Transcript CHAPTER 6
CHAPTER 6 MOTION IN 2 DIMENSIONS PROJECTILE MOTION • Projectile- any object that is shot or thrown through the air. (bullet, ball, drop of water, etc.) – After the initial thrust (force on the object) the only force acting on the object is the force of gravity, if air resistance is ignored. – The path of all projectiles is going to be a parabola; generally moves upward for a distance reaches a maximum height and the travels downward. Trajectory- the path a projectile takes through space; If the initial force of thrust is known you can calculate the trajectory. Important parts of an object’s trajectory. • Maximum height • Distance object travels- range- X component of the motion- horizontal motion • Flight time Independence of motion • The parabolic path of an object is the combination of 2 separate motions: Vertical Horizontal + What if we viewed the 2 people from the previous slide are viewed from behind the person throwing the ball? What would the path of the ball look like? It would appear as if the ball went straight up and then down. What if the 2 people were viewed from directly above? What would the path of the ball look like? It would appear to be moving from one player to the other in a straight line at constant speed A ball is dropped and another is given a horizontal velocity of 2.0 m/s. What is similar about he trajectories of the balls? Why do the balls behave this way? The horizontal motion of the ball is uniform because the velocity is constant. There is no acceleration because there is no horizontal force acting on the ball; The horizontal velocity is always going to be equal to the initial velocity. The vertical motion is shown to be accelerating because of the force of gravity Combining the 2 motions created from the velocity vectors gives the parabolic pathway. PROBLEM SOLVING 1. Separate the projectile motion problem into a vertical motion problem and a horizontal motion problem 2. The vertical motion problem is exactly like an object being dropped straight up or down. 1 2 df di vi (tf ) a (tf ) 2 • a is equal to -9.8 m/s2 A pool ball leaves a 0.60-meter high table with an initial horizontal velocity of 2.4 m/s. Predict the time required for the pool ball to fall to the ground and the horizontal distance between the table's edge and the ball's landing location. 1 df di vi (tf ) a (tf 2 ) 2 .6= ½ (9.8)t2 t2= .122 t= .35 s d= vt d= (2.4) .35s d= .84 m A soccer ball is kicked horizontally off a 22.0-meter high hill and lands a distance of 35.0 meters from the edge of the hill. Determine the initial horizontal velocity of the soccer ball. 1 df di vi (tf ) a (tf 2 ) 2 22m= 0 + (0)t + ½(9.8)t2 22 = -4.9 t2 t2= 4.49 t= 2.11 s d= vt 35 m= v (2.11) v = 16.6 m/s HOMEWORK pg. 150 1-3 Projectiles launched at an angle • When a projectile is launched at an angle it has a vertical component and a horizontal component • As the object is moving up it is slowing down • As the object comes down the object is speeding up • The velocities at each point in the vertical direction are the same just in different directions • Maximum Height is when the velocity is equal to 0 m/s the horizontal component of an object launched at an angle Equations for projectile motion problems Time of Flight Total distance travelled or Range Maximum Height A projectile is fired at 12.5 m/s at an angle of 53.1° with the horizontal from a point 75.0 m above the ground. a) How long does it take to reach the ground? b) What maximum height does it reach? c) What horizontal distance does it travel before striking the ground? d) With what velocity does it strike the ground?