Two-Dimensional Motion
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- Lenard Clarence Bridges
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1 Two-Dimensional Motion Objects don't always move in a straight line. When an object moves in two dimensions, we must look at vector components. The most common kind of two dimensional motion you will encounter is projectile motion. Science of NFL Football: Projectile Motion & Parabolas 1
2 Projectiles A projectile is any object which once set into motion continues in motion by its own inertia and is influenced only by the downwards force of gravity. There are a variety of examples of projectiles. In all cases the influence of air resistance must be negligible. an object dropped from rest an object thrown vertically upwards an object thrown upwards at an angle an object launched with an intial horizontal velocity from an elevated position Since there is only one force acting on a projectile, gravity, all projectiles have the same free body diagram: 2
3 The Path of a Projectile First Law of Motion Second Law of Motion If an object is influenced by an unbalanced force, it will accelerate in the direction of that force. Simulation Parabolic Path Therefore, if an object is influenced by gravity, it will accelerate downwards and fall below its inertial path and follow a parabolic trajectory (path). 3
4 Drop versus Launch At the same moment that the ball on the left is dropped, the ball on the right is launched horizontally. (Galileo was the first person to suggest that an object launched horizontally will reach the ground in the same amount of time as an object dropped vertically.) Since perpendicular components of motion are independent of each other, gravity does not affect a projectile's horizontal motion. Simulation Perpendicular Components of Motion 4
5 Also, it can be seen that the ball on the right has a constant horizontal velocity since the horizontal displacement is the same for equal time intervals. Since the horizontal displacement is constant and the vertical velocity is increasing, the direction of the total resultant velocity is continually changing which accounts for the object's parabolic motion Since there are no horizontal forces, there can be no horizontal acceleration. The only vertical force is the force of gravity, so the vertical acceleration should be 9.80m/s 2 downward. 5
6 Projectile Motion A B 6
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8 Simluation The Truck and The Ball The Truck and The Ball Summary The ball follows a parabolic path. As the ball rises towards its peak, it slows down because it undergoes a downward acceleration. As the ball falls toward the ground, it speeds up because it undergoes a downward acceleration. The time it takes the ball to reach its peak is the same amount of time it will take to return to its initial height. The ball is always located directly above the launcher from which it was projected. We must assume the influence of air resistance is negligible. 8
9 a) the initial horizontal velocity must be horizontal component of the initial velocity vector d) which is a) b) the initial vertical velocity must be the vertical component of d) which is c) c) as with horizontally launched projectiles, there are no horizontal forces acting on the projectile launched at an angle; therefore the horizontal velocity is constant and the horizontal acc. is zero...none of the diagrams meet this requirement d) the vertical acc. must be directed downward due to the only force acting on the projectile gravity...b) e) the net force is equal to the force of gravity acting on the projectile to slow it down as it rises vertically to its highest point and again acting on it as it falls back to the ground...if up is a positive force and down is a negative force, the net force is zero...none of these diagrams meet this requirement. 9
10 10
11 Projectile Motion A B where it was launched Simluation The Truck and The Ball parabola 11
12 20 m/s 9.8 m/s 20 m/s 19.6 m/s 20 m/s 29.4 m/s 20 m/s 39.2 m/s s 8.2 s 12
13 Horizontal Motion 13
14 Vertical Motion 14
15 Calculations Gravity only affects the vertical motion of the balls (both balls accelerate downwards). We will have to use the kinematic equations to determine a projectile's vertical velocity and displacement. No forces affect the projectile's horizontal motion. We will have to use v = d/t for the projectile's horizontal velocity and displacement. Both motions take place during the same time interval. The time interval links the two motions. The horizontal distance the projectile travels is called its range. Projectile Charts Horizontal Motion Vertical Motion v x v yi 0 m/s d x v yf a x t 0 m/s 2 d y a y 9.8 m/s 2 t 15
