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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 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 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