Physics

1. SUPPORT MATERIAL FOR INDEPENDENT LEARNING ENGAGEMENT (SMILE)
12
General Physics 1
First Quarter – Module 4: Week 4
Motion in Two and Three Dimensions
Jeovanny A. Marticion
GOVERNMENT PROPERTY
NOT FOR SALE
A Joint Project of the
SCHOOLS DIVISION OF DIPOLOG CITY
and the
DIPOLOG CITY GOVERNMENT

2. GENERAL PHYSICS 1 - Grade 12
Alternative Delivery Mode
Quarter 1 – Module 4: Motion in Two and Three Dimensions
First Edition, 2020
Republic Act 8293, section 176 states that: No copyright shall subsist in any work of
the Government of the Philippines. However, prior approval of the government agency or office
wherein the work is created shall be necessary for exploitation of such work for profit. Such
agency or office may, among other things, impose as a condition the payment of royalties.
Borrowed materials (i.e., songs, stories, poems, pictures, photos, brand names,
trademarks, etc.) included in this module are owned by their respective copyright holders. Every
effort has been exerted to locate and seek permission to use these materials from their
respective copyright owners. The publisher and authors do not represent nor claim ownership
over them.
Published by the Department of Education
Secretary: Leonor Magtolis Briones
Undersecretary: Diosdado M. San Antonio
Printed in the Philippines by ________________________
Department of Education –Region IX
Office Address: ____________________________________________
____________________________________________
Telefax: ____________________________________________
E-mail Address: ____________________________________________
Development Team of the Module
Writers: Jeovanny A. Marticion
Editors:
Reviewers: Cyrus A. Ratilla
Illustrator:
Layout Artist:
Management Team: Virgilio P. Batan -Schools Division Superintendent
Jay S. Montealto -Asst. Schools Division Superintendent
Amelinda D. Montero -Chief Education Supervisor, CID
Nur N. Hussien -Chief Education Supervisor, SGOD
Ronillo S. Yarag -Education Program Supervisor, LRMS
Leo Martinno O. Alejo-Project Development Officer II, LRMS

3. What I Need to Know
The previous module studied the motion in a straight line. We will now extend
the idea of motion in a curved path lying in a plane. Few examples of these are objects
thrown, battered baseball, bullet fired from a gun shot, motion of an object tied
through a chord and whirled in a circle and even moons and satellites of planetary
bodies. Hence, the displacement, velocity and acceleration doesn’t anymore lie in the
same single line. We now utilize vector algebra to understand the idea of these
motions. Recall how we studied vector algebra in Module 2.
This module is divided into two lessons:
Lesson Learning Competency At the end of the module, you
should be able to:
Lesson 1 Relative
Velocities
Describe motion using the concept of
relative velocities in 1D and 2D
STEM_GP12KIN-Ic-20
1. Define relative velocity
2. Solve problems involving relative
velocities
Lesson 2 Projectile
Motion
Deduce the consequences of the
independence of vertical and horizontal
components of projectile motion
STEM_GP12KIN-Ic-22
Calculate range, time of flight, and
maximum heights of projectiles
STEM_GP12KIN-Ic-23
Solve problems involving two
dimensional motion in contexts such
as, but not limited to ledge jumping,
movie stunts, basketball, safe locations
during firework displays, and Ferris
wheels STEM_GP12KIN-Ic-26
1. Identify situations undergoing
projectile motion.
2. Describe the vertical and
horizontal component of object’s
motion when undergoing
projectile motion.
3. Calculate range, time of flight
and maximum heights of
projectiles
4. Solve problems involving two
dimensional motion in contexts
such as, but not limited to ledge
jumping, movie stunts,
basketball, safe locations during
firework displays
Lesson 3 Circular
Motion
Infer quantities associated with
circular motion such as tangential
velocity, centripetal acceleration,
tangential acceleration, radius of
curvature STEM_GP12KIN-Ic-25
Solve problems involving two
dimensional motion in contexts such
as, but not limited to ledge jumping,
movie stunts, basketball, safe locations
during firework displays, and Ferris
wheels STEM_GP12KIN-Ic-26
1. Infer quantities associated with
circular motion such as
tangential velocity, centripetal
acceleration, tangential
acceleration, radius of curvature
2. Solve problems involving two
dimensional motion in contexts
such as Ferris wheel

4. What I Know
Choose the letter of the best answer. Write the chosen letter on a separate sheet of
paper.
1.Which of the following is TRUE when a projectile is launched at an angle above the
horizontal and reaches its maximum height?
a.accelerations in x and y components are zero
b.y-acceleration is 9.8 m/s2 and zero along x
c.x-acceleration is 9.8 m/s2 and zero along y
d.both x and y-acceleration is 9.8 m/s2
2.A player threw two balls with the same force wherein one thrown at 45° and the
other at 60°. Between the two balls, the one that will have a longer horizontal range is
__.
a.45°
b.60°
c.both
d.either 45° or 60°
3.The horizontal velocity component of a projectile (ignoring air resistance__.
a.remains the same
b.continously increases
c.zero
d.decreases
4.A bullet is fired horizontally and at the same instant velocity a second bullet is
dropped from the same height. Ignoring air resistance, which is true?
a.fired bullet hits first
b.they hit the same time
c.dropped bullet hit first
d.incomplete info
5.The acceleration due to gravioty in the Moon is only 1/6th of the Earth. If you hit a
baseball with the same effort that you would on Earth, the ball would land
a.one-sixth as far
b.same distance
c.6 times as far
d.36 times as far.
6.The angle for a ball to be thrown to reach the maximum horizontal distance is____.
a.0°
b.30°
c.45°
d.90°

