1. Projectiles at an Angle Hardest thing yet.

2. A progression: A)Freefall Review: A ball drops for 3 s. How far does it fall? How fast is it going after 3 s? vi = 0 m/s t = 3s a = -10 m/s 2 d = ? d = v i t + ½ at 2 d = ½ at 2 d = ½ (-10m/s 2 )(3s) 2 d = ½ (-10m/s 2 )(9s 2 ) d = 45 m vf = vi + at vf = at vf = (-10 m/s 2 )(3s) vf = 30 m/s 0 0

3. Next… B) Throwing upward: A ball is thrown upward with an initial velocity of 30 m/s. How far does the ball travel up in the air? How long before it hits the ground? Divide the motion into two halves- up and down Symmetry: time up = time down v i up = -v f down d up = - d down v i up = -v f down 30 m/s = -30 m/s a = -10 m/s 2 v i = 0 m/s vf 2 = vi 2 + 2ad 0 2a d = vf 2 2a d = (-30 m/s) 2 2(-10 m/s 2 ) d = (900 m 2 /s 2 ) (-20 m/s 2 ) d = 45 m

4. v i up = -v f down 30 m/s = -30 m/s a = -10 m/s 2 v i = 0 m/s d = 45 m t=? vf = vi + at vf = at 0 aa t = vf a t = -30 m/s -10 m/s 2 t = 3 s

5. Next… C) Horizontal Projectiles: A balls rolls off a table with a velocity of 6 m/s and lands 2 m away from the table. How long did it take the ball to hit the ground? How high is the table? xy v ix = 6m/s v iy = 0m/s v fx = 6m/s v fy = a x = 0m/s 2 a y = - 10m/s 2 d x = 2m d y =.45m time =.3 s

6. The hardest… D) Projectiles launched at an angle: A ball is launched with a velocity of 100 m/s at an angle of 17.4576º above the horizon. How long did it take for the ball to hit the ground? What is the range of the ball? What was the maximum height of the ball? θ Range Height

7. Our projectile is no longer launched horizontally so it no longer has an v iy = 0 m/s The v i is divided between the x and y directions- we need to do something called “Resolving the x and y components” θ = 17.4576º Range Height Vi = 100m/s

8. Basically the v i given at a certain angle is the resultant velocity of the horizontal (x) and vertical (y) velocity vectors. We have to work backwards. We need to use trigonometry. y x Before: Now:

9. Trigonometry- finding the side or angles of a triangle SOHCAHTOA v i = 100 m/s v ix v iy h o a θ θ = 17.4576º sin Ɵ = o h cos Ɵ = a h tan Ɵ = o a v iy

10. Let’s find v ix (a) knowing v i (h): Vi = 100 m/s v ix v iy h o a θ θ = 17.4576º cos Ɵ = a h cos Ɵ = v ix v i v ix = v i cos Ɵ

11. Let’s find v iy (o) knowing v i (h): v i = 100 m/s v ix v iy h o a θ θ = 17.4576º sin Ɵ = o h sin Ɵ = v iy v i v iy = v i sin Ɵ

12. Let’s use these new formulas to find the v ix and v iy for our example: v ix = v i cos Ɵ v ix = 100 m/s cos 17.4576° v ix = 100 m/s (.95) v ix = 95 m/s v iy = v i sin Ɵ v i = 100 m/s v ix v iy θ = 17.4576º v iy = 100 m/s sin 17.4576° v iy = 100 m/s (.3) v iy = 30 m/s v ix = ? v iy = ?

13. Stop and Check: 100 m/s 30 m/s 95 m/s θ a 2 + b 2 = c 2 (30 m/s) 2 + (95 m/s) 2 = c 2 (900 m 2 /s 2 ) + (9025 m 2 /s 2 ) = c 2 √9925 m 2 /s 2 = c ~100 m/s = c

14. Now we can take this information and put it into a new form of our table: v i = θ = xy v ix = v fx = a x = d x = v iy = v fy = a y = d y = ½ t = t = Don’t forget about symmetry: time up = time down v i up = -v f down d up = - d down

15. Revisit our problem: A ball is launched with a velocity of 100 m/s at 17.4576º above the horizon. How long did it take for the ball to hit the ground? What is the range of the ball? What was the maximum height of the ball? v i = 100 m/s θ = 17.4576º xy v ix = 95 m/s v fx = 95 m/s a x = 0 m/s d x = v iy = 30 m/s v fy = - 30 m/s a y = -10 m/s 2 d y = ½ t = t = v iy = 30 m/s v ix = 95 m/s Vi = 100 m/s v iy = 30 m/s