16 Projectiles Launched Horizontally Sample Problem A soccer ball is kicked horizontally off a 22.0 m high hill and lands a distance of 35.0 m from the edge of the hill. Determine the initial horizontal velocity of the soccer ball. Horizontal Motion Vertical Motion v x v yi 0 m/s d x v yf a x 0 m/s 2 d y 9.8 m/s 2 t a y t 16
17 Projectiles Launched Horizontally Sample Problem A soccer ball is kicked horizontally off a 22.0 m high hill and lands a distance of 35.0 m from the edge of the hill. Determine the initial horizontal velocity of the soccer ball. Horizontal Motion Vertical Motion v x? v yi 0 m/s d x 35 m v yf a x t 0 m/s 2 d y 22 m a y 9.8 m/s 2 t To use V = d/t, we need to know t...we can obtain time from the vertical information: This can now be used to find the horizontal velocity The initial horizontal velocity of the soccer ball is 16.5 m/s 17
18 Sample Problem A pool ball leaves a 0.60 m high table with an initial horizontal velocity of 2.4 m/s. What is the pool ball's velocity just before it hits the ground? Horizontal Motion Vertical Motion v x v yi 0 m/s d x v yf a x t 0 m/s 2 d y a y 9.8 m/s 2 t 18
19 Sample Problem A pool ball leaves a 0.60 m high table with an initial horizontal velocity of 2.4 m/s. What is the pool ball's velocity just before it hits the ground? Horizontal Motion Vertical Motion v x d x a x t 2.4 m/s v yi v yf d y 0 m/s m 9.8 m/s 2 a y t 0 m/s? Since we know the horizontal velocity of the pool ball when it leaves the table, and this does not change while the ball is in the air, we only need to know the time the ball is in the air to find the final vertical velocity. This time can be calculated from the vertical motion. The ball's vertical velocity just before it hits the ground is 3.4 m/s and its horizontal velocity remains at 2.4 m/s 19
20 Simulation The Plane and The Package Sample Problem h = 785 m Horizontal Motion Vertical Motion v x v yi 0 m/s d x v yf a x t 0 m/s 2 d y a y 9.8 m/s 2 t 20
21 Simulation The Plane and The Package Sample Problem h = 785 m Horizontal Motion Vertical Motion v x 53.5 m/s v yi 0 m/s d x a x t? v yf d y 0 m/s m/s 2 a y t 785 m In order to find the horizontal distance before the drop point, we need to know the time it will take for the supplies to fall. The co pilot should drop the supplies at a distance of 674 meters from the drop point. 21
22 A movie stunt driver drives a car off of a cliff that is 70.0 m high. If the car has a speed of 90.0 km/h, how far away does the car land from the base of the cliff? 22
23 A ball thrown horizontally at 12.0 m/s from the roof of a building lands 36.0 m from the base of the building. How high is the building? 23
24 A movie stunt driver drives a car off of a cliff that is 70.0 m high. If the car has a speed of 90.0 km/h, how far away does the car land from the base of the cliff? 24
25 25
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27 Propel Bottle Project 27
28 28
29 Projectiles Launched at Angles 29
30 Parabolic Projectile Path 30
31 Projectiles Launched at an Angle If a projectile is launched at an angle to the horizontal, then the initial velocity of the projectile has both a horizontal and vertical component. Reminder Vector resolution is the process of taking a single vector at an angle and separating it into two perpendicular components. v xi = v cos θ v yi = v sin θ The time for a projectile to rise to its maximum height can be found by using the kinematic equation: For a projectile which lands at the same height at which it started, the total time of flight is twice the time to rise to the peak. The trajectory of a projectile is symmetrical about its maximum height. 31
32 The initial launching velocity, v, must be resolved into vertical and horizontal components, v yi and v xi. v v yi Max Height v xi RANGE (horizontal displacement) Time If it takes a projectile 4.0 s to rise to its peak, then it will take a total of 8.0 s to move through the air from start to finish. t total = 2(t up ) 32
33 Projectile Motion A Level Physics 33
34 Example 1 A football is kicked with a speed of 21.0 m/s at an angle of 37.0 o to the ground. a. How much later does it hit the ground? b. How far away does it land? 34
35 Sample Problems A ball is shot at a speed of 25.0 m/s from a cannon at an angle of 30.0 o above the horizontal. What is the ball's range? 30.0 o Horizontal Motion Vertical Motion v x v yi d x a x t v yf 0 m/s 2 d y a y 9.8 m/s 2 t 35
36 At the 1968 Olympic Games in Mexico City, Bob Beamon set a new world record in the long jump. His initial velocity was m/s at an angle of 40.0 o to the horizontal. a) What was his maximum height? b) What was his range? Horizontal Motion Vertical Motion v x v yi d x a x t v yf d y 0 m/s m/s 2 a y t 36
37 Video Clip Men's Long Jump, Tokyo 1991 part 4, Mike Powell and Carl Lewis World Record Long Jump 37
38 Attachments part 4, Mike Powell and Carl Lewis World Record Long Jump Science of NFL Football: Projectile Motion & Parabolas Projectile Motion A Level Physics
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