5. 7.This refers to an object launched and follows a curved path while influenced by
gravity.
a.object
b.particles
c.projectile
d.free falling bodies
For nos. 8-10, refer to the problem below:
A shell is fired at a velocity of 300 m/s at an angle of 30° above the horizontal.
8.How far does it go?
a.7 953 m
b.4 591 m
c.26.51 m
d.4 592 m
9.What is its time of flight?
a.61 s
b.31 s
c.15 s
d.25 s
10.What is its maximum altitude?
a.1 148 m
b.7.65 m
c.15.31 m
d.2 296 m
11.Which of the following statements is FALSE about the cannon ball’s path thrown
horizontally.
a.it has a uniform velocity along the x component
b.velocity along the y component increases with respect to time
c.velocity along the y component decreases with respect to time
d.the range depends on its initial velocity
12.The ball was released off the table. How much time it takes to reach the ground?
a.
ℎ
𝑣
b.
1
2
ℎ
𝑣
c.√
2ℎ
𝑔
d.√
𝑔
2ℎ
13.A ball is tied on a string and swung in a vertical circular motion. When it reaches
the peak, its acceleration vector is represented by:
a. b. c. d.
14.A ball rotates at a speed of 3 m/s tied on 1.2 m string. What is the centripetal
acceleration of the object? a.1.2 m/s2 b.3.0 m/s2 c.7.5 m/s2 d.3.2 m/s2
15.A girl whirls the ball at the end of a string. Which of the following statement is
TRUE?
a.speed is not constant
b.velocity is not constant
c.radius is constant
d.acceleration varies

6. Lesson
1 Relative Motion
What’s In
When we studied the motion of any object, we always describe it with reference
to a coordinate system. Car’s velocity would always mean its velocity with respect to
Earth or ground. However, let’s say we are throwing a ball when we are riding in a
moving train. Would the velocity change? How will an observer in the train describe its
motion? How will an observer in the ground describe the motion of the ball? This time,
we will regard the perspective of various observers on describing motions.
What’s New
The idea of relative motion can be explained
through a boat crossing a river. The wind can
influence the movement of a boat moving from one to
the opposite side of the river. The movement will be
affected by the river’s flow. This implies that the boat
will not reach the shore directly across from the point
where it started. The boat will be carried downstream.
Although speedometer of the boat will read a certain
value of its speed, the observer on the other side of
the shore will be reading greater than its value.
What is It
Let’s have a long train of flat cars moving to the right along the path of a level
track. Suppose a motorist dares to drive his motorcycle above the flat cars. In any time
interval, the displacement of the motorcycle relative to the Earth is just the sum of
ImageSource:https://www.google.com/search?q=relative+velocity+of+a+boat&tbm=isch&ved
=2ahUKEwiQ36uZ3rjrAhWUAqYKHcTZBa8Q2-
cCegQIABAA&oq=relative+velocity+of+a+boat&gs_lcp=CgNpbWcQA1CrlQJYjqkCYNKrAmgAc
AB4AIABAIgBAJIBAJgBAKABAaoBC2d3cy13aXotaW1nwAEB&sclient=img&ei=ekNGX5C6O
pSFmAXEs5f4Cg&bih=527&biw=616#imgrc=ADCqfYHtkzO5kM

7. displacements relative to the flatcars and their (flatcars and motorcycle) displacements
relative to the Earth. Hence,
𝑋𝑚𝑜𝑡𝑜𝑟𝑐𝑦𝑐𝑙𝑒/𝐸𝑎𝑟𝑡ℎ = 𝑋𝑚𝑜𝑡𝑜𝑟𝑐𝑦𝑐𝑙𝑒/𝑓𝑙𝑎𝑡 𝑐𝑎𝑟𝑠 + 𝑋𝑓𝑙𝑎𝑡 𝑐𝑎𝑟𝑠/𝐸𝑎𝑟𝑡ℎ
𝑑𝑥𝑚/𝐸
𝑑𝑡
=
𝑑𝑋𝑚/𝑓
𝑑𝑡
+
𝑑𝑋𝑓/𝐸
𝑑𝑡
, thus:
𝒗𝒎/𝑬 = 𝒗𝒎/𝒇 + 𝒗𝒇/𝑬
If the motorist is moving at 13 m/s and the train at 30 m/s to the right, then
the velocity m/E (motorist with respect to Earth) is moving at 43 m/s to the right. If
the motorist is moving at same velocity but this time to the left, the velocity m/E
(motorist with respect to Earth) is moving at 17 m/s to the left. It is just the algebraic
sum of the velocities.
If you notice we used subscripts on their velocities. For example, m/E means
velocity of motorcycle with respect to Earth. In writing this equations, make sure that
the first subscript on the left side of the equation is the first subscript on the first term
of the equation while the second subscript at the right side of the equation is the
found at the right side of the second term in the equation.
Problem You drive north on a straight-two lane road at a constant 88 km/h. A
truck in other lane approaches you at a constant 104 km/h.
What is
asked?
Find the truck’s velocity relative to you. 𝑣𝑡𝑟𝑢𝑐𝑘/𝑦𝑜𝑢 =?
What is
given?
𝑣𝑡𝑟𝑢𝑐𝑘/𝑒𝑎𝑟𝑡ℎ = 104 𝑘𝑚/ℎ 𝑣𝑦𝑜𝑢/𝐸𝑎𝑟𝑡ℎ = 88 𝑘𝑚/ℎ
Diagram
Strategy There are three perspectives in the given problem: truck, you and Earth.
Solution 𝑣𝑡𝐸 = 𝑣𝑡𝑦 + 𝑣𝑦𝐸
𝑣𝑡𝑦 = 𝑣𝑡𝐸 − 𝑣𝑦𝐸
= −
104𝑘𝑚
ℎ
− (
88𝑘𝑚
ℎ
) = −192 𝑘𝑚/ℎ
Answer Therefore, the truck’s velocity with respect to your perspective is
moving at -192 km/h
What is
asked?
Your velocity relative to truck
What is
given?
-192 km/h
Strategy Your velocity can always be the same with your perspective to any object
except that you and the truck were moving in opposite directions.
Solution
𝑣𝑡/𝑦 = −𝑣𝑡/𝑦 = − (−192
𝑘𝑚
ℎ
) = 192 𝑘𝑚/ℎ
v = 30 m/s
ImageSource:https://www.google.com/search?q=flatcars+and+their+(flatcars+and+motorcycle)+displacements+relative+to+the+Earth&tbm=isch&ved=2ahUKEwigicTa4bjrAhUL5JQKHQyPAN4Q2-
cCegQIABAA&oq=flatcars+and+their+(flatcars+and+motorcycle)+displacements+relative+to+the+Earth&gs_lcp=CgNpbWcQA1Cu9gNYrvYDYIv8A2gAcAB4AIABAIgBAJIBAJgBAKABAaoBC2d3cy13aXotaW1nwAEB&sclient=img&ei=KEd
GX6DVO4vI0wSMnoLwDQ&bih=527&biw=616#imgrc=DhgTCnZKrn7EJM

