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Republic of the Philippines

Department of Education Regional Office IX, Zamboanga Peninsula

9 Zest for Progress

Zeal of Partnership

Science Grade 9

Quarter 4 - Module 1 UAM and Projectiles

Launched Horizontally

Name of Learner:

Grade & Section:

Name of School:

1

Module 3

What I Need to Know

In your previous science lessons, you have gained a more in-depth understanding of Newton's Three Laws of Motion: Law of Inertia, Law of Acceleration, and the Law of Interaction. You were able to do experiments and activities on the effects of forces and motion and use your knowledge in your life experiences.

You were able to quantify non-uniform motion from your previous grade levels.

In this module, you will mathematically describe the horizontal and vertical dimensions of Uniformly Accelerated Motion. Once you have mastered the concepts of uniformly accelerated motion, you will be able to describe the horizontal and vertical motions of a projectile (S9FE-Iva-34). A projectile is the most common example of an object moving in two dimensions; that is, it has a horizontal and vertical motion that is independent of each other.

After doing all the activities in this module, you should be able to:

• Define projectile.

• Explain uniformly accelerated motion.

• Use kinematic equations in solving for the variables related to the horizontal and vertical motions of a projectile.

What’s In

Activity 1: Scalar or Vector Identify whether the following terms are scalar or vector quantities. Put an S

before the number for scalar quantities, and V for vector quantities. After which connect the terms to their respective SI units. ______1. Distance a. m/s ______2. Speed b. m/s2 ______3. Acceleration c. s ______4. Velocity d. m ______5. Time

Activity 2: Remember me Identify the words being described in the following statements. Choose from the box below.

______________1. A push or pull that causes that causes an object to move faster or

slower. ______________2. Direction and distance of the shortest path between an initial and final position. ______________3. Speed in a given direction.

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Gravity Force Acceleration Velocity Displacement

UAM and Projectiles Launched

Horizontally

2

Stack of

books Board/plank

40 cm

80 cm 120 cm

______________4. Force that attracts all objects towards the center of the Earth. ______________5. The rate of change in velocity.

What’s New

Activity 3: Roll, roll and away!

Objectives:

• Solve for the slope of the can rolling down an inclined plane.

• Describe the motion of an object moving downward. Materials Needed:

• board/plank (at least 200 cm long)

• timing device (stopwatch)

• tin can

• stack of books

• protractor Procedure: 1. Create an inclined plane by placing one end of the plank on top of the pile of books. Put a mark on the plank for every 40 centimeters that you measured starting from the lowest end. Your marks should be 40 cm, 80 cm and 120 cm accordingly.

2. Release the tin can from each of the points that you labeled and start with the 40 cm. Click the timer as you release the tin can and stop the timer when the tin can reach the bottom of the plank. 3. Ask someone (could be any family member) to record the time (t) taken by the tin can to travel each distance (d) down the plane. Perform three trials from each mark. Use the table below for your data. 4. Solve for the slope by dividing distance by the average time.

TABLE 1. Data on the Motion of a Rolling Tin Can

Distance, d (cm)

Time, t (s)

Slope (distance/average

time) Trial 1 Trial 2 Trial 3

Average time

[(Trial 1 + Trial 2 +

Trial 3)/3]

40 cm

80 cm

120 cm

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3

Answer the following questions: Q1. What is the relationship between distance and time of travel of the rolling can? __________________________________________________________________________________ Q2. What have you noticed with the slope as the distance increases? __________________________________________________________________________________ Q3. What do you think the slope suggests? __________________________________________________________________________________

What Is It

To fully understand and describe a projectile’s motion – objects moving in two dimensions, let us first consider objects moving in one dimension (horizontally or vertically). This module will focus on the motion of objects described to be moving at a constant velocity (having an acceleration of 0 m/s2) or moving at a constant acceleration. Objects that are moving at a constant acceleration are said to be in a Uniformly Accelerated Motion (UAM). In other words, the acceleration does not change; it is uniform throughout. In the activity, you related distance and time. In computing the slope, you divided distance by time which is actually the speed of the can. These quantities are essential in the derivation of formulas that will solve problems relating to uniformly accelerated motion.

Table 2. Summary of Uniformly Accelerated Motion (UAM) Formulae

Uniformly Accelerated Motion

Formulae

1. vf = at + vi

2. d = vit + at2 2

3. d = vf + vi 2 4. Vf

2 = vi2 + 2ad

where, SI unit vf = final velocity m/s vi = initial velocity m/s a = acceleration m/s2 t = time s d = displacement m

t

4

position

time

velocity 5m/s 10m/s 15m/s

Uniformly Accelerated Motion (UAM): Horizontal Dimension If a body maintains a constant change in its velocity in a given time

interval along a straight line, then the body is said to have a uniform acceleration.

