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Projectile Motion

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Projectile Motion

Chapter 3.3

- 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

- 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!

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

- 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

- 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

- Use the pythagorean theorem to find the resultant velocity using the components (Vx and Vy)
- Use SOH CAH TOA to find the direction

- 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= ?????

Δx= 2.2 m

Δy= -1.0m (bc the cat falls down)

What are we looking for??

1.0 m

2.2m

- 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.

- 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.

- Rearrange and solve for Vx:
- Cat’s Speed is 4.89 m/s

- 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?

- Δy= ?a=-9.81 m/s2
- t = 6.39 sVx=?
- Vy,i = 0 Δx= 68 m

The cliff is 200 m high

- 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

- 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

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?

- 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?

- 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

- 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

- Vy,i=+26 m/s
- a= -9.81 m/s2
- t = 2 seconds

- 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