one for its vertical motion. Thus, the three equations
above are transformed into two sets of three
equations. For the horizontal
components of motion, the equations are
Of these three equations, the top equation is the most
commonly used. An application of projectile concepts to
each of these equations would also lead one to conclude
that any term with a_{x} in it would cancel out
of the equation since
a_{x} = 0
m/s/s.
For the vertical
components of motion, the three equations are
In each of the above equations, the
vertical acceleration of a projectile is known to be 10
m/s/s (the acceleration of gravity). Furthermore, for
the special case of the first type of
problem (horizontallylaunched projectile problems),
v_{iy} = 0 m/s; thus, any term with
v_{iy} in it will cancel out of the equation.
The above sets of three equations are the kinematic
equations which will be used to solve projectile motion
problems.
To illustrate the usefulness of the
above equations in making predictions about the motion of
a projectile, consider their use in the solution of the
following problem.
Example
A pool ball leaves
a 0.60meter high table with an initial
horizontal velocity of 2.4 m/s. Predict the time
required for the pool ball to fall to the ground
and the horizontal distance between the table's
edge and the ball's landing
location.

The solution of this problem begins with declaring the
values of the known information in terms of the symbols
of the kinematic equations  x, y, v_{ix},
v_{iy}, a_{x}, a_{y}, and
t. In this case, the following
information is either given or implicit in the problem
statement:
Horizontal
Information

Vertical
Information

x = ???
v_{ix} = 2.4 m/s
a_{x} = 0 m/s/s

y = 0.60 m
v_{iy} = 0 m/s
a_{y} = 10 m/s/s

As indicated in the table, the unknown quantity is the
horizontal displacement (and the time of flight) of the
pool ball. The solution of the problem now requires the
selection of an appropriate strategy for using the
kinematic equations and the known
information to solve for the unknown quantities. It will
almost always be the case that such a strategy demands
that one of the vertical equations
be used to determine the time of flight of the projectile
and then one of the horizontal
equations be used to find the other unknown
quantities (or vice versa  first use the horizontal and
then the vertical equation). A careful listing of known
quantities (as in the table above) provides cues for the
selection of the strategy. For example, the table above
reveals that there is more vertical information known
than horizontal information. Thus, it would be reasonable
that a vertical equation be used with the vertical values
to determine time and then the horizontal equations be
used to determine "x." The first
vertical equation (y = v_{iy}*t
+0.5*a_{y}*t^{2}) will allow for the
determination of the time. Once the appropriate equation
has been selected, the physics problem becomes
transformed into an algebra problem. By substitution of
known values, the equation takes the form of
0.60 m = (0
m/s)*t +
0.5*(10
m/s/s)*t^{2}
Since the first term on the right side of the equation
reduces to 0, the equation can be simplified to
0.60 m = (5.0
m/s/s)*t^{2}
If both sides of the equation are divided by 5.0
m/s/s, the equation becomes
0.12 s^{2} =
t^{2}
By taking the square root of both sides of the
equation, the time of flight can then be
determined.
t = 0.35
s (rounded from 0.3464 s)
Once the time has been determined, a horizontal
equation can be used to determine the horizontal
displacement of the pool ball. Recall from the given
information, v_{ix} = 2.4 m/s and
a_{x} = 0 m/s/s. The first horizontal equation (x
= v_{ix}*t + 0.5*a_{x}*t^{2}) can
then be used to solve for "x." With the equation
selected, the physics problem once more becomes
transformed into an algebra problem. By substitution of
known values, the equation takes the form of
x = (2.4 m/s)*(0.3464 s) + 0.5*(0
m/s/s)*(0.3464 s)^{2}
Since the second term on the right side of the
equation reduces to 0, the equation can then be
simplified to
x = (2.4 m/s)*(0.3464
s)
Thus,
x = 0.83
m (rounded from 0.8313 m)
The answers to the stated problem
is that the pool ball is in the air for 0.35 seconds and
lands a horizontal distance of 0.83 m from the edge of
the pool table.
The following procedure summarizes the
above problemsolving approach.
 Carefully read the problem and list known and
unknown information in terms of the symbols of the
kinematic equations. For convenience sake, make a
table with horizontal information on one side and
vertical information on the other side.
 Identify the unknown quantity which the problem
requests you to solve for.
 Select either a horizontal or vertical equation to
solve for the time of flight of the projectile.
 With the time determined, use one of the other
equations to solve for the unknown. (Usually, if a
horizontal equation is used to solve for time, then a
vertical equation can be used to solve for the final
unknown quantity.)
One caution is in order: the sole reliance upon 4 and
5step procedures to solve physics problems is always a
dangerous approach. Physics problems are usually just
that  problems! And problems can often be simplified by
the use of short procedures as the one above. However,
not all problems can be solved with the above procedure.
While steps 1 and 2 above are critical to your success in
solving horizontallylaunched projectile problems, there
will always be a problem which doesn't "fit the mold."
Problemsolving is not like cooking; it is not a mere
matter of following a recipe. Rather, problemsolving
requires careful reading, a firm grasp of conceptual
physics, critical thought and analysis, and lots of
disciplined practice. Never divorce conceptual
understanding and critical thinking from your approach to
solving problems.
Check
Your Understanding
A soccer ball is kicked horizontally off a 22.0meter
high hill and lands a distance of 35.0 meters from the
edge of the hill. Determine the initial horizontal
velocity of the soccer ball.
