# 2-D Projectile Motion Chapter 3-3 2nd edition ```2-D PROJECTILE MOTION
PHYSICS
CHAPTER 6 SECTIONS 1 AND 2
PAGES 150-160
DO NOW
WHAT IS 2-DIMENSIONAL MOTION?MOTION
 Motion of objects moving in 2 dimensions which are under the
influence of gravity.
 What are the 2 dimensions? Rise and Fall (y-axis) and a movement
to the left or right (x-axis)
 What are the forces acting on the object in 2 dimensions? Intial
Force of the Object (Kinetic Energy, Elastic Potential Energy) and
Gravity. Neglect Air Resistance
PROJECTILE VS FREE-FALL
 Objects thrown or launched into the air and whose motion is
subject to gravity’s pull
 Object that is not thrown meaning it has no initial velocity
TRAJECTORY
 Path that gravity causes an object to move
through space.
 It is characterized by a curve downward
 You can determine the path a projectile will
take if you know:
•
Its initial velocity
•
The angle at which an object is launched
PROJECTILES FOLLOW A PARABOLIC PATH WHERE AS
FREE-FALL OBJECTS DO NOT
Explain the difference between these 2 paths: How
would the object in free-fall look?
Launched
horizontally
Launched at an
angle
PROJECTILES FOLLOW A PARABOLIC PATH WHERE AS
FREE-FALL OBJECTS DO NOT
 Projectile motion is essentially free fall motion with
an initial horizontal velocity OR initial horizontal and
vertical velocity.
Launched
horizontally
Launched at an
angle
2 TYPES OF PROJECTILE MOTION
1. Projectiles launched horizontally
 NO initial upward or downward motion
 All motion will be horizontal (x) and down (y)
2 TYPES OF PROJECTILE MOTION
2. Projectiles launched at an angle
 Use x and y components to analyze motion
 Have initial upward angle of motion (could be initially downward as well)
 Have initial horizontal motion
CONCEPT CHECK: PROJECTILE OR NOT? WHY
Directions (15 minutes) :
whether the scenarios are scenarios of Projectiles
Your Group will be assigned different Scenarios:
Will Byers’ Group: Scenarios 1, 2 and 4
Dustin’s Group: Scenarios 6 and 7
Eleven’s Group: 3 and 5
Lucas’ Group: Scenarios 4,5,6
if so state which type of projectile:
Free-Fall, Horizontal Projectile, Projectile
Launched at an angle.
Sketch what you expect the trajectory to look like
EQUATIONS FOR HORIZONTALLY
LAUNCHED PROJECTILES
Horizontal Motion
Δx = VxΔt
Vx = Vxi = constant
Vertical Motion
Δy = +&frac12; g •(Δt)2
Vyf = +g •Δt
Vyf2 = +2g •Δy
Overall Final Velocity
Vf2 = Vx2 + Vyf2
PRACTICE PROBLEMS FOR PROJECTILES
Grab your Worksheet as we will complete number one together as a class.
 You throw a stone horizontally at a speed of 5.0 m/s from the top of a cliff that is
78.4 m high
 a. How long does it take the stone to reach the bottom of the cliff?
 b. How far from the base of the cliff does the stone hit the ground?
 c. What are the horizontal and vertical components of the stone’s velocity just
before it hits the ground?
END OF DAY 1
 Use the remainder of class time to complete practice problems on Projectile
motion
PROJECTILE MOTION DAY 2
REVIEW OF PROJECTILE MOTION
 What is Projectile Motion?
 Trajectory of a Projectile
 Calculation of Projectile Motion
FEATURES OF PROJECTILE MOTION?
Thrown into the Air
2-D Motion
Parabolic Path
Affected by Gravity
Determined by Initial Velocity
DEFINITION: PROJECTILE MOTION
Projectile motion refers to the 2-D motion of
an object that is given an initial velocity and
projected into the air at an angle.
The only force acting upon the object is
gravity. It follows a parabolic path
determined by the effect of the initial velocity
and gravitational acceleration.
TRAJECTORY (PATH) OF A PROJECTILE
TRAJECTORY (PATH) OF A PROJECTILE
 The path of a projectile is the result of the
simultaneous effect of the H (x) &amp; V(y) components
of its motion
 H component  constant velocity motion
 V component  accelerated downward motion
 H &amp; V motions are independent
 H &amp; V motions share the same time t
 The projectile flight time t is determined by the V
component of its motion
TRAJECTORY (PATH) OF A PROJECTILE
 H velocity(Horizontal component) is constant
 V velocity (vertical component) is changing
 vy = - g t
 H distance:
dx = v0 t
 V distance:
dy = 1/2 g t2
PROJECTILE MOTION AT AN ANGLE
AIM: HOW CAN WE FULLY UNDERSTAND
PROJECTILE MOTION AT AN ANGLE?
 Basically, we’re going to break the projectile
motion up into its basic components
 To understand projectile motion, let’s look at
the motion of a ball thrown into the air and
follow its trajectory (Note: assume that there are
no forces acting on the ball besides gravity.
EQUATIONS FOR PROJECTILES LAUNCHED AT AN
ANGLE
Horizontal Component
Δx = Vi (cosθ) •Δt
Vxi = Vi (cosθ)  remains constant
Vertical Component
Vyi = Vi(sinθ)
Δy = Vi(sinθ) • Δt + &frac12; g •(Δt)2
Vyf = Vi(sinθ) + g •(Δt)
Vyf2 = Vi2(sinθ)2 + 2g •(Δy)
Overall Final Velocity
Vf2 = Vx2 + Vyf2
LETS COMPLETE A HORIZONTALLY LAUNCHED
PROJECTILE
 Scenario #1
 You throw a stone horizontally at a speed of 5.0 m/s from the top of a cliff that
is 78.4 m high
 C. What are the horizontal and vertical components of the stone’s velocity just
before it hits the ground?
EXAMPLE PROBLEM LETTER C
WHAT IF YOU DO NOT HAVE THE ANGLE?
 If you do not have the angle utilize the following equation:
 θ = tan-1 (diy/dix)
SAMPLE QUESTION #1
A cannonball is fired from ground level at an angle of 60&deg; with the ground at a speed
of
72 m/s. What are the vertical and horizontal components of the velocity at
the time of launch?
SAMPLE QUESTION #1
Given
vi = 72 m/s
θ = 60&deg;
Missing:
viy = ??
vix = ??
Remember the formula for finding the x and y components of
velocity? Also: x component is horizontal and y component is
vertical (think of a coordinate graph!)
vix = vi cosθ
vix = 72 m/s cos60&deg;
vix = 36 m/s
viy = vi sinθ
viy = 72 m/s sin60&deg;
viy = 62.35 m/s
SAMPLE QUESTION #2
You kick a soccer ball at an angle of 40&deg; above the ground with a velocity of 20 m/s.
What is the maximum height the soccer ball will reach?
SAMPLE QUESTION #2
Given
vi = 20 m/s
θ = 40&deg;
Missing:
viy = ???
dy = ???
For this question, you must find the initial vertical velocity in
order to find the maximum height. Then plug it into a second
equation noting that vfy = 0 m/s at maximum point.
viy = vi sinθ
viy = 20m/s sin40&deg;
viy = 12.86 m/s
vfy2 = viy2 – 2gdy
(0m/s)2 = (12.86 m/s)2 – 2(9.81 m/s2)dy
-165.38 m2/s2 = (-19.62m/s2)dy
8.43 m = dy
END OF DAY 2
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