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Projectile Motion PowerPoint PPT Presentation

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.

Projectile Motion

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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

• 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)

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

• 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

• 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!!!!!!

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

• 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?

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?

• 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

• 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

• Rearrange and solve for Vx:

• Cat’s Speed is 4.89 m/s

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?

• Δy= ?a=-9.81 m/s2

• t = 6.39 sVx=?

• Vy,i = 0 Δx= 68 m

The cliff is 200 m high

What is the initial velocity of the boulder?

• The boulder rolls off the cliff horizontally

• Therefore, we are looking for Vx

Vi

Vi = Vx

Vy,i

θ

Vx,i

Vi

Vy,i

θ

Vx,i

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?

25°

42.0 m

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 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?

• Δx= 90 m

• Θ=60°

• Total time= 6 s

• Horizontal velocity is constant: Vx

Vi

Vy,i

θ

Vx,i

Vi

Vy,i

θ

Vx,i

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?

• Vy,i=+26 m/s

• a= -9.81 m/s2

• t = 2 seconds

Important Concepts for Projectiles Launched at an Angle

• 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