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PROJECTILE MOTION
by: JANET BRIGIDA A. CATIPON
MHS Science 9 Teacher
LEARNING OBJECTIVES
• Define Projectile Motion
• Explain Projectile Motion
• Identify the types of Projectile Motion
• Differentiate the types of Projectile Motion
• Explain and summarize all the kinematics equation
in solving Projectile Motion problems
• Solve problems involving the types of Projectile
Motion
WHAT IS PROJECTILE?
Projectile -Any object which projected by some means and continues
to move due to its own inertia (mass).
PROJECTILES MOVE IN
TWO DIMENSIONS
A projectile moves in 2 -
dimensions, therefore, it has
2 components just like a
resultant vector.
HORIZONTAL “VELOCITY”
COMPONENT
• It NEVER changes, covers equal
displacements in equal time periods. This
means the initial horizontal velocity equals
the final horizontal velocity
In other words, the horizontal
velocity is CONSTANT. BUT WHY?
Gravity DOES NOT work
horizontally to increase or
decrease the velocity.
VERTICAL “VELOCITY”
COMPONENT
• Changes (due to gravity), does NOT cover equal
displacements in equal time periods.
Both the MAGNITUDE and DIRECTION
change. As the projectile moves up the
MAGNITUDE DECREASES and its direction
is UPWARD. As it moves down the
MAGNITUDE INCREASES and the direction
is DOWNWARD.
COMBINING THE
COMPONENTS
These components
produce what is called a
TRAJECTORY or path.
This path is PARABOLIC
in nature.
Component Magnitude Direction
Horizontal Constant Constant
Vertical Changes Changes
HORIZONTALLY
LAUNCHED PROJECTILES
Projectiles which have NO upward trajectory and NO
initial VERTICAL velocity.
0 /oyv m s
constantox xv v 
HORIZONTALLY LAUNCHED PROJECTILES
To analyze a projectile in 2 dimensions we need 2
equations. One for the “x” direction and one for the
“y” direction. And for this we use kinematic #2.
oxx v t
Remember, the velocity is
CONSTANT horizontally, so
that means the acceleration
is ZERO!
21
2
y gt
Remember that since the
projectile is launched
horizontally, the INITIAL
VERTICAL VELOCITY is equal
to ZERO.
HORIZONTALLY LAUNCHED PROJECTILES
Example:
A plane traveling with a
horizontal velocity of 100 m/s is
500 m above the ground. At
some point the pilot decides to
drop some supplies to
designated target below. (a)
How long is the drop in the air?
(b) How far away from point
where it was launched will it
land?
What do I
know?
What I want to
know?
vox=100 m/s t = ?
y = 500 m x = ?
voy= 0 m/s
g = -9.8 m/s/s
2 2
2
1 1500 ( 9.8)
2 2
102.04
y gt t
t t
    
   10.1 seconds
(100)(10.1)oxx v t   1010 m
VERTICALLY LAUNCHED PROJECTILES
Component Magnitude Direction
Horizontal Constant Constant
Vertical Decreases up, 0
@ top, Increases
down
Changes
Horizontal Velocity
is constant
Vertical
Velocity
decreases on
the way
upward
Vertical Velocity
increases on the
way down,
NO Vertical Velocity at the top of the trajectory.
VERTICALLY LAUNCHED PROJECTILES
Since the projectile was launched at a angle, the
velocity MUST be broken into components!!!
cos
sin
ox o
oy o
v v
v v




vo
vox
voy

VERTICALLY LAUNCHED
PROJECTILES
There are several things you
must consider when doing
these types of projectiles
besides using components. If it
begins and ends at ground
level, the “y” displacement is
ZERO: y = 0
VERTICALLY LAUNCHED PROJECTILES
You will still use kinematic #2, but YOU MUST use
COMPONENTS in the equation.
cos
sin
ox o
oy o
v v
v v




vo
vox
voy

oxx v t 21
2oyy v t gt 
EXAMPLE
A place kicker kicks a football with a velocity of 20.0 m/s and at
an angle of 53 degrees.
(a) How long is the ball in the air?
(b) How far away does it land?
(c) How high does it travel?
cos
20cos53 12.04 /
sin
20sin53 15.97 /
ox o
ox
oy o
oy
v v
v m s
v v
v m s