Consider an airplane on a runway preparing for takeoff.

Figure 1. An airplane preparing for take-off. What have you noticed with the velocity of the airplane in the figure above? In the first second, the velocity is 5m/s, in the next time it becomes 10m/s, and on the third second it becomes 15m/s. Notice that the velocity of the airplane as it runs on the runway is increasing. When the velocity changes every second, it means that the airplane is accelerating. How much is the change in velocity for every second? If you try calculating for the change in velocity from 1s to 2s, that is 10m/s – 5m/s = 5m/s. The change in velocity from 2s to 3s is 15m/s – 10m/s = 5m/s. Notice that the change in velocity for every second is 5m/s, it is not changing, it remains the same or uniform for every second. The figure shows that the airplane is moving faster at a constant acceleration. This motion is referred to as Uniformly Accelerated Motion (UAM).

Let’s try solving problems! Sample Problem1 An airplane from rest accelerates on a runway at 5.50 m/s2 for 20.25s until it finally takes off the ground. What is the distance covered before takeoff? Note: If the airplane started from rest, it means to say that the initial velocity is zero. Solution:

1. Write the given or what information is given in the problem. Given: Find: a = 5.50 m/s2 d = ? t = 20.25 s vi = 0 m/s

2. Next, find out which formula to use in solving the problem. Use the formula

checklist table below to help you find out which formula to use. Note: Eliminate the formula that does not contain the variable that you need to find. In this case, we are looking for d (displacement), so eliminate the formula that does not contain the variable d.

0m 5m

1s

10m 20m 45m 30m

2s 3s

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Table 3. Formula Checklist

Given variables

Formulae variables Check ( ) if the variable has a value and put an X if the variable has a missing value (?)

Which formula has one missing variable?

a = 5.50 m/s2

t = 20.25 s

vi = 0 m/s d = ?

1. vf = vi + at

vf a t vi

eliminate this formula since it doesn’t have the variable d

Formula No. 2 then we use formula no. 2 2. d = vit + at2

2 d vi t a

3. d = vf + vi 2

d vf vi

t

4. Vf2 = vi

2 +2ad vf vi

a d

3. Solve the problem using the chosen formula. d = vit + at2

2 d = (0 m/s)(20.25 s) + (5.50 m/s2) (20.25 s)2

2 d = 0 + (5.50 m/s2) (410.0625 s2)

2

d = 1128 m

Sample Problem 2

A motorcyle from rest accelerates uniformly over a time of 3.25 seconds and covers a distance of 15 m. What is the acceleration of the motorcycle?

Given: Find: vi = 0 m/s a = ? t = 3.25 s 2 d = 15 m

t

d = vit + at2

21 5 m = (0 m/s) (3.25 s)+ a (3.25 s)2

2 15 m = 0 + a (10.5625 s2)

2 15 m = a (5.28 s2)

a = 15 m 5.28 s2

a = 2.8 m/s2

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Try solving this… A car accelerates to a speed of 15 m/s over 200 m distance. Determine the acceleration (assume uniform) of the car.

Uniformly Accelerated Motion (UAM): Vertical Dimension

You have been introduced to the concept of gravity in Grade 8. You have learned that gravity acts on all objects by pulling it towards the center of the Earth. So, on Earth when you throw something up, it will go down. Things thrown upward always fall at a constant acceleration (ag) which has a magnitude of 9.8 m/s2. This means that the velocity of an object in free fall changes by 9.8 m/s every second of

fall. Consider a stone dropped from a cliff as shown in Figure 2. For equal time

interval, the distance travelled increases quadratically.

Figure 2. Motion of the stone dropped from a hill.

Sample Problem 1 Ben is playing with his ball on top of a building, but the ball fell and hits the ground after 2.6 seconds, what is the final velocity of the ball just before it hits the ground and how high is the building? Note: The initial velocity of an object that is falling or dropped from a height will

always be equal to 0 m/s, since the object started from rest. Replace the variable d (displacement) with h (height) since we are referring to the position of an object from top to bottom. Given: Find: ag = -9.8 m/s2 t = 2.6 s vf = ? assume vi = 0 m/s h = ?

