Two-Dimensional Motion and Vectors

1. Two-Dimensional Motion and Vectors Chapter 3

2. Introduction to Vectors • A scalar is a quantity that can be completely specified by its magnitude • A vector is a physical quantity that has both magnitude and direction. • Some examples of vector quantities are velocity, acceleration, displacement and forces. • Vectors can be added graphically, but when adding vectors you must be sure that they have the same units and that the vectors describe similar quantities. • The answer found by adding vectors is called the resultant.

3. Vector operations The magnitude and direction of the resultant of two perpendicular vectors can be calculated using the Pythagorean Theorem and the tangent function. Magnitude of vector → Pythagorean Theorem Direction of vector → Tangent function

4. Basic Vector OperationsBoth a magnitude and a direction must be specified for a vector quantity, in contrast to a scalar quantity which can be quantified with just a number. Any number of vector quantities of the same type (i.e., same units) can be combined by basic vector operations.

5. Addition of 2 Vectors

6. Addition of 3 Vectors

7. Set # 9P/3A football player runs 15 m down the field, and then turns to the left at an angle of 15º from his original direction and running an additional 10 m before getting tackledFind the magnitude and direction By 15º 10m Bx 15 m

8. P/5 Set # 9A person walk’s the path shown below, what’s is the person resultant displacement measured from the starting person 100 m Resultant Vector ? 300 m 30° 200 m 150 m 60°

9. Example 4:A plane flies from city A to city B. City B is 1,540 km west and 1,160 km south of city A. What is the total displacement and direction of the plane? • 1,930 km, 43° south of west • 1,850 km, 43° south of west • 1,850 km, 37° south of west • 1,930 km, 37° south of west

10. A plane flies from city A to city B. City B is 1,540 km west and 1,160 km south of city A. What is the total displacement and direction of the plane? 1,540 km A 1,160 km B

11. Example # 5During a rodeo, a clown runs 8 m north, turns 35º east of north and runs 3.5 m. Then after awaiting for the bull to come near, the clown turns due east and runs 5m to exit the arena.Find the clown’s displacement Resultant Vector 5 m 3.5 m 35º 8 m

12. Example 6:A duck waddles 2.5 m east and 6 m north. What is the magnitude and direction of the duck’s displacement with respect to its original position? • 3.5 m at 19° north of east • 6.3 m at 67° north of east • 6.5 m at 67° north of east • 6.5 m at 72° north of east

13. Physics Set # 10 Oct/2/2007Student Name: ____________________________________ Class Period: _______ 300 m 20º 120 m 60º 150 m 30º 150 m 500 m 250 m 25º 160 m 125 m 15º

14. Physics Set # 11 Oct/3/2007Student Name: ____________________________________ Class Period: _______ 5 yds 30º 6 yds 7 yds 50º 14 yds 12 yds 4 yds 6 yds 20º 10 yds

15. Physics Set # 12 Oct/4/2007Student Name: ____________________________________ Class Period: _______ 8 yds 20º 3 yds 40º 4 yds 6 yds 60º 16 yds 11 yds 5 yds 75º 7 yds 40º 15 yds

16. Physics Set # 13 Oct/4/2007Student Name: ____________________________________ Class Period: _______ 18 yds 40º 4 yds 6 yds 60º 12 yds 8 yds 5 yds 65º 7 yds 40º 12 yds

17. 250 m 20° 20º 50 m 60 m 60º 30º 500 m 180 m 250 m 25º 160 m 125 m 15º

18. Vectors • The study of vectors allows for breaking a single vector into components: • the vertical component • the horizontal component. • By breaking a single vector into two components, or resolving it into its components, an object’s motion can sometimes be described more conveniently in terms of directions, such as north to south, or up and down. vi vy vx

19. Essay: Wernher von Braun2 pages/ 12 font Chunk Paragraph:topic sentence + 2 concrete details +4 commentaries + concluding sentence • The early childhood • Family • His achievements in Germany • His involvement in the Nazi Party • How did he ended up in U.S.A.? • His achievements in U.S.A.

20. Projectile Motion

21. Projectile Motion • A projectile is any object upon which the only force acting is gravity. • Projectiles travel with a parabolic trajectory due to the influence of gravity, • There are no horizontal forces acting upon projectiles and thus no horizontal acceleration, • The horizontal velocity of a projectile is constant (never changing in value), • There is a vertical acceleration caused by gravity; its value is -9.81 m/s², down. • The vertical velocity of a projectile changes by – 9.81 m/s each second. • The horizontal motion of a projectile is independent of its vertical motion.