 

 
  53
EXAMPLE
A place kicker kicks a
football with a velocity
of 20.0 m/s and at an
angle of 53 degrees.
(a) How long is the ball
in the air?
What I know What I want
to know
vox=12.04 m/s t = ?
voy=15.97 m/s x = ?
y = 0 ymax=?
g = - 9.8
m/s/s
2 2
2
1 0 (15.97) 4.9
2
15.97 4.9 15.97 4.9
oyy v t gt t t
t t t
t
    
    
 3.26 s
EXAMPLE
A place kicker kicks a
football with a velocity
of 20.0 m/s and at an
angle of 53 degrees.
(b) How far away does it
land?
What I know What I want
to know
vox=12.04 m/s t = 3.26 s
voy=15.97 m/s x = ?
y = 0 ymax=?
g = - 9.8
m/s/s
(12.04)(3.26)oxx v t   39.24 m
SAMPLE PROBLEM:
A place kicker kicks a
football with a velocity
of 20.0 m/s and at an
angle of 53 degrees.
(c) How high does it
travel?
CUT YOUR TIME IN HALF!
What I know What I want
to know
vox=12.04 m/s t = 3.26 s
voy=15.97 m/s x = 39.24 m
y = 0 ymax=?
g = - 9.8
m/s/s
2
2
1
2
(15.97)(1.63) 4.9(1.63)
oyy v t gt
y
y
 
 
 13.01 m
BASICS STUDENTS SHOULD KNOW
1. What is a Projectile Motion?
2. What is a Projectile?
3. What is aTrajectory?
4. Why is Horizontal Velocity is constant all throughout in Projectile
Motion?
5. Why is Vertical velocity is zero at maximum height?
6. What is changing in Projectile Motion?
7. What is the difference between Half Projectile Motion and Full
Projectile Motion?
8. What is the difference Half-Time and Hang-Time?
9. Is there an acceleration along the horizontal in Projectile Motion?
10. Is there an acceleration along the vertical in Projectile Motion?
What is it?
HALF PROJECTILE MOTION
FULL PROJECTILE MOTION
PROJECTILE MOTION
HORIZONTAL
ax = o, Vox=Vx =
constant
Half projectile:
R= Voxt
Full Projectile:
X = Xo + Voxt
R = VoxT
VERTICAL
Half Projectile:
Voy=0
Y=1/2 ag t², use ag = -9.8 m/s²
Full Projectile:
@max pt/ht:
Vy=0, use ag = -9.8 m/s²
Y = Yo + Voyt + ½ agt²
OTHER KINEMATICS EQUATIONS TO
BE USED IN PROJECTILE MOTION
1. Vox = Vo cos ø
2. Voy = Vo sin ø
3. V = √Vx² + Vy²
4. Ø = tanˉ¹ (Voy/Vox) or Vy/Vx
5. Vy² = Voy² + 2 agY
6. Vy = Voy + agt
MORE EXAMPLES
1. A slingshot is used to launch a stone horizontally
from the top of a 20.0 meter cliff. The stone lands
36.o meters away.
a. At what speed was the stone launched? (17.82 m/s)
b. What is the speed and angle of impact? ( 26.64
m/s, -47.98 degrees)
2. A cannon fires a cannonball 500.0m downrange when
set at 45 degree angle. At what velocity does the
cannonball leave the cannon? (Answer: 70.0m/s)
EVALUATION
1. A punter in a football game kicks a ball from the goal
line at 60 degrees from the horizontal at 25.0 m/s
a. What is the hang time of the punt? (Ans: 4.41 s)
b. How far downfield does the ball land? (Ans: 55.2m)
2. A skier leaves the horizontal end of a ramp with a
velocity of 25.0m/s and lands 70.0 m from the base of
the ramp. How high is the end of the ramp from the
ground? (Answer: 38.5 m)
ASSIGNMENT
1. What is a Momentum
2. What is an Impulse
3. Bring the following
a. Block of Wood
b. Masking Tape
c. Protractor
d. Ruler/Meter Stick
QUOTE TO LIVE BY…
“Project, launch yourself
and be discovered…”
-YOURSTRULY-

