Analysis

Projectile motion is a special case of two-dimensional motion. A projectile is

an object upon which the only force acting is gravity. The motion of the object is

called the projectile motion. A projectile is a particle that moves in a vertical plane

with some initial velocity but its acceleration is always free-fall acceleration,

which the acceleration due to gravity. 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. An object dropped from rest is a projectile,

provided that the influence of air resistance is negligible. An object that is thrown

vertically upward is also a projectile, provided that the influence of air resistance is

negligible. And an object which is thrown upward at an angle to the horizontal is

also a projectile. A projectile is any object that once projected or dropped

continues in motion by its own inertia and is influenced only by the downward

force of gravity.

A projectile has a single force that acts upon it - the force of gravity. If there

were any other force acting upon an object, then that object would not be a

projectile. Thus, the free-body diagram of a projectile would show a single force

acting downwards and labeled force of gravity. You can see in figure 4, the

different types of projectiles.

The path of motion of the projectile is called the trajectory which is a

parabolic curve. Once the projectile is launched, the only force affecting its motion

would be gravity. Since gravity acts along the y-axis, there is no force on the

projectile along the x-axis. In figure 1 you can see the different types of projectiles.

Figure 1

In the experiment, we will able to analyze the motion of a projectile.

Specially, we 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. To

simplify, we will be able to resolve the position, velocity, and acceleration into

horizontal and vertical components, because these components are perpendicular,

meaning they are independent to each other

The objectives of the experiment is to analyze the motion of a projectile and

to compute ranges if projectile launch at different angles particularly, 60 and 30

degrees. The materials of the experiment are projectile launcher, metal ball (25 mm

in diameter), meter stick, plumb line, iron stand clamp, target board, bond paper,

and carbon paper. The projectile launcher can be very dangerous because if the

projectile launcher loaded with metal ball is fired to a person or an object, it may

hurt the person or it may destroy things, so for safety purposes, we only fired the

projectile launcher after the area that it may cover is cleared.

From Newton’s second law of motion, we can say that the vertical motion of

the projectile is accelerated due to gravity. The motion of the projectile is simply

that of a body of free fall. The horizontal motion, in the absence of force along the

x-axis, is uniform and with constant velocity.

In the previous experiment, we learned that there are three kinematic

equations that are also applicable in two-dimensional motion.

The first equation is the change in velocity is equal to the integral of the

acceleration with respect to time.

v=v0+at

The second equation is the change in position is equal to the integral of velocity

with respect to time.

x=x0+v0t +12

a t2

The third equation is the change in velocity is equal to the integral of acceleration

with respect to position.

v2=v02+2a(x−x0)

Using these three equations, we can derive the equations for projectile

motion, as shown in Table 1. In the next part, the derivation of equations for

projectile motion will also be discussed.

Table 1. The equations for Projectile Motion

x-axis y-axis

acceleration 0 a=-9.8m/s2

velocity vx=v0cosϴ vy=v0sinϴ

position x=x0+

v0cosϴt

y=y0+ v0sinϴt+1/2(gt2)

The path of a projectile that is launched at x0=0 and y0=0, with an initial

velocity v0.The initial velocity and the velocities at various points along its path are

shown, along with their components. Note that the horizontal velocity component

remains constant but the vertical velocity component changes continuously. The

initial velocity can be resolved into the x and y components.

The horizontal velocity:

v0 x=v0cos θ

The vertical velocity:

v0 y=v0 sin θ

The range R is the horizontal distance the projectile has traveled when it

returns to its launch height. As shown in Figure 2.

Figure 2

The acceleration along the x-axis is zero, the horizontal velocity is constant

all throughout the motion. Along the y-axis, the velocity changes because of the

acceleration due to gravity. The velocity of the projectile at any given time is given

by

v fy=v0 y+¿after substituting t, we get

v fy=√v0 y2+2gy

where t is the time travelled and y is the vertical distance travelled.

In horizontal motion, there is no acceleration in the horizontal direction. The

horizontal component vx of the projectile’s velocity remains unchanged from its

initial value v0x throughout the motion. At any time t, the projectile’s horizontal

displacement x - x0 from an initial position x0 can be derive by substituting

v0 cos θ to v0x, the equation is given by x=v0 cosθ t

If the projectile starts and ends along the same vertical level, the horizontal

distance traveled called the range R, can be computed from the equation

R=v0

2sin 2 θg

For the vertical distance travelled at any given time, the following equation

can be use

y=v0 sinθ t+ 12

g t 2

The maximum vertical distance reached by the projectile can be computed

from the equation

ymax=(v0 sinθ)2

2 g

The experiment is divided into three parts. The first part is getting the initial

velocity of the projectile. The second part is determining the range of the

projectile. The third part is determining the maximum height of the projectile.

The first part of the experiment is getting the initial velocity of the projectile.

First we set-up the materials carefully. The launcher is attached to the iron stand.

We placed the launcher setup in a flat surface and we set the angle indicator to 0

degrees so that the launcher us in horizontal position. Like I’ve said earlier, the

launchers path should be cleared to avoid damages and injuries. Then we measured

the vertical distance from the floor to the crosshairs on the side of the launcher. We

then place the metal ball in the launcher and used a ramrod to set into long-range

setting. Then we fired a test trial so that we can know where we can place the bond

paper and carbon paper. After setting-up all the required materials, we launched

the metal ball, and doing it with 5 trials and we measure the horizontal distance in

meters. The set-up is shown in Figure 3.