22. Projectile Motion A projectile is an object upon which the only force acting is gravity. There are a variety of examples of projectiles • An object dropped from rest is a projectile (provided that the influence of air resistance is negligible). • An object which is thrown vertically upwards is also a projectile (provided that the influence of air resistance is negligible). • An object which is thrown upwards at an angle is also a projectile (provided that the influence of air resistance is negligible). • A projectile is any object which once projected continues in motion by its own inertia and is influenced only by the downward force of gravity.

24. Projectile Motion • By definition, a projectile has only one force acting upon - the force of gravity. If there were any other force acting upon an object, then that object would not be a projectile. Thus, the free-body diagram of a projectile would show a single force acting downwards and labeled "force of gravity" (or simply Fg). • This is to say that regardless of whether a projectile is moving downwards, upwards, upwards and rightwards, or downwards and leftwards, the free-body diagram of the projectile is still as depicted in the following diagram

25. Projectile Motion

26. Projectile Motion

27. Projectile Motion

28. Projectile Motion Many projectiles not only undergo a vertical motion, but also undergo a horizontal motion. That is, as they move upward and/or downward they are also moving horizontally. There are the two components of the projectile's motion - horizontal and vertical motion. And since "perpendicular components of motion are independent of each other," these two components of motion can (and must) be discussed separately. • The goal of this part of the lesson is to discuss the horizontal and vertical components of a projectile's motion; specific attention will be given to the presence/absence of forces, accelerations, and velocity.

29. Projectile Motion

30. Projectile Motion • In this course, we are going to study 2 different situations about projectiles: • 1. Projectiles launched horizontally • 2. Projectiles launched at an angle θ

31. Projectiles launched horizontally Horizontal motion of a projectile ∆x = vx ∙∆t vx = vi,x= constant Vertical Motion of a projectile ∆y = ½ a(∆t)² vy,f= a ∆t vy,f ²= 2 a∆y

32. Projectiles launched at an angle v vx,i = vi cos Ө vy,i= vi sin Ө vy Horizontal motion of a projectile ∆x = vi ∙cos θ∙∆t vx =vi ∙cosθ = constant Vertical Motion of a projectile ∆y = vi∙sinθ ∙∆t +½ a(∆t)² vy,f = vi∙sin θ + a∙∆t vy,f² = vi² (sin θ)²+2a∙∆y vx Ө

33. Example #1The New York Yankees, in their quest for an unprecedented 27th. World Series Championship, have an extraordinary pitching rotation. One of our best pitchers is Mike Mussina,aka Moose, who is capable of throwing his fastball at 92 mph. If he pitched to the home-plate (horizontally) (60 ft away), how far will the ball fall vertically by the time it reaches home plate?.

34. Example # 1Moose, who is capable of throwing his fastball at 92 mph. If he pitched to the home-plate (horizontally) (60 ft away), how far will the ball fall vertically by the time it reaches home plate?. Given: vix = 92 mi/h → vix = 41 m/s ∆x = 60 ft → ∆x = 18.29 m Unknowns: ∆t =?, ∆y=? Solution: • First find the time that it takes for the baseball to reach home plate: ∆x = vix∙∆t ∆t = ∆x/vix ∆t = 18.29 m / 41 m/s = 0.44 s • Now find the vertical fall: ∆y = ½ a∙(∆t)² = ½ (-9.81 m/s²) (0.44 s)² = - 0.95 m

35. Example # 2A plane wants to drop a cargo package into a drop area. The plane is 120 m above the ground, and traveling at a velocity of 40 m/s in the positive x direction.Find how long will it take the cargo load to reach the ground, and how much horizontal distance will it take to drop it in the landing area. • Given: ∆y = -120 m a= -9.81 m/s2 vx = 40 m/s • Unknowns: ∆t, ∆x 120 m ∆x Drop Zone

