ANALYSIS

Physics is not all about the Resolution of Forces and Kinematics. One of
the topic in Physics is Projectile Motion. Projectile Motion is a special case of
two-dimensional motion. Gravity is the only considered external force acting
on it while an object is airborne. Projectile is the moving body in this kind of
motion. It refers to any object thrown, launched or otherwise projected so that
once released, if air resistance is neglected, its path is affected only by the
Earth’s gravity. As fired at an angle, it is influenced by its horizontal inertia
and vertical gravity. The projectile creates a parabolic curve. The curved path
is known as the trajectory. We are assuming here that there is no or very little
air resistance. The equations of motion for one dimension are also valid for
two dimensions. To simplify our analysis, we will resolve the position,
velocity, and acceleration into horizontal and vertical components. Because
these components are perpendicular, they are independent to each other.

In this experiment, you will analyse the motion of a projectile.

Specifically, you will be able to explain the effects of variable launch angles
and initial speeds to the positions of the projectile along the x-axis and y-axis.
The objective of this experiment is to analyse the motion of projectile and to
compare the ranges of projectiles launched at different angles. In this, the
explanation on the effects of variable launch angles and initial speeds to the
positions of the projectile along the x-axis and y-axis will be obtained. The
restrictions of this experiments is to never aim the projectile launcher to any
person or breakable objects and to make sure that the launcher’s base is
securely clamped to the iron stand.
 From Newton’s first Law of Motion, it states that an object will
continue to a straight line with constant velocity if there is no external factor
that will cause its stop or change its direction. If an object is fired in
horizontal orientation, the object will go in to the direction perfectly with
constant velocity. It is also called the “free-way path.” It is shown in Figure 1.

Figure 1. Free-way Path

As to consider gravity as external force acting on the object and a factor

to change the state of the object, the motion creates a trajectory or simply a
projectile motion. Since gravity is force downward, the object fired on the
straight line or in angle, ϴ, will tend to go to the direction of the force,
downward just as shown in Figure 2.

Figure 2

The trajectory made by the projectile is a parabolic curve. Considering

the only force acting on the object is gravity, the object slows as it approaches
the peak or the maximum height or when velocity is equal to zero. It starts to
go down because gravity affects the vertical terms of the projectile. In the
absence of force in horizontal terms, the velocity in horizontal direction is
constant. To aid visualization, see Figure 3.

Figure 3

As said earlier, the three kinematic equations in one-dimensional

motion is also applicable in two dimensions. Those are velocity as a function
of time (1), position as a function of time (2) and velocity as a function of
position (3). Then, from these equations, we can derive the equations for
projectile motion as shown in Table 1.

v=v 0 + at (1)

1
x=x 0 +v 0 t+ a t 2 (2)
2

v 2=v 02 +2 a( x−x 0 ) (3)

 Table 1. Equations for Projectile Motion
x-axis y-axis
acceleration 0 a=-9.8m/s2
velocity vx=v0cosϴ vy=v0sinϴ
position x=x0+ y=y0+ v0sinϴt+1/2(gt2)
v0cosϴt

In this experiment, we used a projectile launcher (Fig. 4), an iron stand

with clamp, and target board (Fig. 5), a metal ball and a plumb line (Fig. 6), a
meter stick and projectile launcher’s rod (Fig. 7). We also used bond papers
and carbon paper to determine where the metal ball will fall. To avoid
inaccuracy, we put a lot of tape on the projectile launcher and also on the
papers. We are going to observe the trajectory of the projectile record the total
distance traveled by it. Each method requires a specific angle for each launch.
Values gathered throughout this method will be considered as the
experimental value and it will be compared to the computed value.

Figure 4. Projectile Launcher Figure 5. Target Board and Iron Stand

 Figure 6. Metal Ball & Plumb Line Figure 7. Projectile Launcher’s Rod

The first part of the experiment is

the “Determination of the Initial Velocity
of the Projectile” as shown in Figure 8.
Upon setting up the materials needed, we
began the experiment. We set the angle of
the projectile launcher to 0 degrees
therefore making it at a horizontal
position. The vertical distance from the
crosshair which is located on the side of
the projectile launcher to the surface is
measured as 0.265m.

We fired a sample to know where

to put the bond paper and the carbon
Figure 8 Determination of the Initial
paper. So the metal ball is placed inside Velocity of the Projectile's Set-Up
the projectile launcher and use the ramrod
to set it into mid-range, different from the long range that is instructed in the
manual. We pulled the cord to launch the ball. Upon seeing where the ball
land, we place there the papers to mark the distance of the ball. We taped a
carbon paper over the blank sheet to mark where the ball landed. After five
trials, we measure the distance from the mark of the ball in the carbon paper
to the projectile launcher. After measuring, we encircled or cross out the mark
done by the ball to avoid confusion in the next experiment.