8. What’s More
Solve the problem below. Write your answers on the provided answer sheet.
An airplane’s compass indicates that the is headed due north and is airspeed indicator
shows that it is moving through the air at 240 km/h. If there’s a 100 km/h wind from
west to east, what is velocity of the airplane relative to earth?
What I Have Learned
Fill in the blanks. Write your answers on the provided answer sheet.
Suppose a woman is driving a car. She was spotted by a patrol officer when she
was driving beyond the road’s limit. Prior to pursuit, the woman has a velocity with
respect to ______________________. On the other hand, motorist has a velocity with
respect to ___________________ and ____________________. ___________________ velocity
refers to velocity of one’s body relative to other.
What I Can Do
The idea of relative motion has been considered as today’s complicated concept
in Physics. There are 3 main problems here: how the event took place, how observers
in relative motion see it? And when do they see this event? Provide a situation where
these questions could be applied. Write your answers on the provided answer sheet.

9. Lesson
2 Projectile Motion
What’s In
The previous module discussed the velocities observed from various
perspectives. In this lesson, we will learn bodies moving in a curved path and how its
components behave as the objects follows this path.
What’s New
The concept of projectile motion is widely used in various
sports. In playing soccer, when you kick the ball it follows a path
similar to a parabola. As the ball reaches the highest peak, it goes
back to the ground as influenced by the gravity and air
resistance. Gravity tries to bring back any object in
projectile moti on to the ground while air
resistance slows down the projectile’s flight.
What is It
A body that is given an initial velocity and then follows a path determined
entirely by the effects of gravitational acceleration and air resistance is undergoing
projectile motion. The path it follows which accelerates sideways is called trajectory.
In this case, air resistance is ignored and curvature of Earth and z-dimension is
objects undergoing projectile motion are bullet shot from a rifle, object dropped from
an airplane, batted baseball and a thrown football. Free falling bodies are special cases
of projectile motion in which trajectory is vertical straight line.
Forces involved on a projectile are drag, object’s weight, wind and lift due to
spinning motion. In most cases, we deal with ideal projectile model where objects with
ImageSource:https://www.google.com/search?q=projectile+motion+sports&tbm=isch&ved=2ahUKEwiLiozU4rjrAhUEXJQKHUCUA2wQ2-
cCegQIABAA&oq=projectile+motion+sports&gs_lcp=CgNpbWcQA1DgwgFYh8oBYPzLAWgAcAB4AIABAIgBAJIBAJgBAKABAaoBC2d3cy13aXotaW1nwAEB&sclient=img&ei
=J0hGX8vaN4S40QTAqI7gBg&bih=527&biw=616&rlz=1C1JZAP_enPH705PH706#imgrc=H1sh2AHDqJeW1M

10. range
y
x
vx
vy
vx
vy
vx
vy
vx
reasonable mass moves with lower speed. On the other hand, objects with higher
speed and forces becomes a significant factor, the idealized model for projectile model
could not anymore fit. This is also true for lighter objects due to influence of external
forces which could affect the movement of the ball. Therefore, idealized model
considers weight as the only force.
The projectile motion is a combination of horizontal and vertical components of
motions with constant acceleration. These are independent from each other and we
will analyze them separately. These motions are just superimposed from each other.
The figure below shows the direction of velocity components along x and y-axis.
Acceleration vectors can then be shown
using the figure below. Since velocity
along x-component is constant, the
acceleration is zero. On the other hand,
velocity along y-component is moving at
constant acceleration. The value of
acceleration is equal to 9.8 m/s2.
Along x –axis
𝒗𝒙 = 𝒗𝟎𝒙 (1)
𝒙 = 𝒙𝒐 + 𝒗𝟎𝒙𝒕 (2)
Along y –axis; where a = -g
𝒗𝒚 = 𝒗𝟎𝒚 − 𝒈𝒕 (3)
𝒚 = 𝒚𝟎 + 𝒗𝟎𝒚𝒕 −
𝟏
𝟐
𝒈𝒕𝟐
(4)
range
y
x
vx
vy
vx
vy
vx
vy
vx

11. The previous case is the simplest example for a projectile motion. This is when you
usually release a ball in a trajectory path or a bullet fired from a horizontal gun and
leaves it moving in trajectory path. The figure below shows another case of projectile
motion thrown at some angle. The projectile trajectory is a parabola. The velocity along
y component decreases and becomes zero when it reaches the highest peak. When it
goes back to Earth’s ground, the velocity along y component increases but directed in
opposite direction. The velocity along x component, on the other hand, is constant
throughout the path.
From equation 2,
we substitute vox
𝑥 = 𝑥𝑜 + 𝑣0𝑥𝑡
Where, 𝑣0𝑥 = 𝑣0𝑐𝑜𝑠𝜃
𝒙 = 𝒙𝒐 + 𝒗𝟎𝒄𝒐𝒔𝜽𝒕 (7)
From equation 4,
we substitute voy
𝑦 = 𝑦0 + 𝑣0𝑦𝑡 −
1
2
𝑔𝑡2
Where 𝑣0𝑦 = 𝑣0𝑠𝑖𝑛𝜃
𝒚 = 𝒚𝟎 + 𝒗𝟎𝒔𝒊𝒏𝜽𝒕 −
𝟏
𝟐
𝒈𝒕𝟐
(8)
distance r of the projectile from
the
𝒓 = √𝒙𝟐 + 𝒚𝟐 (9)
The speed of a projectile at any
given time
𝒗 = √𝒗𝒙
𝟐 + 𝒗𝒚
𝟐 (10)
The direction of projectile in terms
of angle
𝒕𝒂𝒏 Ɵ =
𝒗𝒚
𝒗𝒙
(11)
Along x -axis
𝑣𝑥 = 𝑣0𝑥
𝑤ℎ𝑒𝑟𝑒 𝑣0𝑥 = 𝑣0𝑐𝑜𝑠𝜃
Then, 𝒗𝒙 = 𝒗𝟎𝒄𝒐𝒔𝜽 (5)
𝑣𝑥
𝑣𝑦
𝑣
Ɵ
y
x
vy
vx
vy
vx
vx
vy
vx
vy
vx
maximum height
Along y-axis
𝑣𝑦 = 𝑣0𝑦 − 𝑔𝑡
Where 𝑣0𝑦 = 𝑣0𝑠𝑖𝑛𝜃
Then, 𝒗𝒚 = 𝒗𝟎𝒔𝒊𝒏𝜽 − 𝒈𝒕 (6)
Similarly, in the
vector resolution
lesson, you need to
analyze the vector
using right triangles.
Ɵ

