Presentation on theme: "Projectile Motion Chapter 3.3. Objectives Recognize examples of projectile motion Describe the path of a projectile as a parabola Resolve vectors into."— Presentation transcript:
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 jumping, the person is moving in two dimensions Therefore, the velocity has two components. V y V x While running, the person is only moving in one dimension Therefore, the velocity only has one component. V
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!!!!!!
Vertical and Horizontal Motion Horizontal MotionVertical Motion Velocity = V x Displacement = Δx Velocity = V y Displacement = Δy Because gravity does not act in the horizontal direction, V x is always constant! Gravity acts vertically, therefore a = -9.81 m/s 2
Equations for projectiles launched horizontally Horizontal MotionVertical Motion Δx= V x t V x is constant! a=0 V y,i =0 (initial velocity in y direction is 0)
Finding the total velocity Use the pythagorean theorem to find the resultant velocity using the components (V x and V y ) Use SOH CAH TOA to find the direction Vx Vy V
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 is the cat’s velocity just before it hits the ground?
What do we know and what are we looking for? 2.2m Δx= 2.2 m Δy= -1.0m (bc the cat falls down) 1.0 m Vx= ????? What are we looking for??
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? – What is the velocity of the boulder just as it strikes the ground?
How high is the cliff? Δy= ?a=-9.81 m/s 2 t = 6.39 sV x =? V y,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
Important Concepts for Projectiles Launched Horizontally Horizontal ComponentsVertical Components Horizontal Velocity is constant throughout the flight Horizontal acceleration is 0 Initial vertical velocity is 0 but increases throughout the flight Vertical acceleration is constant: -9.81 m/s 2
Projectiles Launched at An Angle Projectiles Launched Horizontally Projectiles Launched at an Angle V x is constant Initial V y is 0 V x is constant Initial V y is not 0 ViVi V x,i V y,i θ V i = V x
Components of Initial Velocity for Projectiles Launched at an angle Use soh cah toa to find the V x,i and V y,i ViVi V x,i V y,i θ
Revise the kinematic equations again Horizontal MotionVertical Motion
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? Δx= 42.0 m θ= 25° V i = 23.0 m/s V y at top = 0 Δt=? Δy=? 42.0 m 25°
What can we use to solve the problem? Find t using the horizontal eqn: Δx=v x Δt = v i cos(θ)t How to find Δy? – V y,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
What is the initial vertical velocity of the ball? Vx,i Vy,i θ Vi
What is the total initial velocity of the ball? Vx,i Vy,i θ Vi
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? V y,i =+26 m/s a= -9.81 m/s 2 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 V y is 0 – V x is constant throughout the flight – Horizontal acceleration is always 0 – Vertical acceleration is always -9.81 m/s 2