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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

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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!!!!!!**Revised Kinematic Eqns for projectiles launched horizontally****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 s Vx=? • 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 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**• 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 do we know?**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 What is the initial vertical 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?**• 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

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