12. Just like the previous example, the observation for acceleration vectors can
then be shown using the figure. Since velocity along x-component is constant, the
acceleration is zero. On the other hand, velocity along y-component is moving at
constant acceleration. The value of acceleration is equal to 9.8 m/s2.
The equations can also prove that we are dealing with a parabola.
𝑥 = 𝑥𝑜 + 𝑣0𝑥𝑡, where 𝑥𝑜 = 0. Thus: 𝑥 = 𝑣0𝑥𝑡
𝑦 = 𝑦0 + 𝑣0𝑦𝑡 −
1
2
𝑔𝑡2
, where 𝑦0 = 0. Thus: 𝑦 = 𝑣0𝑦𝑡 −
1
2
𝑔𝑡2
Deriving t from 𝑥 = 𝑣0𝑥𝑡, we get 𝑡 =
𝑥
𝑣0𝑥
Substitute t by
𝑥
𝑣0𝑥
in 𝑦 = 𝑣0𝑦𝑡 −
1
2
𝑔𝑡2
𝑦 = 𝑣0𝑦(
𝑥
𝑣0𝑥
) −
1
2
𝑔 (
𝑥
𝑣0𝑥
)
2
𝑦 = (
𝑣0𝑦
𝑣0𝑥
)𝑥 −
𝑔
2
(
𝑥2
𝑣0𝑥
2)
𝑦 = (
𝑣0𝑦
𝑣0𝑥
)𝑥 −
1
2
(
𝑔
𝑣0𝑥
2) 𝑥2
→ the equation follows the general eq’n for parabola 𝑦 = 𝑎𝑥2
+ 𝑏𝑥
The angle Ɵ can produce various parabolic trajectory when launched at speed v0
y
x
maximum height
ag
ag
ag
ag
ag
Max range = 45°
Ɵ= 60°
Ɵ= 30°
Ɵ= 40°
Ɵ= 50°

13. For a projectile launched with initial velocity v0 with an angle Ɵ from the horizontal,
we can derive general expression for the maximum height.
𝑣𝑓
2
= 𝑣𝑜
2
− 2𝑎𝛥𝑥
Since we are dealing with motion along the y-axis, then initial velocity along y is given
by 𝑣𝑜𝑠𝑖𝑛Ɵ , the acceleration is equal to g and displacement is represented by h,
maximum height.
𝑣𝑓
2
= (𝑣𝑜𝑠𝑖𝑛Ɵ)2
− 2𝑔ℎ
The velocity along y axis at maximum height is zero while x component is still
constant. Then,
0 = (𝑣𝑜𝑠𝑖𝑛Ɵ)2
− 2𝑔ℎ
2𝑔ℎ = (𝑣𝑜𝑠𝑖𝑛Ɵ)2
ℎ =
(𝑣𝑜𝑠𝑖𝑛Ɵ)2
2𝑔
Derive general expression for the maximum horizontal range R
𝑥 = 𝑥𝑜 + 𝑣0𝑐𝑜𝑠𝜃𝑡
𝑅 = 𝑥𝑜 + 𝑣0𝑐𝑜𝑠𝜃𝑡, when 𝑥𝑜 = 0
𝑡 =
𝑅
𝑣0𝑐𝑜𝑠𝜃
𝑦 = 𝑦0 + 𝑣0𝑠𝑖𝑛𝜃𝑡 −
1
2
𝑔𝑡2
= 0
𝑦 = 𝑦0 + 𝑣0𝑠𝑖𝑛𝜃 (
𝑅
𝑣0𝑐𝑜𝑠𝜃
) −
1
2
𝑔 (
𝑅
𝑣0𝑐𝑜𝑠𝜃
)
2
= 0
𝑣0𝑠𝑖𝑛𝜃 (
𝑅
𝑣0𝑐𝑜𝑠𝜃𝑡
) −
1
2
𝑔 (
𝑅
𝑣0𝑐𝑜𝑠𝜃𝑡
)
2
= 0
(
𝑣0𝑠𝑖𝑛𝜃𝑅
𝑣0𝑐𝑜𝑠𝜃
) −
1
2
𝑔 (
𝑅2
𝑣0
2𝑐𝑜𝑠2𝜃
) = 0
(
1
𝑅
) (
𝑣0𝑠𝑖𝑛𝜃𝑅
𝑣0𝑐𝑜𝑠𝜃
) =
1
2
𝑔 (
𝑅2
𝑣0
2𝑐𝑜𝑠2𝜃
) (
1
𝑅
)
(
𝑣0𝑠𝑖𝑛𝜃
𝑣0𝑐𝑜𝑠𝜃
) =
1
2
𝑔 (
𝑅
𝑣0
2𝑐𝑜𝑠2𝜃
)
(
𝑠𝑖𝑛𝜃
𝑐𝑜𝑠𝜃
) =
1
2
𝑔 (
𝑅
𝑣0
2𝑐𝑜𝑠2𝜃
)
𝑣0
2
𝑐𝑜𝑠2
𝜃 (
𝑠𝑖𝑛𝜃
𝑐𝑜𝑠𝜃
) =
1
2
𝑔 (
𝑅
𝑣0
2𝑐𝑜𝑠2𝜃
)(𝑣0
2
𝑐𝑜𝑠2
𝜃)
𝑣0
2
𝑐𝑜𝑠2
𝜃 (
𝑠𝑖𝑛𝜃
𝑐𝑜𝑠𝜃
) =
1
2
𝑔 (
𝑅
𝑣0
2𝑐𝑜𝑠2𝜃
)(𝑣0
2
𝑐𝑜𝑠2
𝜃)
𝑣0
2
𝑐𝑜𝑠𝜃𝑠𝑖𝑛𝜃 =
1
2
𝑔𝑅
(
2
𝑔
) (𝑣0
2
𝑐𝑜𝑠𝜃𝑠𝑖𝑛𝜃) =
𝑔
2
𝑅 (
2
𝑔
)
𝑣0
2
2𝑐𝑜𝑠𝜃𝑠𝑖𝑛𝜃
𝑔
= 𝑅
𝑣0
2
𝑠𝑖𝑛2𝜃
𝑔
= 𝑅
Range along x-
axis
solve for t
Range along y-
axis
substitute
"𝑡 "𝑏𝑦
𝑅
𝑣0𝑐𝑜𝑠𝜃
eliminate “R”
eliminate “𝑣0" on the
left side of the
equation
solve for R
since
𝑠𝑖𝑛2𝜃 = 2𝑐𝑜𝑠𝜃𝑠𝑖𝑛𝜃