0 m/s

-9.8 m/s

-19.6 m/s

-29.4 m/s

-39.2 m/s

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h = vit + agt2

2 -h = [ (0 m/s)(2.6 s) ] + (-9.8 m/s2)(2.6 s)2

2 -h = 0 + (-9.8 m/s2)(6.76 s2)

2 -h = -66.248m

2

h = 33 m

Try solving this…

The acceleration of gravity on the moon is 1.62 m/s2. If a ball is dropped on the moon from a height of 2 m. Determine the time for the ball to fall to the surface of the moon.

Motion in Two Dimensions Have you seen a soccer player kick a ball? Or a basketball player shoots a ball into the ring? What have you noticed with the path the balls travel? If you noticed, the balls travel a curve path. The curve naturally happens when an object, called a projectile, moves in two dimensions – having both horizontal and vertical motion components, acted by gravity only. In physics, this is called projectile motion. Understanding the Physics of projectile helps players enhance their game skills and experience. Projectile motion can happen in two situations – projectiles launched horizontally and projectiles launched at an angle. This module will describe motion of projectiles launched horizontally.

Projectiles Launched Horizontally A projectile launched horizontally has no initial vertical velocity (vix = 0). Thus, its vertical motion is similar to that of a dropped object.

Projectiles are only acted upon by gravity. Therefore projectiles launched horizontally only have downward acceleration (ay), which is equivalent to -9.8 m/s2 and has no horizontal acceleration (ax = 0).

Projectiles have horizontal (x) and vertical (y) components that are independent of each other. Thus, we will recreate the Uniformly Accelerated Motion Formulae according to the horizontal and vertical components of a projectile.

Figure 3. Velocity component

vector diagram for horizontally-fired projectile.

vf = vi + agt vf = 0 + (-9.8 m/s2)(2.6 s)

vf = -25 m/s

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Table 4. Horizontal and Vertical Component Formulae for Projectiles Launched Horizontally

ax =horizontal acceleration ay = vertical acceleration (ag) vfx = final horizontal velocity vfy = final vertical velocity vix = initial horizontal velocity vix = initial vertical velocity x = horizontal displacement y = vertical displacement Sample Problem 1

A marble is thrown horizontally from a table top with a velocity of 1.50 m/s. The marble falls 0.70 m away from the table’s edge. A) How high is the lab table? B) What is the marble’s velocity just before it hits the floor?

Before you can find the height of the lab table, you must determine first the time it took the marble to reach the ground. Given/Find:

Horizontal (x) component

Vertical (y) component

vix = 1.50 m/s x = 0.70 m

ay = -9.8 m/s2

y = ?

vfy = ?

t = ?

Note: You can solve for time using equations from Horizontal or Vertical Component depending on the given. In this problem, we cannot solve for time using the Y component since the variables are lacking. If you look at the X Component, Vix and X are given; therefore, we can use these variables in solving for time.

Solve for time using this formula: X =vixt X = vixt

0.70 m = (1.50 m/s) t t = _0.70 m_

1.50 m/s t = 0.47 s

Uniformly

Accelerated Motion

Formulae

Horizontal (x)

component Formulae

ax = 0

Vertical (y)

component Formulae

viy = 0

1. vf = at + vi

vf x= axt + vix vf x = vix

vf y= ayt + viy vf y = ayt

2. d = vit + at2

2 X = vixt + axt2 X=vixt 2

y= viy t + ay t2 y =ayt2

2 2 3. d = vf + vi 2

X = vfx + vix_ 2

y = vfy_ 2

4. Vf2 = vi

2 + 2ad

Vfx2 = vix

2 + 2axx Vfx

2 = vix2

Vfy2 = viy

2 +2ayy Vfy

2 = 2ayy

t t t

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Since we already have the value for time = 0.47s, we can now solve for y (vertical displacement) or the height of the lab table. We can use this formula in solving for y: y = ayt2

2 y = ayt2 2

-y = (-9.8 m/s2) (0.47 s)2

2 -y = (-9.8 m/s2) (0.2209 s2)

2 -y = -2.16482 m

2 -y = -1.08 m

y = 1.08 m

We can also solve for vfy using our value for time = 0.47 s. We can use

this formula : vf y = ayt vf y = ayt

vf y = (-9.8 m/s2) (0.47 s)

vf y = 4.61 m/s

What’s More

Activity 4: Solve the Problems

Directions: Solve the following problems. Show complete solution.

1. Maria dropped a coin in the wishing well and it hits the bottom after 3 seconds. Determine the final velocity of the coin as it hits the bottom of the well and find out the height of the wishing well.

Given/Find (3 points)

Formula/e (2 points)

Solution (4 points)

Given: ag=________ vi = _______ t = ________ Find: vf = ? h = ?

a._______________ b. ______________

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y takes a negative sign since we are dealing with downward displacement

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2. An airplane from rest accelerates on a runway at 6.5 m/s2 for 15 s until it finally takes off the ground. What is the distance covered before takeoff?