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PROJECTILE MOTION

  • 1. PROJECTILE MOTION by: JANET BRIGIDA A. CATIPON MHS Science 9 Teacher
  • 2. LEARNING OBJECTIVES • Define Projectile Motion • Explain Projectile Motion • Identify the types of Projectile Motion • Differentiate the types of Projectile Motion • Explain and summarize all the kinematics equation in solving Projectile Motion problems • Solve problems involving the types of Projectile Motion
  • 3. WHAT IS PROJECTILE? Projectile -Any object which projected by some means and continues to move due to its own inertia (mass).
  • 4. PROJECTILES MOVE IN TWO DIMENSIONS A projectile moves in 2 - dimensions, therefore, it has 2 components just like a resultant vector.
  • 5. HORIZONTAL “VELOCITY” COMPONENT • It NEVER changes, covers equal displacements in equal time periods. This means the initial horizontal velocity equals the final horizontal velocity In other words, the horizontal velocity is CONSTANT. BUT WHY? Gravity DOES NOT work horizontally to increase or decrease the velocity.
  • 6. VERTICAL “VELOCITY” COMPONENT • Changes (due to gravity), does NOT cover equal displacements in equal time periods. Both the MAGNITUDE and DIRECTION change. As the projectile moves up the MAGNITUDE DECREASES and its direction is UPWARD. As it moves down the MAGNITUDE INCREASES and the direction is DOWNWARD.
  • 7. COMBINING THE COMPONENTS These components produce what is called a TRAJECTORY or path. This path is PARABOLIC in nature. Component Magnitude Direction Horizontal Constant Constant Vertical Changes Changes
  • 8. HORIZONTALLY LAUNCHED PROJECTILES Projectiles which have NO upward trajectory and NO initial VERTICAL velocity. 0 /oyv m s constantox xv v 
  • 9. HORIZONTALLY LAUNCHED PROJECTILES To analyze a projectile in 2 dimensions we need 2 equations. One for the “x” direction and one for the “y” direction. And for this we use kinematic #2. oxx v t Remember, the velocity is CONSTANT horizontally, so that means the acceleration is ZERO! 21 2 y gt Remember that since the projectile is launched horizontally, the INITIAL VERTICAL VELOCITY is equal to ZERO.
  • 10. HORIZONTALLY LAUNCHED PROJECTILES Example: A plane traveling with a horizontal velocity of 100 m/s is 500 m above the ground. At some point the pilot decides to drop some supplies to designated target below. (a) How long is the drop in the air? (b) How far away from point where it was launched will it land? What do I know? What I want to know? vox=100 m/s t = ? y = 500 m x = ? voy= 0 m/s g = -9.8 m/s/s 2 2 2 1 1500 ( 9.8) 2 2 102.04 y gt t t t         10.1 seconds (100)(10.1)oxx v t   1010 m
  • 11. VERTICALLY LAUNCHED PROJECTILES Component Magnitude Direction Horizontal Constant Constant Vertical Decreases up, 0 @ top, Increases down Changes Horizontal Velocity is constant Vertical Velocity decreases on the way upward Vertical Velocity increases on the way down, NO Vertical Velocity at the top of the trajectory.
  • 12. VERTICALLY LAUNCHED PROJECTILES Since the projectile was launched at a angle, the velocity MUST be broken into components!!! cos sin ox o oy o v v v v     vo vox voy 
  • 13. VERTICALLY LAUNCHED PROJECTILES There are several things you must consider when doing these types of projectiles besides using components. If it begins and ends at ground level, the “y” displacement is ZERO: y = 0