14. Problem A motorcycle stunt rider rides off the edge of a cliff. Just as the edge
of its velocity is horizontal with magnitude of 9.0 m/s. Find the
position, distance from the edge of cliff and
of 0.5 s after it leaves the edge of the cliff.
What is
asked?
x =?, y = ? at 0.5 s,
What is
given?
initial velocity of 9.0 m/s
Strategy? use equations 𝑦𝑥 = 𝑥𝑜 + 𝑣0𝑥𝑡 and 𝑦 = 𝑦0 + 𝑣0𝑦𝑡 −
1
2
𝑔𝑡2
Solution
𝑥 = 𝑥𝑜 + 𝑣0𝑥𝑡 = (
9.0𝑚
𝑠
) (0.5𝑠) = 4.5 𝑚
𝑦 = 𝑦0 + 𝑣0𝑦𝑡 −
1
2
𝑔𝑡2
= −
1
2
𝑔𝑡2
= −
1
2
(9.8
𝑚
𝑠
)(0.5𝑠)2
= −1.225 𝑚
Answer The stunt rider is located (4.5 m, -1.225 m) at t = 0.5s
What is
asked?
velocity at t = 0.5 s
What is given? initial velocity of 9.0 m/s
Strategy? use equations 𝑦 = 𝑦0 + 𝑣0𝑡 +
1
2
𝑎𝑔𝑡2
and 𝑣 = 𝑣0 + 𝑎𝑔𝑡
Solution solving for x component
of velocity
𝑣𝑥 = 𝑣0𝑥 = 9.0 𝑚/𝑠
solving for y component
of velocity
𝑣𝑦 = 𝑣0𝑦 − 𝑔𝑡 = −𝑔𝑡
= (−9.8
𝑚
𝑠2
) (0.5𝑠) = −4.9 𝑚/𝑠
solving for magnitude
𝑣 = √𝑣𝑥
2 + 𝑣𝑦
2
= √(9
𝑚
𝑠
)
2
+ (−
4.9𝑚
𝑠
)
2
= 10.24 𝑚/𝑠
solving for direction
∝ = tan−1
𝑣𝑦
𝑣𝑥
= tan−1
−4.9 𝑚/𝑠
9 𝑚/𝑠
= 28.6°
Answer The stunt rider has a velocity of 10.24 m/s before it reachers
the ground directed at 26°
Problem A batter hits a baseball so that it leaves the bat at a speed of 𝑣0 = 37.0 𝑚/𝑠
at an angle of 53.1°.
What is
asked?
Find the position of the ball and its velocity at t = 2 s
What is given? 𝑣0 = 37.0
𝑚
𝑠
Ɵ = 53.1°
Solution we use eq’n 5 to
solve x
component
𝑣𝑥 = 𝑣0𝑐𝑜𝑠𝜃 = (37
𝑚
𝑠
) (cos 53.1°) = 22.2 𝑚/𝑠
we plug-in the
value of vx in
eq’n 7
𝑥 = 𝑥𝑜 + 𝑣0𝑥𝑡 = (
22.2𝑚
𝑠
) (2𝑠) = 44.4 𝑚
we use eq’n 6 to
solve for
position at y
𝑣0𝑠𝑖𝑛𝜃 = (37
𝑚
𝑠
) (sin 53.1°) = 29.6 𝑚/𝑠
𝑦 = 𝑦0 + 𝑣0𝑦𝑡 −
1
2
𝑔𝑡2
= (29.6
𝑚
𝑠
) (2𝑠)2
= 39.6 𝑚

15. Answer Therefore, the ball is located at (44.4 m, 39.6m) at t = 2s.
What is
asked?
Find the time when the ball reaches the highest point
What is given? The velocity along y component when it reaches the highest point is zero,
vy = 0 m/s with its initial velocity along y axis of 29.6 m/s.
Solution solving for t 𝑣𝑦 = 𝑣0𝑦 − 𝑔𝑡
0
𝑚
𝑠
=
29.6𝑚
𝑠
− (9.8
𝑚
𝑠2) 𝑡
𝑡 = 3.02 𝑠
Answer Therefore, the ball reaches the highest peak at t = 3.02 s
What is
asked?
Find the horizontal range and the ball’s velocity just before it hits the
ground.
What is given? 𝑣0 = 37.0
𝑚
𝑠
Ɵ = 53.1°
Solution From the previous solution, we were
able to solve for 3.02 s as travel time
to reach the highest peak. When it
goes back to ground it takes another
3.02 s.
𝑇 = 2𝑡 = 2 (3.02𝑠) = 6.04 𝑠
𝑥 = 𝑥𝑜 + 𝑣0𝑥𝑡
= (22.2
𝑚
𝑠
) (6.04 𝑠)
= 134.088 𝑚
Solving for velocity along x, we will
use eq’n 5. Take note velocity along x
component is always constant
throughout the motion. Hence, initial
velocity along x is equal to its final
velocity.
𝑣𝑥 = 𝑣0𝑐𝑜𝑠𝜃
= (37
𝑚
𝑠
) 𝑐𝑜𝑠53.1°
= 22.2 𝑚/𝑠
Solving for velocity along y, we
will use eq’n 6.
𝑣𝑦 = 𝑣0𝑦 − 𝑔𝑡
= 29.6 𝑚. 𝑠 − (9.8
𝑚
𝑠2
) (6.04 𝑠)
= −29.592 𝑚/𝑠
Negative sign refers to its downward
direction.
We solve for the resultant
velocity
𝑣 = √𝑣𝑥
2 + 𝑣𝑦
2
= √(22.2
𝑚
𝑠
)2 + (−29.592
𝑚
𝑠
)
2
= 37 𝑚/𝑠
Answer Therefore, the ball is 134.00 m from its origin point and has a final
velocity of 37 m/s when it reaches the ground.
What’s More
Solve the given problem. Write your answers on the provided answer sheet.
A tennis ball rolls off the edge of a table top 1.00 m above the floor and strikes the
floor at a point 2.60 m horizontally from the edge of the table. Find the time of the
flight. Find the magnitude of the initial velocity. Find the magnitude the ball’s velocity
just before it strikes the floor. Find the direction of the ball’s velocity just before it
strikes the floor.