Given/Find (3 points)

Formula/e (1 point)

Solution (2 points)

Given: vi = ________ a = ________ t = _________ Find: d = ?

a._______________

3. A motorcycle from rest accelerates uniformly over a time of 5 seconds and covers a

distance of 20 m. Determine the acceleration of the motorcycle.

Given/Find (4 points)

Formula/e (1 point)

Solution (2 points)

Given: vi = ________ t = _________ d = ________ Find: ___________

a._______________

4. A cannonball is launched horizontally from the top of a mountain with a velocity of

3.5 m/s. The cannonball falls 50 m away from the edge of the mountain. a) How high is the mountain? b) What is the cannonball’s velocity just before it hits the ground? Given/Find:

Horizontal (x) component (2 pts) Vertical (y) component (3 pts)

vix = _______ x =________

ay = ________

Find: ________ ________

t = ?

Solution: (6 pts)

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20

What I Have Learned

Activity 5: How much do you know?

Direction: Complete the table below with kinematic equations.

Uniformly Accelerated Motion Formulae

Horizontally Launched Projectiles Vertical

Component Formulae

Horizontally Launched Projectiles Horizontal Component Formulae

Activity 6: Complete me!

Direction: Complete the statements below with the correct words. Choose from the box below.

1. Projectiles have vertical and horizontal components that are ___________ of each other.

2. The only force acting upon a projectile is _________________________.

3. The ball rolling down an incline has a velocity that is ______________________.

4. Projectiles do not have horizontal forces, therefore horizontal acceleration is _________.

5. There are two kinds of problems related to projectiles namely _________________________ and ________________________________.

What I Can Do

Activity 7: Think Like a Projectile! Directions: Solve the following problems. Show complete solution. Round off your final answers to 2 decimal places. (ex. 2.4545697 -→ 2.45) 1. A moving airplane dropped a cargo (ignore the effect of air resistance). If the plane from which the cargo was dropped was flying at a height of 600 m, and the cargo lands 300 m horizontally from the initial dropping point, a. how fast was the plane flying when the bottle was released? (vix = ?)

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gravity increasing projectles lauched at an angle projectiles launched horizontally zero Independent

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b. what was the velocity of the bottle right before it hit the ground? (vfy = ?) Solve for time first using this equation: t = 2y ay

c. Draw a figure or diagram that illustrates the given problem.

2. Which of the two balls thrown horizontally has greater initial velocity? (vix = ?)

Assessment

TEST I. Multiple Choice Directions. Choose the letter of the correct answer. Write the letter on space provided before each number. _____1. If a freely falling ball is somehow equipped with a speedometer, by how much would its speed reading increase for every second?

a. 0 m/s b. 9.8 m/s c. 10 m/s d. 20 m/s _____2. A soccer ball is hit vertically upward by a player. What is its acceleration after 1 second?

a. 0 m/s2 b. 1 m/s2 c. 9.8 m/s2 d. -9.8 m/s2

25 m

5 m

15 m

5 m

Draw the figure here Solution

20

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_____3. The motion of an object with constant acceleration is also known as

a. Motion c. constant motion

b. uniform motion d. uniformly accelerated motion _____4. A toy is dropped from the top of a 100 m building. What is the ball’s initial velocity?

a. 0 m/s b. 9.8 m/s c. 10 m/s d. 15 m/s _____5. What is the only force that acts upon projectiles?

a. Motion b. gravity c. inertia d. velocity _____6. Which illustrates projectile motion?

a. Running

b. Driving a car

c. Kicking a soccer ball

d. Dropping a stone from the top of the building _____7. What is the horizontal distance covered by the ball thrown horizontally at 10 m/s if its time of flight is 3.0 s?

a. 10 m b. 20 m c. 30 m d. 35 m _____ 8. A man drives his motorcycle horizontally off the end of a cliff and lands in the water 2 s later. How high is the cliff?

a. 19.6 m b. 15.6 m c. 17.9 m d. 20 m _____9. A marble is launched horizontally from a table at 12 m/s and lands 5 m away from the edge of the table. What is the time of flight of the marble?

a. 0.10 s b. 0.42 s c. 0.5 s d. 0.45 s _____10. What is the direction of acceleration for any projectile?

a. Up b. down c. left d. right Test II. Matching Type. Direction: Match Column A with Column B. Write the letter of your answer on the space provided before each number.

