PROJECTILE MOTION by: JANET BRIGIDA A. CATIPON MHS Science 9 Teacher
PROJECTILE MOTION
by: JANET BRIGIDA A. CATIPONMHS 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 PROJECTILESTo 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!
212y gt
Remember that since the projectile is launched horizontally, the INITIAL VERTICAL VELOCITY is equal to ZERO.
HORIZONTALLY LAUNCHED PROJECTILESExample:
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/s2 2
2
1 1500 ( 9.8)2 2102.04
y gt t
t t
10.1 seconds(100)(10.1)oxx v t 1010 m
VERTICALLY LAUNCHED PROJECTILES
Component Magnitude DirectionHorizontal Constant ConstantVertical 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
sinox o
oy o
v vv 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.
cossin
ox o
oy o
v vv v
vo
vox
voy
oxx v t 212oyy 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?
cos20cos53 12.04 /sin
20sin53 15.97 /
ox o
ox
oy o
oy
v vv m sv v
v m s
v o=20
.0 m/s
53
EXAMPLEA 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) 4.9215.97 4.9 15.97 4.9
oyy v t gt t t
t t tt
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 svoy=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 svoy=15.97 m/s x = 39.24 my = 0 ymax=?g = - 9.8 m/s/s
2
2
12
(15.97)(1.63) 4.9(1.63)
oyy v t gt
yy
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 MOTIONHORIZONTALax = o, Vox=Vx = constantHalf projectile:R= VoxtFull Projectile:X = Xo + VoxtR = VoxT
VERTICALHalf Projectile:Voy=0Y=1/2 ag t², use ag = -9.8 m/s²Full Projectile:@max pt/ht:Vy=0, use ag = -9.8 m/s²Y = Yo + Voyt + ½ agt²
OTHER KINEMATICS EQUATIONS TO BE USED IN PROJECTILE MOTION1. Vox = Vo cos ø2. Voy = Vo sin ø3. V = √Vx² + Vy²4. Ø = tanˉ ¹ (Voy/Vox) or Vy/Vx5. Vy² = Voy² + 2 agY6. 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 Momentum2. What is an Impulse3. Bring the following
a. Block of Woodb. Masking Tapec. Protractord. Ruler/Meter Stick
QUOTE TO LIVE BY…
“Project, launch yourself and be discovered…”
- YOURS TRULY-