16. ImageSource:https://www.stevespanglerscience.com/la
b/experiments/the-coin-drop-sick-science/
What I Have Learned
Fill in the blanks. Write your answers on the provided answer sheet.
An object traveling in a curved path is called ___________________. The motion is
called _________________________. The path it travels is called _____________________.
When an object is thrown and follows this motion, the vertical component of its
acceleration is equal to ___________ while the horizontal component of its acceleration
is equal to ___________________. This is due to the fact that horizontal component of its
velocity is always ____________ while its vertical component is ____________. The vertical
component of velocity ____________ when it goes upward. It becomes _________ when it
reaches the peak. Upon returning to the ground, the velocity _________________ and
directed __________________.
What I Can Do
You will be observing two coins dropping from a table. Coin A will be dropped from the
table while the other one will be projected from the table. You can use a phone camera
to take a video of the demonstration
1. Place two coins on the edge of the table at the same distance above the floor.
2. Drop the two coins at the same time and listen to the sound as they strike the
floor. Coin A must be dropped directly while
Coin B must be thrown in a projectile. You can
do this by flicking the coin across the table to
strike the first coin at the table’s edge. Try to
aim “off center”, this will drop the coin straight
down while projecting the coin with some
horizontals peed.
3. Record the time it takes for the two coins to
drop the floor. You may ask assistance of two
person in your house or someone to record the time.
Record your data on the provided answer sheet.
Calculate the time it will take for the coin to drop using equations for free-falling
bodies and projectile motion. Assume a zero initial velocity for both coins.
Compute for the percentage difference between calculated time and the experiments.
Show your solutions on the provided answer sheet.

17. Lesson
3 Circular Motion
What’s In
The previous module discussed how the components of object’s motion changes
when it undergoes projectile motion. The motion is uniformly accelerated throughout
its journey due to influence of Earth’s gravity. This lesson will discuss circular motion
which is also influenced under constant acceleration although it moves at constant
speed.
What’s New
Satellites are considered as projectiles. These satellites are objects where the
only force experienced is the Earth’s gravity. When a projectile is launched with a
enough speed, it will orbit around the Earth. If the launch speed is too small, this will
lead the satellite to fall towards Earth. Hence, speed must be calculated carefully to
ensure it would not fall back toward Earth and will just maintain its height
throughout its journey.
What is It
When a particle moves in a circular path at
constant speed, the motion of the object is said to
undergo uniform circular motion. The acceleration
is not parallel to the path. It is always directed
towards the center. We will derive the relation.
Acceleration is always perpendicular to the velocity
vector and it changes direction continuously.
a
a
ImageSource:https://www.google.com/search?q=circular+motion&rlz=1C
1JZAP_enPH705PH706&hl=fil&sxsrf=ALeKk02NOLCJhRgLmVaS0Q5rCe8
5WbBJLw:1598442875215&source=lnms&tbm=isch&sa=X&ved=2ahUKE
wiQjJXe57jrAhUxGqYKHRIOD4UQ_AUoAXoECA0QAw&biw=616&bih=52
7#imgrc=jmZZGPp4vrOStM

18. We derive relationship between two
similar triangles
∆𝑣
𝑣1
=
∆s
𝑅
∆𝑣 =
𝑣1
𝑅
∆𝑠
The magnitude of average acceleration
during time interval is given by the
equation a = v/t. Then, we substitue v
from the relationship of two similar
triangles.
𝑎𝑎𝑣𝑒 =
∆𝑣
∆𝑡
=
𝑣1
𝑅
∆𝑠
∆𝑡
he magnitude of instantaneous acceleration
is the limit of this expression.
𝑎 = lim
∆𝑡→∞
𝑣1
𝑅
∆𝑠
∆𝑡
=
𝑣1
𝑅
lim
∆𝑡→∞
∆𝑠
∆𝑡
However, the limit of Δs/Δt is just the speed. So, we represent it
with equation of centripetal acceleration. The symbol ⫠ means the
acceleration is normal to the velocity vector. The word centripetal
means seeking the center. In this case, acceleration is always
directed towards the center.
𝑎⫠ =
𝑣2
𝑅
The magnitude of acceleration in uniform acceleration can also be
expressed in terms of period τ as the time for one revolution. If a
particle travels a full circle, it covers a distance of 2πr, in a time τ,
its speed is given by
𝑣 =
2𝜋𝑟
𝜏
Thus, acceleration is
𝑎⫠ =
𝑣2
𝑅
=
4𝜋2
𝑅
𝑟2
Problem A car travelling at a constant speed of 20 m/s rounds at a curve of radius
100 m. What is its acceleration?
What is
asked?
a = ?
What is
given?
v = 20 m/s, R = 100 m
Strategy solve for a using 𝑎⫠ =
𝑣2
𝑅
Solution
𝑎⫠ =
𝑣2
𝑅
=
(20 𝑚/𝑠)2
100 𝑚
= 4.0 𝑚/𝑠
Answer Therefore, the acceleration of the car is 4 m/s directed at center
Problem In a carnival ride, the passengers travel in a circle at radius 5.0 m.
making complete circle in 4s. What is its acceleration?
What is
asked?
a = ?
What is
given?
r = 5.0 m, τ = 4s
Strategy solve for a using 𝑣 =
2𝜋𝑟
𝜏
and 𝑎⫠ =
𝑣2
𝑅
Solution
𝑣 =
2𝜋𝑟
𝜏
=
2𝜋 (5.0 𝑚)
4𝑠
= 7.85 𝑚/𝑠
𝑎⫠ =
𝑣2
𝑅
=
(7.85 𝑚/𝑠)2
5.0 𝑚
=
61.62 𝑚2
/𝑠2
5.0 𝑚
= 12.32 𝑚/𝑠
Answer Therefore the acceleration of the ride is 12.3 m/s2
Δs
Ɵ
R
ΔƟ
Δv
ImageSource:https://www.google.com/search?q=circular+motion&rlz=1C1JZAP_enPH705PH70
6&hl=fil&sxsrf=ALeKk02NOLCJhRgLmVaS0Q5rCe85WbBJLw:1598442875215&source=lnms&t
bm=isch&sa=X&ved=2ahUKEwiQjJXe57jrAhUxGqYKHRIOD4UQ_AUoAXoECA0QAw&biw=616&
bih=527#imgrc=jmZZGPp4vrOStM