A ______1. A free falling object has an acceleration that is _______. ______2. The vertical and horizontal components of projectiles are ________ of each other. ______3. Projectiles launched horizontally will have zero ________ acceleration.

______4. An object that moves in two dimensions. ______5. The ball rolling down an incline has a velocity that is ______. ______6. ___________ acceleration of projectiles launched horizontally is constant. ______7. Projectiles launched horizontally has a horizontal _________ that is uniform or constant. ______8. Things thrown upward always fall at a constant _________. ______9. The initial velocity of a free falling object is _______. ______10. Refers to the path the projectiles make.

B

a. Velocity

b. Vertical

c. Constant

d. Horizontal

e. Projectile

f. Curve

g. Increasing

h. Zero

i. Acceleration

j. independent

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Answer Key – Gr9Q4W1 Science

What I have Learned

Activity 5: How much do you know?

What’s In

Activity 1 Activity 6: Complete me!

1.S , d 1. independent

2.S, a 2. gravity

3.V, b 3. increasing 4.V, a 4. zero

5.S, c 5. Projectiles launched horizontally, projectiles launched at

Activity 2 an angle

1.Force

2.Displacement Assessment 3.Velocity

4.Gravity

5.Acceleration

What’s New

Students’ answers may vary based on the experiment

What’s More

1.Given: ag = -9.8m/s2

vi = 0 m/s

t = 3 s

Formulae: vf = vi + agt, h = vit + agt2

2

Soln: vf = 0 m/s + (-9.8m/s2) (3 s) h = (0 m/s) (3 s) + (-9.8m/s2) (3 s)2

vf = 29.4 m/s 2

h = 44.1 m

2.Given: a =6.5 m/s2

t = 15 s

vi = 0 m/s

Formula: d = vit + at2 d= (0 m/s) (15 s) + (6.5 m/s2) (15 s)2

2 2

d= 731.25 m

3.Given: vi = 0 m/s Formula: d = vit + at2/2 Soln: 20 m = 0 + a (5 s)2

t = 5 s a = 1.6 m/s2 d = 20 m

4.

Horizontal (x) component (2 pts) Vertical (y) component (3 pts)

vix = 3.5 m/s

x = 50 m

ay = -9.8 m/s2 Find: y = ?

Vf = ?

Soln:

X = vixt y = ayt2/2 vf = ayt

50 m = (3.5 m/s) t -y= -9.8 m/s2 (14.29 s)2/2 vf = -9.8 m/s2(14.29 s)

t= 14. 29 s y = 1000.6 m vf = 140.04 m/s

UAM Formulae: Horizontal Component Vertical

vf = at + vi vfx = vix vfy = ayt

d = vit + at2 x =vixt y = ayt2/2

2 x = [(vfx + vix)/2] t y = (vfy/2)t

d = [(vf + vi)/2] t vfx2 = vix2 vfy = 2ayy vf2 = vi2 + 2ad

TEST I TEST II

1.B 1. c

2.D 2. j 3.D 3. d

4.A 4. e

5.B 5. g

6.C 6. b

7.C 7. a 8.A 8. i

9.B 9. h

10.B 10. f

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References

Books: Alvarez, Liza A., Dave G. Angeles, Herman L. Apurada, Ma. Pilar P. Carmona,

Oliver A. Lahorra, Judith F. Marcaida, Ma. Regaele A. Olarte, Estrella C. Osorio, Digna C. Paningbatan, Marivic S. Rosales, and Ma. Teresa B. Delos Santos. “Unit 4 Force, Motion and Energy.” In Science – Grade 9 Learner’s

Module, First Edition, 229-254. Pasig City, Philippines: Department of Education, 2013.

Electronic Resources:

Henderson, Tom. “What is a Projectile.“ Physicsclassroom.com.

https://www.physicsclassroom.com/class/vectors/Lesson-2/What-is-a-

Projectile#:~:text=characteristic%20of%20projectiles.-

,A%20projectile%20is%20an%20object%20upon%20which%20the%20on

ly%20force,in%20motion%20at%20constant%20velocity (accessed

December 30, 2020).

Development Team Region IX Hymn

Writer: Michelle F. Baguio Teacher III Zamboanga del Sur NHS

Editor: Irmina C. Calibo Teacher III

Reviewer: Mila P. Arao EPS - Science Management Team:

Danny B. Cordova, EdD CESO VI SDS

Ma. Colleen L. Emoricha, EdD, CESE ASDS

Maria Diosa Z. Peralta CID Chief

Ma. Madelene P. Mituda, EdD EPS-LRMDS

Mila P. Arao EPS -Science

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