  • 14. VERTICALLY LAUNCHED PROJECTILES You will still use kinematic #2, but YOU MUST use COMPONENTS in the equation. cos sin ox o oy o v v v v     vo vox voy  oxx v t 21 2oyy v t gt 
  • 15. EXAMPLE A place kicker kicks a football with a velocity of 20.0 m/s and at an angle of 53 degrees. (a) How long is the ball in the air? (b) How far away does it land? (c) How high does it travel? cos 20cos53 12.04 / sin 20sin53 15.97 / ox o ox oy o oy v v v m s v v v m s           53
  • 16. EXAMPLE A place kicker kicks a football with a velocity of 20.0 m/s and at an angle of 53 degrees. (a) How long is the ball in the air? What I know What I want to know vox=12.04 m/s t = ? voy=15.97 m/s x = ? y = 0 ymax=? g = - 9.8 m/s/s 2 2 2 1 0 (15.97) 4.9 2 15.97 4.9 15.97 4.9 oyy v t gt t t t t t t            3.26 s
  • 17. EXAMPLE A place kicker kicks a football with a velocity of 20.0 m/s and at an angle of 53 degrees. (b) How far away does it land? What I know What I want to know vox=12.04 m/s t = 3.26 s voy=15.97 m/s x = ? y = 0 ymax=? g = - 9.8 m/s/s (12.04)(3.26)oxx v t   39.24 m
  • 18. SAMPLE PROBLEM: A place kicker kicks a football with a velocity of 20.0 m/s and at an angle of 53 degrees. (c) How high does it travel? CUT YOUR TIME IN HALF! What I know What I want to know vox=12.04 m/s t = 3.26 s voy=15.97 m/s x = 39.24 m y = 0 ymax=? g = - 9.8 m/s/s 2 2 1 2 (15.97)(1.63) 4.9(1.63) oyy v t gt y y      13.01 m
  • 19. BASICS STUDENTS SHOULD KNOW 1. What is a Projectile Motion? 2. What is a Projectile? 3. What is aTrajectory? 4. Why is Horizontal Velocity is constant all throughout in Projectile Motion? 5. Why is Vertical velocity is zero at maximum height? 6. What is changing in Projectile Motion? 7. What is the difference between Half Projectile Motion and Full Projectile Motion? 8. What is the difference Half-Time and Hang-Time? 9. Is there an acceleration along the horizontal in Projectile Motion? 10. Is there an acceleration along the vertical in Projectile Motion? What is it?
  • 22. PROJECTILE MOTION HORIZONTAL ax = o, Vox=Vx = constant Half projectile: R= Voxt Full Projectile: X = Xo + Voxt R = VoxT VERTICAL Half Projectile: Voy=0 Y=1/2 ag t², use ag = -9.8 m/s² Full Projectile: @max pt/ht: Vy=0, use ag = -9.8 m/s² Y = Yo + Voyt + ½ agt²
  • 23. OTHER KINEMATICS EQUATIONS TO BE USED IN PROJECTILE MOTION 1. Vox = Vo cos ø 2. Voy = Vo sin ø 3. V = √Vx² + Vy² 4. Ø = tanˉ¹ (Voy/Vox) or Vy/Vx 5. Vy² = Voy² + 2 agY 6. Vy = Voy + agt
  • 24. MORE EXAMPLES 1. A slingshot is used to launch a stone horizontally from the top of a 20.0 meter cliff. The stone lands 36.o meters away. a. At what speed was the stone launched? (17.82 m/s) b. What is the speed and angle of impact? ( 26.64 m/s, -47.98 degrees) 2. A cannon fires a cannonball 500.0m downrange when set at 45 degree angle. At what velocity does the cannonball leave the cannon? (Answer: 70.0m/s)
  • 25. EVALUATION 1. A punter in a football game kicks a ball from the goal line at 60 degrees from the horizontal at 25.0 m/s a. What is the hang time of the punt? (Ans: 4.41 s) b. How far downfield does the ball land? (Ans: 55.2m) 2. A skier leaves the horizontal end of a ramp with a velocity of 25.0m/s and lands 70.0 m from the base of the ramp. How high is the end of the ramp from the ground? (Answer: 38.5 m)
  • 26. ASSIGNMENT 1. What is a Momentum 2. What is an Impulse 3. Bring the following a. Block of Wood b. Masking Tape c. Protractor d. Ruler/Meter Stick
  • 27. QUOTE TO LIVE BY… “Project, launch yourself and be discovered…” -YOURSTRULY-