19. What’s More
Solve the given problem. Write your answers on the provided answer sheet.
A Ferris wheel with radius 15.0 m is turning about a horizontal axis through its
center; the linear speed of the passenger rim is constant and equal to 9 m/s.
What are the magnitude and direction of the acceleration of a passenger as she passes
through the lowest point in her circular motion?
How much time does it take the Ferris wheel to make one revolution?
What I Have Learned
Fill in the blanks. Write your answer on the provided answer sheet.
An object undergoing uniform circular motion has a acceleration pointed at the
_____________ although the speed is ___________________. The time it takes to complete
one full circle is called _________________________.
What I Can Do
A viral video of a Physics professor in Silliman University is currently circulating
because of his demonstration on circular motion. He securely tied the handle of a pail
through a string. Then, he filled it with water. He reminded the students that he will
swirl the system of objects and assured they will not get wet. Everyone was screaming
and when he was finished, the class was shocked no amount of water was spilled
throughout the demonstration. Why do you think this happened? Write your answer
on the provided answer sheet.

20. Assessment
1.What is the projectile’s horizontal accelerations when it was thrown at an angle of 30
degrees above its horizontal?
a.zero b.9.81 m/s2 c.it varies d.insufficient information
2.A player kicks the ball with a velocity of 25 m/s directed 53 degrees above the
horizontal. What is the vertical component of its initial velocity?
a.15 m/s b.20 m/s c.33 m/s d.25 m/s
3.At what other angle will the football be kicked to travel 50 yards if its initial velocity
was the same with the ball kicked at 25 degree and travels 50 yards?
a.90 degrees b.45 degrees c.55 degrees d.65 degrees
4.Two balls were thrown horizontally from the same height. Ball A has speed of 0.4
m/s while ball B has a speed of 20 m/s. The time takes for Ball B to reach the ground
compared to Ball A is
a.same b.twice c.half d.four times
5.The ball was fired initially at 12 m/s from a cannon facing northwards. The cannon
moves eastward at 24 m/s. Which of the vectors represent the resultant velocity of the
ball?
a. b. c. d.
6.An arrow was thrown at angle of 45 degrees while the other arrow was thrown at 60
degrees. Compared to arrows fired at 60 degrees, the arrow fired at 45 degrees is
______.
a.longer time of flight and range c.longer time of flight and shorter range
b.shorter time of flight and longer range d.shorter time of flight and range
7.A ball is thrown at 38 degrees. What happens to its velocity?
a.it decreases then increases c.it decreases then remains the same
b.it increases then decreases d.it increases the remains the same
8. The diagram at the right is a setup for
demonstration of motion. When we release the
lever, the rod releases ball B while the ball A will
be thrown horizontally. Which of the following
statements is true?
a.Ball A travels with constant velocity
b.Ball A and Ball will hit the table top at the
same time
c.Ball B will touch the table top before Ball A
d.Ball B will have an increasing acceleration ImageSource:https://www.google.com/search?q=compressed+spring+lever+suppot+rod+support+base+&tbm=isch&ved=2
ahUKEwi149Xu4rjrAhUHvJQKHbMZDKMQ2-
cCegQIABAA&oq=compressed+spring+lever+suppot+rod+support+base+&gs_lcp=CgNpbWcQA1Dx-
iRYrtElYNfUJWgAcAB4AIABAIgBAJIBAJgBAKABAaoBC2d3cy13aXotaW1nwAEB&sclient=img&ei=X0hGX_W3J4f40gSzs7C
YCg&bih=527&biw=616&rlz=1C1JZAP_enPH705PH706#imgrc=mzF5oVrWuKoB7M

21. 9.A machine throws tennis balls at 25 degrees in 14 m/s. The ball then returns to
level ground. Which of the following must be done to produce an increased time of
flight for the second launch?
a.decrease angle and speed c.increase angle and speed
b.decrease angle, increase speed d.increase angle, decrease speed
10.What is the initial horizontal velocity of the ball?
a.
𝐷
√2ℎ
𝑔
⁄
b.2Dh/g c.D√2ℎ 𝑔
⁄ d.Dhg
11.An object was thrown at an initial angle with the
horizontal. Which of the pairs of graph represents its motion?
A C
B D
12.A ball rotates at a speed of 3 m/s tied on 1.2 m string. What is the centripetal
acceleration of the object? a.1.2 m/s2 b.3.0 m/s2 c.7.5 m/s2 d.3.2 m/s2
13.A girl whirls the ball at the end of a string. Which of the following statement is
true?
a.speed is not constant c.radius is constant
b.velocity is not constant d.acceleration varies
14.The longest displacement in vertical axis of any projectile is
a.
𝑣𝑠𝑖𝑛2𝜃
𝑔
b.(𝑣𝑠𝑖𝑛𝜃)2
𝑔 c.
(𝑣𝑠𝑖𝑛𝜃)2
2𝑔
d.
2𝑣𝑠𝑖𝑛2𝜃
𝑔
15.A cannon ball was fired and travels on a parabolic path. Air resistance is neglected.
Which of the following describes the speed of the cannon ball?
a.𝑣𝑎 < 𝑣𝑏 < 𝑣𝑐 b. 𝑣𝑎 > 𝑣𝑏 < 𝑣𝑐 c. 𝑣𝑎 < 𝑣𝑏 = 𝑣𝑐 d.𝑣𝑎 > 𝑣𝑏 = 𝑣𝑐
y
t
a
t
y
t
a
t
y
t
a
t
y
t
a
t
A
B
C