In Table 2, the launcher is set on top of the table with its angle indicator
set at 0°. The y represents the vertical distance from the crosshair of the
launcher and the surface of the table while t represents the time of travel of
the projectile. We can see that the vertical distance ( x ) is directly proportional
to its initial velocity ( v o). After all 5 trials, we have computed its average
initial velocity to being 3.4556m/s. The results are shown in Figure 9.

Table 2. Getting the Initial Velocity of the

Projectile
y=0.3 m t=0.2 5 s
Trial x vo
1 0.64 m 2.56 m/s
2 0.48 m 1.92 m/s
3 0.52 m 2.08 m/s
4 0.59 m 2.36 m/s
5 0.66 m 2.64 m/s
Average = 2.31 m/s

FIGURE 9. Result
The second part of the experiment is to
determine the range of the projectile when the
projectile launcher, just like shown in Figure 10, is set to 30° and 60°. First
is to set up the angle of the projectile launcher at 30°. We again fired a
sample to know where to estimate where the ball is going to land. Upon
knowing it, we again put a carbon paper over a blank sheet. We launched the
ball five times. We measured the distance with the use of meter stick, from
the mark of the ball in the paper to the projectile launcher. We again crossed
out the mark of the ball in the paper.

After that, we set-up again the projectile launcher at 60° and same
procedure is again applied. We also make five trials and measure again its
distance, from the mark of the ball in the paper to the projectile launcher. Five
trials is also done and we measure the distance from the mark of the ball in
the paper to the projectile launcher. We then, change the blank paper because
it already has too many marks, and it may confuse us when measuring the
distance, thus, the result will become inaccurate.

In Table 3, we are now required to do trials with two required angles.

The launcher is now set at the floor and facing the long table. The computed
value of Range is supposed to be acquired with this formula,

V o2 sin 2θ
R= ,
g

Instead, we were given a much more suitable formula for this

procedure so that we won’t get erroneous results, R=X o +V o cos θ t f

Where t f can be solved by using this

formula,

V o sin θ+ √ V o2 sin2 θ+ 2 gy
tf =
g

Based on the results as shown in Figure

11 too, we could say that projectiles launched at
30° reached a farther Range than those launched
in 60°. The 30° projectile reached a greater

FIGURE 11. Result

Range because it is directed more to the horizontal axis while 60° is directed
more to the vertical axis.

Table 3. Determining the Range of the Projectile

m
V o =2.31
s
launch angle=30 ° launch angle=60 °
R ( computed value )=0.47 m R ( computed value )=0.47 m
Tria PERCENTAGE PERCENTAGE
R(EV) R(EV)
l DIFFERENCE DIFFERENCE
1 0.980 m 0.70% 1.10 m 0.80%
2 0.920 m 0.64 % 1.05 m 0.76 %
3 1.070 m 0.77 % 1.16 m 0.84 %
4 1.075 m 0.78 % 0.96 m 0.68 %
5 1.07 m 0.77 % 1.07 m 0.77 %
Lastly, the third experiment is the “Determination of the Maximum
Height of the Projectile”. In the third part of our
experiment, we required to find the maximum
height of the projectile when the projectile
launcher is at 30° and 60°. We put compute for
the distance from the target board to the
projectile launcher. The next thing we did is we
set-up the projectile launcher at 30°. We make a
practice launch to ensure that the ball will hit the
target board. We cover the target board with a
carbon paper over a blank sheet. We make 5
trials for it. We measure the height of the ball Figure 12. Determination of the
Maximum Height of the
to the ground (see picture c). We also make Projectile

sure to cross out the mark of the ball on the

latter experiment.
 After measuring the maximum height of the ball at 30°, we change the
angle from 30° to 60° and same procedure is applied. Again, we make a
practice launch to estimate how high the ball is going to hit. We then place
the target board and perform 5 trials.

In Table 4, we will now make use of the target board and find the
maximum height of the projectile. Procedures are similar to Table 3 and we
are still required to use 2 angles. The formula
Figure 13. Result
to be used to get the computed value of the
maximum height is given as,

(V o sin θ)2
Y max =
2g

But to prevent from getting erroneous results, we are given another

more suitable formula,

1
Y max =Y o +V o sin θ t m− g t m2
2

V o sin θ
t m=
g

Here the projectile launched at 60° has the bigger maximum height than
those launched at 30°. The reason to this is the same as the one in table 2
about the 60° projectile ball being more directed to vertical axis than the 30°
ball.

ss

Table 4. Determining the Maximum Height of the Projectile

m
V o =2.31
s
launch angle=30 ° launch angle=60 °
Y max ( computed value )=0. 07 m Y max ( computed value )=0. 20 m
 Tria Ymax(EV) PERCENTAGE Ymax(EV) PERCENTAGE
l DIFFERENCE DIFFERENCE
1 0.330 m 1.30 % 0.921 m 1.28 %
2 0.340 m 1.31 % 0.920 m 1.28 %
3 0.410 m 1.41 % 0.923 m 1.28 %
4 0.325 m 1.29 % 0.915 m 1.28 %
5 0.295 m 1.23 % 0.931 m 1.29 %

CONCLUSION

Based on our observation and obtained data, the ball that is launched at
30° reached the ground first as compared to the ball that is launched at 60°.
But the ball that is launched at 60° had the highest height before falling. The
peak height of a projectile is determined by the initial value of the vertical
velocity component. The greater the initial value of v with respect to y, the
higher that a projectile will rise. The projectile launched at 60-degrees has the
greatest vy, and as such the greatest peak height.