36. Example 2:A plane wants to drop a cargo package into a drop area. The plane is 120 m above the ground, and traveling at a velocity of 40 m/s in the positive x direction.Find how long will it take the cargo load to reach the ground, and how much horizontal distance will it take to drop it in the landing area. • Given: ∆y = -120 m a= -9.81 m/s2 vi,x = 40 m/s • vi,y = 0 m/s • Unknowns: ∆t, ∆x • Solution: ∆y =½ a ∆t² → ∆t² = 2∆y/a ∆t² = (2) (-120 m)/ (- 9.81 m/s²) ∆t² = 24.464 → ∆t = √ 24.64 ∆t = 4.96 s ∆x = vi,x ∙∆t ∆x = (40 m/s) (4.96 s) ∆x = 198.40 m

37. Example # 3People in movies often jump from buildings into pools. If a person jumps from a 10th. floor(30 m) to a pool that is 5 m away from the building, with what initial horizontal velocity must the person jump? • Given: • Δy = - 30 m Δx = 5 m a = - 9.81 m/s² 30 m 5 m

38. Example 4:A zookeeper finds an escaped monkey hanging from a light pole. Aiming his tranquilizer gun at the monkey, the zookeeper kneels 12 m from the light pole which is 6 m high. The tip of his gun is 1 m above the ground. The monkey tries to trick the zookeeper by dropping a banana, then continues to hold onto the light pole. At the moment the monkey releases the banana, the zookeeper shoots. If the tranquilizer dart travels at 50 m/s, will the dart hit the monkey or the banana? 6 m 1m 12 m

39. Steps • 1. Find the angle : Θ = tan‾¹ (5/12) = 23º • 2. Find the time it takes to reach the target: ∆x = vi ∙cos θ∙∆t ∆t = ∆x / vi ∙cos θ ∆t =

40. Example 3:A ball is fired from the ground with an initial speed of 1,700 m/s at an initial angle of 30° to the horizontal. Neglecting air resistance, find:a) The ball’s horizontal rangeb) The amount of time the ball was in motion • Given: vi = 1,700 m/s Θ = 30° a = -9.81 m/s² • Unknowns: ∆t, ∆x, ∆y • Steps: ∆x = vi cos Θ ∆t → ∆x = (1,700 m/s) cos 30°∙ ∆t ∆x = 1,472.24 m/s ∆t ∆y = vi sin Θ∆t + ½ a (∆t)² → ∆y = (1,700 m/s) sin 30° ∆t + ½ (- 9.81 m/s²) ∆t² ∆y = 850 m/s ∆t – 4.91 ∆t² ∆y = 0 (because it ends at the same vertical position) 4.91 ∆t²- 850 ∆t = 0 4.91 ∆t- 850= 0 ∆t = 173 s ∆x = (1,472.24 m/s) (173 s) ∆x = 254,697.52 m

41. Golf and PhysicsWhen you play golf, you have a variety of golf clubs to hit for different distances. Most people use the driver to tee off on the par 4’s, and then a 7 iron from the middle of the fairway. The driver has typically an angle on 10˚, and the 7 iron an angle of 35˚, Calculate the horizontal and vertical displacement of a golfer that hits golf balls at an average initial velocity of 120 mi/h for the initial 10 s after impact. www,explore science.com Y axis Height X axis (0,0) (t,0) Time

42. Driver 10˚vi = 120 mi/h (53.63 m/s)

43. 7 iron 35˚vi = 120 mi/h (53.63 m/s)

44. Something’s wrong??? • The average displacement for a 7 iron is 150 yds!! • Calculate the angle for: • Vi = 120 mi/h → 53.63 m/s • ∆x = 150 yds → 137.16 m • ∆t = 6.3s ? • θ = ?????

45. BONUS POINTSThe great pyramid of Giza is approximately 150 m high and has a square base approximately 230 m on a side. What is the approximate area of a horizontal cross section of the pyramid taken 50 m above its base?(Determine this by using 3 different methods!!) • 5,880 m² • 11,760 m² • 23,510 m² • 35,270 m²

46. A F B E C G D

47. CD = 230 mAG = 150 mGF = 50 m A F B E C G D

48. CD = 230 mAG = 150 mGF = 50 mAF = 100 m A AG = AF CD BE BE = 153.33 m BE² = 23,511 m² 100 m F B E 50 m C G D 230 m

49. BONUS POINTSUsing a protractor, a student measures the sum of the interior anglesin a triangle and obtains 176°. What is the percent error of this measurement? • 0.04% • 2.22% • 2.27% • 4.00%

50. BONUS POINTSFor what value of c will the function below have exactly one verticalasymptote? y = ____4______ x² – cx + 9 • 0 • 4 • 6 • 9