16. 1. Find the maximum height (d y ): θ = 17.4576º Range Height Vi = 100m/s HEIGHT WILL LEAD YOU TO TIME

17. Height is a vertical variable- Need to use info from the y direction only: v i = 100 m/s θ = 17.4576º xy v ix = 95 m/s v fx = 95 m/s a x = 0 m/s d x = v iy = 30 m/s v fy = - 30 m/s a y = -10 m/s 2 d y = ½ t = t = Can we use: v fy 2 = v iy 2 + 2a y d y

18. We can only use this equation if we analyze just half of the projectiles motion and make it like a drop problem: v iy = 0 m/s v fy 2 = v iy 2 + 2a y d y d y = ? θ = 17.4576º Range Vi = 100m/s V top = V f =

19. θ = 17.4576º Range Vi = 100m/s V top = V f = For time also just analyze half the motion. Now this becomes like a horizontal projectile problem. Remember t up = t down 2. Finding time:

20. v iy = 0 m/s d y = v iy t + ½ a y t 2 t = ? v i = 100 m/s θ = 17.4576º xy v ix = 95 m/s v fx = 95 m/s a x = 0 m/s d x = v iy = 30 m/s v fy = - 30 m/s a y = -10 m/s 2 d y = 45 m ½ t = t =

21. 3. Finding the range (d x ): d x = v ix t + ½ a x t 2 d x = ? v i = 100 m/s θ = 17.4576º xy v ix = 95 m/s v fx = 95 m/s a x = 0 m/s d x = v iy = 30 m/s v fy = - 30 m/s a y = -10 m/s 2 d y = 45 m ½ t = 3s t = 6s

22. v i = 100 m/s θ = 17.4576º xy v ix = 95 m/s v fx = 95 m/s a x = 0 m/s d x = 570 m v iy = 30 m/s v fy = - 30 m/s a y = -10 m/s 2 d y = 45 m ½ t = 3s t = 6s Completed Chart v i = 100 m/s Θ = 17.4576°

23. 4. Finding components using a ruler and a protractor: Used when the #s aren’t easy and you don’t have a calculator 1 st thing to do is make an x-y axis using your protractor and ruler that is a 90° Set a scale- 1 cm = 10 m/s Draw in the vi vector at the correct scaled length with the correct angle Draw in the v iy line- be careful!!! The length of this line, converted using the established scale, is the magnitude of the initial velocity in the y direction (for our example it should be 3 cm) Draw in the v ix line- The length of this line, converted using the established scale, is the magnitude of the initial velocity in the x direction (for our example it should be 9.5 cm) v i = 100 m/s = 10 cm 17.4576° v ix = 95 m/s = 9.5 cm v iy = 30 m/s = 3 cm

24. sincostan 0°010 30°.5.866.577 45°.707 1 60°.866.51.73 90°10____ Common sin/cos/tan to MEMORIZE

25. HW #1: A ball is launched with a velocity of 100 m/s at 30º above the horizon. How long did it take for the ball to hit the ground? What is the range of the ball? What was the maximum height of the ball? θ Range Height

26. Let’s use these new formulas to find the v ix and v iy for our example: v ix = v i cos Ɵ v ix = 100 m/s cos 30° v ix = 100 m/s (.866) v ix = 86.6 m/s v iy = v i sin Ɵ v i = 100 m/s v ix v iy θ = 30º v iy = 100 m/s sin 30° v iy = 100 m/s (.5) v iy = 50 m/s v ix = ? v iy = ?

27. Stop and Check: 100 m/s 50 m/s 86.6 m/s θ a 2 + b 2 = c 2 (50 m/s) 2 + (86.6 m/s) 2 = c 2 (2500 m 2 /s 2 ) + (7500 m 2 /s 2 ) = c 2 √10000 m 2 /s 2 = c 100 m/s = c

28. Revisit our problem: A ball is launched with a velocity of 100 m/s at 30º above the horizon. How long did it take for the ball to hit the ground? What is the range of the ball? What was the maximum height of the ball? v i = 100 m/s θ = 30º xy v ix = 86.6 m/s v fx = 86.6 m/s a x = 0 m/s d x = v iy = 50 m/s v fy = - 50 m/s a y = -10 m/s 2 d y = ½ t = t = v iy = 50 m/s v ix = 86.6 m/s Vi = 100 m/s v iy = 50 m/s

29. 1. Find the maximum height (d y ): θ = 30º Range Height Vi = 100m/s HEIGHT WILL LEAD YOU TO TIME