Figure 3

Table 2. Getting the Initial Velocity of the Projectile

Vertical distance, y=0.3 m time of travel, t=0.25 s

Trial Horizontal distance, x Initial velocity, vo

1 1.89 m 4.0445 m/s

2 1.965 m 4.2050 m/s

3 1.897 m 4.0595 m/s

4 1.925 m 4.1194 m/s

5 1.964 m 4.203 m/s

Average = 4.1263 m/s

The results of the first part of the experiment are shown in Table 2. We have

measure that the vertical distance is equal to 1.07 m. We have also computed the

time of travel t which is equal to 0.4673 s using the formula

t=√ 2 yg

we also computed the initial velocity v0 of each trial, and we get the average

initial velocity which is 4.1263 m/s using the formula

v0=xt

The second part of the experiment is determining the range of the projectile.

To begin this part we set-up the launcher near the edge of the table. We then set the

angle at 30° in the first 5 trials and then switches it to 60° in the next 5 trials. Like

I’ve said earlier, the launchers path should be cleared to avoid damages and

injuries. We then place the metal ball in the launcher and used a ramrod to set into

long-range setting. Then we fired a test trial so that we can know where we can

place the bond paper and carbon paper. After setting-up all the required materials,

we launched the metal ball, and doing it with 5 trials with the angle set to 30° and

then 5 trials with the angle set to 60°, where in each trial we measure the range in

meters. The set-up is shown in Figure 4.

Figure 4

The result in each trial in 30° and 60° in which we measure the experimental

range value is shown in Figure 5.

Figure 5

The results of the second part of the experiment is shown in Table 3. From

the previous part we determined the average initial velocity which is 4.1263 m/s

and we used that value to compute the computed value of range using the formula

R=v0

2sin 2 θg

In the first 5 trials which is set to 30°, we determined the computed

range value which is equal to 1.5046 m.

In the next 5 trials which is set to 60°, determined the computed range value which

is equal to 1.5046 m.

As you can see, the value for the range in 30° and 60 ° angles are equal, this is

because sin2(30) and sin2(60) have the same value.

We have also computed the percentage difference in the 10 trials using the formula

percent difference=|difference of the twovalues|

average of the twovalues

Table 3. Determining the Range of the Projectile

Average initial velocity, V o=4.1263ms

launch angle=30 °

Range (computed value )=1.5046 m

launch angle=60 °

Range (computed value )=1.5046 m

Trial RExpValue

PERCENTAGE

DIFFERENCERExpValue

PERCENTAGE

DIFFERENCE

1 2.022 m 29.34 % 2.008 m 28.66 %

2 1.995 m 28.03 % 2.038 m 30.11 %

3 2 m 28.27 % 2.06 m 31.16 %

4 1.975 m 27.04 % 2.022 m 29.34 %

5 1.00 m 27.78 % 2.056 m 30.97 %

In the third part of the experiment is determining the maximum height of the

projectile. We set-up the launcher and set it to 30° and 60° with each having 5

trials. We placed the target board and made a trial test to ensure that the metal ball

will hit the target board. Like I’ve said earlier, the launchers path should be cleared

to avoid damages and injuries. We then cover the target board with bond paper and

carbon paper. After setting up all the required things we did 5 trials on 30° and 5

trials on 60° angle of the launcher. The set-up is shown in Figure 5.

Figure 5

The result in each trial in 30° and 60° in which we

measure the experimental range value is shown in

Figure 6.

Figure 6

Table 4. Determining the Maximum Height of the Projectile

Average initial velocity, V o=4.1263ms

launch angle=30 °maximum vertical

distance(computed value )

Y max=0.2172

launch angle=60 °

maximum vertical distance

(computed value )

Y max=0.6515 m

Tria

l

Ymax

experimental

value

Percentage

Difference

Ymax

experimental

value

Percentage

Difference

1 0.314 m 36.45 % 0.909 m 33.00 %

2 0.315 m 36.75 % 0.92 m 34.17 %

3 0.326 m 40.06 % 0.923 m 34.49 %

4 0.355 m 48.16 % 0.926 m 34.80 %

5 0.340 m 44.08 % 0.929 m 35.12 %

The results of the third part of the experiment is shown in Table 4. We have

determined in the first part of the experiment that the average initial velocity is

4.1263 m/s and we used it to solve the computed maximum vertical distance for

the 30° and 60° angle of the launcher. To compute the maximum vertical distance

we use the formula

ymax=(v0 sinθ)2

2 gIn the launch angle of 30°, the computed maximum vertical distance

is 0.2172 m.

In the launch angle of 60°, the computed maximum vertical distance is 0.6515 m.

We have also computed the percentage difference in the 10 trials using the formula

percent difference=|difference of the twovalues|

average of the twovalues

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 range of the projectile launched at 30

degrees is almost the same with the projectile launched at 60 degrees. It is

consistent with the theory because force acting on y-axis is independent from the

force acting on x-axis. The formula of range is equal to the square of initial

velocity multiplied by the sine function of twice the angle all over gravity, but sin

2(30°) is √32

and sin 2(60°) is also√32

, so this means that the computed range of the

projectile launched at 30° and the projectile launch at 60° is always equal. The

reason is because 30° and 60° are complementary angles.

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?

Base on the results of the experiment, the projectile launched at 60° has higher

maximum vertical distance compared to the projectile launched at 30°. The

experimental maximum vertical distance show that the maximum vertical distance

compared to the projectile launched at 30° is almost three times the maximum

vertical distance compared to the projectile launched at 60°. This is consistent with

the theory because we are comparing the maximum height of a projectile launched

at two complementary angles, therefore it would give different values because in

the formula, we use sin θ, and we know that sin 30° and sin 60° have different

values, therefore the two launch angles will give different maximum vertical

distance.

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

One possible error of this experiment is the air resistance because we didn’t consider its

effect on the projectile. Another error is the accurate measurements and also the

accurate setting of angles.