A ball launched at 60 degree angle comes with a higher maximum

vertical distance than a 30 degrees angle. It travels much farther up than the
30 degree angle because it's directed upward more. Horizontal and vertical
motions are independent of one another.  The acceleration due to gravity is a
constant regardless of what horizontal velocity an object has.  Thus, the time
for an object to free-fall from a given height is the same no matter what speed
it is moving at in the horizontal direction.

As the net force of an object increases with a constant mass, the

acceleration increases, making the object move faster. When the mass of an
 object increases but the net force stays the same, the acceleration decreases,
making the object take more time to travel. When both the mass of an object
as well as the net force changes with a certain amount, the acceleration still
increases, depending on the difference between the mass and the force. The
net force of an object being pulled as with the experiment is always relative to
the hanging mass and the constant pull of gravity. It was proven in the
experiment that acceleration is always directly proportional to force, and
inversely to mass.

GUIDE QUESTIONS:

1. From the result of your experiment, how would you compare the range of the
projectile launched at 30 degrees to the one launched at 60 degrees? Is this
consistent with the theory? Defend your answer.

Based on the results of the experiment, the Vo which was projected 30 o

from the ground obtained an equal range value with the projectile which
launched Vo 60o from the horizontal. And yes, this is consistent with the theory
that projectiles which are projected in complementary angles obtain the same

V o2 sin ⁡( 2θ)
value for its range. If we take a look with the formula R= we can have
g
it as
V o2 sin ⁡( 2θ) 2V o2 sin θcosθ 2V o2 cos ⁡(90−θ)sin ⁡(90−θ) 2 V o2 sinθcosθ
R= = = =
g g g g

Mathematically, this simply means that if we substitute the complement of

θ in the equation, we will still end up having the same formula for the range.
 Thus, this theory that projectiles launched in complementary angles obtain the
same range holds true. However if this is to be understood in the actual scenario,
in respect to the value of the range, we can regard the factors – time flight and
horizontal component. Now if we have a projectile launched in 0o angle, we
may have the maximum horizontal component, but certainly we obtain zero time
of flight. And that makes 0 range. On the other hand, if we have 90o projectile,
we will obtain zero x-component and we have maximum time flight, but we end
up in the same initial position so still, the range is zero. By these factors – time
flight & horizontal component, it won’t be a surprise already that projectiles of
complementary angles have equal ranges.

2. From the results of your experiment, how would you compare the maximum
vertical distance reached by the projectile launched at 30 degrees to the one
launched at 60 degrees? Is this consistent with the theory?
The maximum vertical distance reached by the projectile, ymax launched at
30o is the product of the projectile at 60o times tan2(30). That is

(V ¿¿ o sin θ)2 [V ¿¿ o sin ⁡( 90−θ)]2

=tan 2 θ ¿ ¿ . Another relationship is that the sum of
2g 2g
the ymax of both projectile equals the square of the initial velocity, Vo divided

v o2
by twice the acceleration due to gravity. That is y max 1 + y max 2= . And yes, this
2g
correlation will always coincide with the theory as long as we are comparing
projectiles that are launched in angle complements. This will hold true because
by virtue of mathematical derivations we can have the following relationship k
and their sum S as:
 (V ¿¿ o sin θ)2
y max 1 2g
o k= = ¿
y max 2 [V ¿¿ o sin ⁡(90−θ)]2 (sinθ)
2
(sinθ)
2
2
= 2
= 2
=(tanθ ) ¿
2g [sin ( 90−θ ) ] ( cosθ)

o S= y max 1 + y max 2=
( V ¿¿ o sin θ)2 [V ¿¿ o sin ⁡( 90−θ)]2
+ ¿¿
2g 2g
v o2 2 2 v o2 2 2 v o2
o¿
2g
[( sinθ) + ( sin ( 90−θ ) ) ] = [ (sinθ) + ( cos θ ) ] =
2g 2g

3. List down the possible sources of errors in this experiment.

Possible errors could have come from human technical errors in
measurement, rounding off values, estimation of the exact measurement of
displacement/distances and calibration of inclination for the projectile launcher.
Other sources of errors could be air resistance and instability of the project
launcher, table, and iron stand with the wooden platform.