Motion in Two Directions Chapter 6: Motion in Two Directions Click the mouse or press the spacebar to continue. Splash

Chapter 6: Motion in Two Dimensions Section 6.1: Projectile Motion Chapter Menu

What is projectile motion? Projectile motion is the curved path that an object follows when thrown, launched, or otherwise projected near the surface of the Earth. It looks like a parabola (quadratic equation thank goodness we spend a month in Alg 2 solving ….YEA WHOOOOO!!!!!) Can you think of some examples of projectile motion?

Yes Examples Non Examples Tennis ball Football Golf ball Basketball Javelin Jumping out of a car Cannon ball Plane Rocket Missile

Projectile Motion No matter what the object is, after a projectile has been given an initial thrust, if you ignore air resistance, it moves through the air only under the force of gravity. The force of gravity is what causes the object to curve downward in a parabolic flight path. Its path through space is called its trajectory. Section 6.1-4

An object moving in projectile motion has two components: Horizontal Vertical

Consider a pitcher throwing a baseball: After the ball leaves the pitcher’s hand, the ball’s horizontal velocity is constant. b) The ball’s vertical velocity increases because gravity causes it to accelerate downward. c) The two motions combine to form a curved path.

Will a projectile fall to the Earth at the same time as an object that is dropped straight down? Let’s Find Out! Let’s Do some physics!!!!

The ball is in free fall vertically and moves at constant speed horizontally!!!

Its all about vectors. I cannot wait to start solving!!!

Projectiles Launched at an Angle The adjoining figure shows the separate vertical- and horizontal- motion diagrams for the trajectory of the ball. Section 6.1-7

Projectile Motion Equations Vertical y-motion experiences acceleration due to gravity y=yo+viyt+½at2 vfy=viy+at vfy2=viy2+2ad Horizontal x-motion is uniform (velocity is constant) Range= ΔX=vxt

Problem Type 1: (Starts at same height as it ends) A ball is launched at 4.5 m/s at 66° above the horizontal. What are the maximum height and flight time of the ball? Section 6.1-13

Step 1: Analyze and Sketch the Problem The Flight of a Ball Step 1: Analyze and Sketch the Problem Section 6.1-14

The Flight of a Ball Establish a coordinate system with the initial position of the ball at the origin. Section 6.1-15

The Flight of a Ball Show the positions of the ball at the beginning, at the maximum height, and at the end of the flight. Section 6.1-15

The Flight of a Ball Draw a motion diagram showing v, a, and Fnet. Section 6.1-16

The Flight of a Ball Identify the known and unknown variables. Known: yi = 0.0 m θi = 66° vi = 4.5 m/s ay = −g Unknown: ymax = ? t = ? Section 6.1-17

Step 2: Solve for the Unknown The Flight of a Ball Step 2: Solve for the Unknown Section 6.1-18

The Flight of a Ball Find the y-component of vi. Section 6.1-19

The Flight of a Ball to the top Substitute vi = 4.5 m/s, θi = 66° Section 6.1-20

The Flight of a Ball to the top Find an expression for time. Section 6.1-21

The Flight of a Ball Substitute ay = −g Section 6.1-22

The Flight of a Ball to the top Solve for t. t= .42 sec Total Flight time will be 2 X .42 = .84sec Section 6.1-23

The Flight of a Ball Solve for the maximum height. Substitute yi = 0.0 m, vyi = 4.1 m/s, vy = 0.0 m/s at ymax, t= .42 sec a=g = 9.80 m/s2 Section 6.1-24

The Flight of a Ball Solve for distance the ball was thrown Section 6.1-24

Splat the cat is tossed horizontally at 20 m/s Problem Type 2: Projectiles Launched Horizontally at a at a different height than landing height (no initial vertical velocity) Splat the cat is tossed horizontally at 20 m/s off a cliff that is 50 m high. How long is the Splat in the air? b) How far from the base of the cliff does Splat land? How fast is the cat moving the Instant before he Splats?

Problems Solving Strategy 1. Write known and unknown variables 2. Write variables for x and y directions What’s missing? vyf = ? ay = g = -9.8m/s2 yi = 0 vyi = 0 ax = 0 vxi = 20 m/s WHY? vxf = 20 m/s vnetf = ? yf = -50 m t = ? xf = ? xi = 0

Problem Type 2 (again): Initial Vx A plane that is moving 200 m/s drops a package. The package takes 15 seconds to reach the ground. A) What height was the package dropped from? B) How far did the plane travel from the time the package was released until the package hit the ground? Is this the same distance the package travels? C) How fast was the package moving when it hit the ground? D) Where will the plane be with respect to the package when the package hits the ground?

Question 1 A boy standing on a balcony drops one ball and throws another with an initial horizontal velocity of 3 m/s. Which of the following statements about the horizontal and vertical motions of the balls is correct? (Neglect air resistance.) Section 6.1-38

Question 1 A. The balls fall with a constant vertical velocity and a constant horizontal acceleration. B. The balls fall with a constant vertical velocity as well as a constant horizontal velocity. C. The balls fall with a constant vertical acceleration and a constant horizontal velocity. D. The balls fall with a constant vertical acceleration and an increasing horizontal velocity. Section 6.1-38

Answer 1 Reason: The vertical and horizontal motions of a projectile are independent. The only force acting on the two balls is the force of gravity. Because it acts in the vertical direction, the balls accelerate in the vertical direction. The horizontal velocity remains constant throughout the flight of the balls. Section 6.1-39

Question 2 Which of the following conditions is met when a projectile reaches its maximum height? A. The vertical component of the velocity is zero. B. The vertical component of the velocity is maximum. C. The horizontal component of the velocity is maximum. D. The acceleration in the vertical direction is zero. Section 6.1-40

Answer 2 Reason: The maximum height is the height at which the object stops its upward motion and starts falling down, i.e. when the vertical component of the velocity becomes zero. Section 6.1-41

Question 3 Suppose you toss a ball up and catch it while riding in a bus. Why does the ball fall in your hands rather than falling at the place where you tossed it? Section 6.1-42

Answer 3 Trajectory depends on the frame of reference. For an observer on the ground, when the bus is moving, your hand is also moving with the same velocity as the bus, i.e. the bus, your hand, and the ball will have the same horizontal velocity. Therefore, the ball will follow a trajectory and fall back into your hands. Section 6.1-43