22. Additional Activities
1.A “moving sidewalk” in an airport terminal moves at 1.0 m/s and is 35.0 m long. If a
woman steps on at one end and walks at 1.5 m/s relative to the moving sidewalk, how
much time does it take her to reach the opposite end if she walks
(a) in the same direction the sidewalk is moving?
(b) In the opposite direction?
2.A canoe has a velocity of 0.40 m/s southeast relative to the earth. The canoe is on a
river that is flowing 0.50 m/s east relative to the earth. Find the velocity (magnitude
and direction) of the canoe relative to the river.
3. A physics book slides off a horizontal tabletop with a speed of 1.10 m/s. It strikes
the floor in 0.480 s. Ignore air resistance. Find
(a) the height of the tabletop above the floor;
(b) the horizontal distance from the edge of the table to the point where the book
strikes the floor;
(c) the horizontal and vertical components of the book’s velocity, and the magnitude
and direction of its velocity, just before the book reaches the floor.
(d) Draw x-t, y-t, vx-t, and vy-t graphs for the motion.
4.A grasshopper leaps into the air from the edge of a vertical cliff. Find (a) the initial
speed of the grasshopper and (b) the height of the cliff.
5.A 76.0-kg rock is rolling horizontally at the top of a vertical cliff that is 20 m above
the surface of a lake. The top of the vertical face of a dam is located 100 m from the
foot of the cliff, with the top of the dam level with the surface of the water in the lake.
A level plain is 25 m below the top of the dam.
(a) What must be the minimum speed of the rock just as it leaves the cliff so that it
will reach the plain without striking the dam?
(b) How far from the foot of the dam does the rock hit the plain?
Answer Key

23. References
Department of Education. (2018). General Physics 1 Reader.
Young, H. and Freedman, R. (2016). University Physics with Modern Physics. Pearson
Sears, F., Zemansky, M. and Young H. (1992). College Physics 7th Edition. Addison-
Wesley Publishing Company.
Zitzewits, Haase and Harper. (2013). PHYSICS Principles and Problems. Phoenix
Publishing House, Inc.
URL:
https://www.google.com/search?q=flatcars+and+their+(flatcars+and+motorcycle)+disp
lacements+relative+to+the+Earth&tbm=isch&ved=2ahUKEwigicTa4bjrAhUL5JQK
HQyPAN4Q2-
cCegQIABAA&oq=flatcars+and+their+(flatcars+and+motorcycle)+displacements+r
elative+to+the+Earth&gs_lcp=CgNpbWcQA1Cu9gNYrvYDYIv8A2gAcAB4AIABAIgB
AJIBAJgBAKABAaoBC2d3cy13aXotaW1nwAEB&sclient=img&ei=KEdGX6DVO4vI
0wSMnoLwDQ&bih=527&biw=616#imgrc=DhgTCnZKrn7EJM
https://www.google.com/search?q=relative+velocity+of+a+boat&tbm=isch&ved=2ahU
KEwiQ36uZ3rjrAhWUAqYKHcTZBa8Q2-
cCegQIABAA&oq=relative+velocity+of+a+boat&gs_lcp=CgNpbWcQA1CrlQJYjqkCY
NKrAmgAcAB4AIABAIgBAJIBAJgBAKABAaoBC2d3cy13aXotaW1nwAEB&sclient=
img&ei=ekNGX5C6OpSFmAXEs5f4Cg&bih=527&biw=616#imgrc=ADCqfYHtkzO5
kM
https://www.google.com/search?q=projectile+motion+sports&tbm=isch&ved=2ahUKE
wiLiozU4rjrAhUEXJQKHUCUA2wQ2-
cCegQIABAA&oq=projectile+motion+sports&gs_lcp=CgNpbWcQA1DgwgFYh8oBYP
zLAWgAcAB4AIABAIgBAJIBAJgBAKABAaoBC2d3cy13aXotaW1nwAEB&sclient=i
mg&ei=J0hGX8vaN4S40QTAqI7gBg&bih=527&biw=616&rlz=1C1JZAP_enPH705
PH706#imgrc=H1sh2AHDqJeW1M
https://www.google.com/search?q=compressed+spring+lever+suppot+rod+support+ba
se+&tbm=isch&ved=2ahUKEwi149Xu4rjrAhUHvJQKHbMZDKMQ2-
cCegQIABAA&oq=compressed+spring+lever+suppot+rod+support+base+&gs_lcp=
CgNpbWcQA1Dx-
iRYrtElYNfUJWgAcAB4AIABAIgBAJIBAJgBAKABAaoBC2d3cy13aXotaW1nwAEB
&sclient=img&ei=X0hGX_W3J4f40gSzs7CYCg&bih=527&biw=616&rlz=1C1JZAP_
enPH705PH706#imgrc=mzF5oVrWuKoB7M
https://www.google.com/search?q=circular+motion&rlz=1C1JZAP_enPH705PH706&h
l=fil&sxsrf=ALeKk02NOLCJhRgLmVaS0Q5rCe85WbBJLw:1598442875215&sour
ce=lnms&tbm=isch&sa=X&ved=2ahUKEwiQjJXe57jrAhUxGqYKHRIOD4UQ_AUo
AXoECA0QAw&biw=616&bih=527#imgrc=jmZZGPp4vrOStM