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2-D PROJECTILE MOTION PHYSICS CHAPTER 6 SECTIONS 1 AND 2 PAGES 150-160 DO NOW https://www.youtube.com/watch?v=tF_zv3TCT1U WHAT IS 2-DIMENSIONAL MOTION?MOTION Motion of objects moving in 2 dimensions which are under the influence of gravity. What are the 2 dimensions? Rise and Fall (y-axis) and a movement to the left or right (x-axis) What are the forces acting on the object in 2 dimensions? Intial Force of the Object (Kinetic Energy, Elastic Potential Energy) and Gravity. Neglect Air Resistance PROJECTILE VS FREE-FALL Objects thrown or launched into the air and whose motion is subject to gravity’s pull Object that is not thrown meaning it has no initial velocity TRAJECTORY Path that gravity causes an object to move through space. It is characterized by a curve downward You can determine the path a projectile will take if you know: • Its initial velocity • The angle at which an object is launched PROJECTILES FOLLOW A PARABOLIC PATH WHERE AS FREE-FALL OBJECTS DO NOT Explain the difference between these 2 paths: How would the object in free-fall look? Launched horizontally Launched at an angle PROJECTILES FOLLOW A PARABOLIC PATH WHERE AS FREE-FALL OBJECTS DO NOT Projectile motion is essentially free fall motion with an initial horizontal velocity OR initial horizontal and vertical velocity. Launched horizontally Launched at an angle 2 TYPES OF PROJECTILE MOTION 1. Projectiles launched horizontally NO initial upward or downward motion All motion will be horizontal (x) and down (y) 2 TYPES OF PROJECTILE MOTION 2. Projectiles launched at an angle Use x and y components to analyze motion Have initial upward angle of motion (could be initially downward as well) Have initial horizontal motion CONCEPT CHECK: PROJECTILE OR NOT? WHY Directions (15 minutes) : Your job is to discuss with your groupmates whether the scenarios are scenarios of Projectiles Your Group will be assigned different Scenarios: Will Byers’ Group: Scenarios 1, 2 and 4 Dustin’s Group: Scenarios 6 and 7 Eleven’s Group: 3 and 5 Lucas’ Group: Scenarios 4,5,6 if so state which type of projectile: Free-Fall, Horizontal Projectile, Projectile Launched at an angle. Sketch what you expect the trajectory to look like EQUATIONS FOR HORIZONTALLY LAUNCHED PROJECTILES Horizontal Motion Δx = VxΔt Vx = Vxi = constant Vertical Motion Δy = +½ g •(Δt)2 Vyf = +g •Δt Vyf2 = +2g •Δy Overall Final Velocity Vf2 = Vx2 + Vyf2 PRACTICE PROBLEMS FOR PROJECTILES Grab your Worksheet as we will complete number one together as a class. You throw a stone horizontally at a speed of 5.0 m/s from the top of a cliff that is 78.4 m high a. How long does it take the stone to reach the bottom of the cliff? b. How far from the base of the cliff does the stone hit the ground? c. What are the horizontal and vertical components of the stone’s velocity just before it hits the ground? END OF DAY 1 Use the remainder of class time to complete practice problems on Projectile motion PROJECTILE MOTION DAY 2 REVIEW OF PROJECTILE MOTION What is Projectile Motion? Trajectory of a Projectile Calculation of Projectile Motion FEATURES OF PROJECTILE MOTION? Thrown into the Air 2-D Motion Parabolic Path Affected by Gravity Determined by Initial Velocity DEFINITION: PROJECTILE MOTION Projectile motion refers to the 2-D motion of an object that is given an initial velocity and projected into the air at an angle. The only force acting upon the object is gravity. It follows a parabolic path determined by the effect of the initial velocity and gravitational acceleration. TRAJECTORY (PATH) OF A PROJECTILE TRAJECTORY (PATH) OF A PROJECTILE The path of a projectile is the result of the simultaneous effect of the H (x) & V(y) components of its motion H component constant velocity motion V component accelerated downward motion H & V motions are independent H & V motions share the same time t The projectile flight time t is determined by the V component of its motion TRAJECTORY (PATH) OF A PROJECTILE H velocity(Horizontal component) is constant V velocity (vertical component) is changing vy = - g t H distance: dx = v0 t V distance: dy = 1/2 g t2 PROJECTILE MOTION AT AN ANGLE AIM: HOW CAN WE FULLY UNDERSTAND PROJECTILE MOTION AT AN ANGLE? Basically, we’re going to break the projectile motion up into its basic components To understand projectile motion, let’s look at the motion of a ball thrown into the air and follow its trajectory (Note: assume that there are no forces acting on the ball besides gravity. EQUATIONS FOR PROJECTILES LAUNCHED AT AN ANGLE Horizontal Component Δx = Vi (cosθ) •Δt Vxi = Vi (cosθ) remains constant Vertical Component Vyi = Vi(sinθ) Δy = Vi(sinθ) • Δt + ½ g •(Δt)2 Vyf = Vi(sinθ) + g •(Δt) Vyf2 = Vi2(sinθ)2 + 2g •(Δy) Overall Final Velocity Vf2 = Vx2 + Vyf2 LETS COMPLETE A HORIZONTALLY LAUNCHED PROJECTILE Scenario #1 You throw a stone horizontally at a speed of 5.0 m/s from the top of a cliff that is 78.4 m high C. What are the horizontal and vertical components of the stone’s velocity just before it hits the ground? EXAMPLE PROBLEM LETTER C WHAT IF YOU DO NOT HAVE THE ANGLE? If you do not have the angle utilize the following equation: θ = tan-1 (diy/dix) SAMPLE QUESTION #1 A cannonball is fired from ground level at an angle of 60° with the ground at a speed of 72 m/s. What are the vertical and horizontal components of the velocity at the time of launch? SAMPLE QUESTION #1 ANSWER AND EXPLANATION Given vi = 72 m/s θ = 60° Missing: viy = ?? vix = ?? Remember the formula for finding the x and y components of velocity? Also: x component is horizontal and y component is vertical (think of a coordinate graph!) vix = vi cosθ vix = 72 m/s cos60° vix = 36 m/s viy = vi sinθ viy = 72 m/s sin60° viy = 62.35 m/s SAMPLE QUESTION #2 You kick a soccer ball at an angle of 40° above the ground with a velocity of 20 m/s. What is the maximum height the soccer ball will reach? SAMPLE QUESTION #2 ANSWER AND EXPLANATION Given vi = 20 m/s θ = 40° Missing: viy = ??? dy = ??? For this question, you must find the initial vertical velocity in order to find the maximum height. Then plug it into a second equation noting that vfy = 0 m/s at maximum point. viy = vi sinθ viy = 20m/s sin40° viy = 12.86 m/s vfy2 = viy2 – 2gdy (0m/s)2 = (12.86 m/s)2 – 2(9.81 m/s2)dy -165.38 m2/s2 = (-19.62m/s2)dy 8.43 m = dy END OF DAY 2