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Projectile Motion. Chapter 3.3. Objectives. Recognize examples of projectile motion Describe the path of a projectile as a parabola Resolve vectors into their components and apply the kinematic equations to solve problems involving projectile motion. Projectile Motion.

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Projectile motion

Projectile Motion

Chapter 3.3


Objectives
Objectives

  • Recognize examples of projectile motion

  • Describe the path of a projectile as a parabola

  • Resolve vectors into their components and apply the kinematic equations to solve problems involving projectile motion


Projectile motion1
Projectile Motion

  • How can you know the displacement, velocity and acceleration of a ball at any point in time during its flight?

  • Use the kinematic equations of course! 


Vector components p 98 running vs jumping
Vector Components p.98 (Running vs Jumping)

While running, the person is only moving in one dimension

Therefore, the velocity only has one component.

V

While jumping, the person is moving in two dimensions

Therefore, the velocity has two components.

Vy

Vx


Definition of projectile motion
Definition of Projectile Motion

  • Objects that are thrown or launched into the air and are subject to gravity are called projectiles

  • Examples?

    • Thrown Football, Thrown Basketball, Long Jumper, etc


Path of a projectile
Path of a projectile

  • Neglecting air resistance, the path of a projectile is a parabola

  • Projectile motion is free fall with an initial horizontal velocity

  • At the top of the parabola, the velocity is not 0!!!!!!





Finding the total velocity

V

Vy

Vx

Finding the total velocity

  • Use the pythagorean theorem to find the resultant velocity using the components (Vx and Vy)

  • Use SOH CAH TOA to find the direction


Example p 102 2
Example p. 102 #2

  • A cat chases a mouse across a 1.0 m high table. The mouse steps out of the way and the cat slides off the table and strikes the floor 2.2 m from the edge of the table. What was the cat’s speed when it slid off the table?


What do we know and what are we looking for

Vx= ?????

What do we know and what are we looking for?

Δx= 2.2 m

Δy= -1.0m (bc the cat falls down)

What are we looking for??

1.0 m

2.2m


How do we find vx
How do we find Vx?

  • Equation for horizontal motion:

  • We have x…so we need t.

  • How do we find how long it takes for the cat to hit the ground?

  • Use the vertical motion kinematic equations.


Vertical motion
Vertical Motion

  • Δy= -1.0m

  • a=-9.81 m/s^2

  • What equation should we use?

  • Rearrange the equation, to solve for t then plug in values.


Horizontal equation
Horizontal equation

  • Rearrange and solve for Vx:

  • Cat’s Speed is 4.89 m/s


Cliff example
Cliff example

  • A boulder rolls off of a cliff and lands 6.39 seconds later 68 m from the base of the cliff.

    • What is the height of the cliff?

    • What is the initial velocity of the boulder?


How high is the cliff
How high is the cliff?

  • Δy= ? a=-9.81 m/s2

  • t = 6.39 s Vx=?

  • Vy,i = 0 Δx= 68 m

The cliff is 200 m high


What is the initial velocity of the boulder
What is the initial velocity of the boulder?

  • The boulder rolls off the cliff horizontally

  • Therefore, we are looking for Vx



Projectiles launched at an angle

Vi

Vi = Vx

Vy,i

θ

Vx,i

Projectiles Launched at An Angle


Components of initial velocity for projectiles launched at an angle

Vi

Vy,i

θ

Vx,i

Components of Initial Velocity for Projectiles Launched at an angle



Example p 104 3
Example p. 104 #3

  • A baseball is thrown at an angle of 25° relative to the ground at a speed of 23.0 m/s. If the ball was caught 42.0 m from the thrower, how long was it in the air? How high was the tallest spot in the ball’s path?


What do we know

25°

42.0 m

What do we know?


What can we use to solve the problem
What can we use to solve the problem?

Find t using the horizontal eqn:

Δx=vxΔt = vicos(θ)t

  • How to find Δy?

    • Vy,f = 0 at top of the ball’s path

    • What equation should we use?


Cliff example1
Cliff example

  • A girl throws a tennis ball at an angle of 60°North of East from a height of 2.0 m. The ball’s range is 90 m and it is in flight for 6 seconds.

    • What is the initial horizontal velocity of the ball?

    • What is the initial vertical velocity of the ball?

    • What is the total initial velocity of the ball?

    • How high above the initial position does the ball get?

    • What is the vertical velocity of the ball 2 seconds after it is thrown?


What is the initial horizontal velocity of the ball
What is the initial horizontal velocity of the ball?

  • Δx= 90 m

  • Θ=60°

  • Total time= 6 s

  • Horizontal velocity is constant: Vx


What is the initial vertical velocity of the ball

Vi

Vy,i

θ

Vx,i

What is the initial vertical velocity of the ball?


What is the total initial velocity of the ball

Vi

Vy,i

θ

Vx,i

What is the total initial velocity of the ball?


How high above the ground does the ball get
How high above the ground does the ball get?

  • At the top of the parabola, Vy is 0…so use the revised kinematic equations

  • Add 2m to get the height above the ground: 36.65 m


What is the vertical velocity of the ball 2 seconds after it is thrown
What is the vertical velocity of the ball 2 seconds after it is thrown?

  • Vy,i=+26 m/s

  • a= -9.81 m/s2

  • t = 2 seconds


Important concepts for projectiles launched at an angle
Important Concepts for Projectiles Launched at an Angle is thrown?

  • At the top of the parabola, neither the object’s velocity nor it’s acceleration is 0!!!!!

    • Only Vy is 0

    • Vx is constant throughout the flight

    • Horizontal acceleration is always 0

    • Vertical acceleration is always -9.81 m/s2


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