30. v i = 100 m/s θ = 30º xy v ix = 86.6 m/s v fx = 86.6 m/s a x = 0 m/s d x = v iy = 50 m/s v fy = - 50 m/s a y = -10 m/s 2 d y = ½ t = t = Make it like a drop problem: v iy = 0 m/s v fy 2 = v iy 2 + 2a y d y d y = ? Height is a vertical variable- Need to use info from the y direction only:

31. θ = 30º Range Vi = 100m/s V top = V f = Now this becomes like a horizontal projectile problem. Find time. Remember t up = t down 2. Finding time:

32. v i = 100 m/s θ = 30º xy v ix = 86.6 m/s v fx = 86.6 m/s a x = 0 m/s d x = v iy = 50 m/s v fy = - 50 m/s a y = -10 m/s 2 d y = 125 m ½ t = t = v iy = 0 m/s d y = v iy t + ½ a y t 2 t = ?

33. 3. Finding the range (d x ): d x = v ix t + ½ a x t 2 v i = 100 m/s θ = 30º xy v ix = 86.6 m/s v fx = 86.6 m/s a x = 0 m/s d x = v iy = 50 m/s v fy = - 50 m/s a y = -10 m/s 2 d y = 125 m ½ t = 5s t = 10s d x = ?

34. 4. Finding components using a ruler and a protractor: Used when the #s aren’t easy and you don’t have a calculator 1 st thing to do is make an x-y axis using your protractor and ruler that is a 90° Set a scale- 1 cm = 10 m/s Draw in the vi vector at the correct scaled length with the correct angle Draw in the v iy line- be careful!!! The length of this line, converted using the established scale, is the magnitude of the initial velocity in the y direction (for our example it should be 5 cm) Draw in the v ix line- The length of this line, converted using the established scale, is the magnitude of the initial velocity in the x direction (for our example it should be 8.6 cm) v i = 100 m/s = 10 cm 30° v ix = 86.6 m/s = 8.6 cm v iy = 50 m/s = 5 cm

35. v i = 200 m/s θ = 20.4873º xy v ix = v fx = a x = d x = v iy = v fy = a y = d y = ½ t = t = HW #2- v i = 200 m/s Θ = 20.4873° cos of 20.4873 =.94 sin of 20.4873 =.35

36. Set up your paper like this- v i = 200 m/s Θ = 20.4873° v i = 200 m/s θ = 20.4873º xy v ix = v fx = a x = d x = v iy = v fy = a y = d y = ½ t = t = Resolving Components: d y =? d x =? t=? FRONT BACK v i = 200 m/s = 10 cm 20.4873° v ix = ? cm = ? m/s v iy = ? cm = ? m/s

37. Answer: v i = 200 m/s θ = 20.4873º xy v ix = 188m/s v fx = 188m/s a x = 0m/s 2 d x = 2632m v iy = 70m/s v fy = -70m/s a y = -10m/s 2 d y = 245m ½ t = 7s t = 14s cos Ɵ = a h cos Ɵ = v ix v i v ix = v i cos Ɵ v iy = v i sin Ɵ v ix = 200 m/s cos 20.4873° v ix = 200 m/s (.94) v ix = 188 m/s v iy = 200 m/s sin 20.4873° v iy = 200 m/s (.35) v iy = 70 m/s

38. HW #3 d y = 125 m d X = 180 m

39. Orange Review Book

41. v i = 200 m/s v ix v iy θ = 30º v ix = v i cos Ɵ v iy = v i sin Ɵ

42. v i = 25 m/s v ix v iy θ = 53º v ix = v i cos Ɵ v iy = v i sin Ɵ

43. v i = 15 m/s v ix v iy θ = 35º v ix = v i cos Ɵ v iy = v i sin Ɵ

44. v i = 60 m/s v ix v iy θ = 37º v ix = v i cos Ɵ v iy = v i sin Ɵ

45. v i = 19.6 m/s v ix v iy θ = 30º v ix = v i cos Ɵ v iy = v i sin Ɵ v i = 19.6 m/s θ = 30º xy v ix = m/s v fx = m/s a x = 0 m/s d x = v iy = m/s v fy = - m/s a y = -10 m/s 2 d y = ½ t = t =

46. v iy = v i sin Ɵ v i = 40 m/s θ = 40º xy v ix = m/s v fx = m/s a x = 0 m/s d x = v iy = m/s v fy = - m/s a y = -10 m/s 2 d y = ½ t = t = v ix = v i cos Ɵ

47. v i = 8.5 m/s θ = 35º xy v ix = m/s v fx = m/s a x = 0 m/s d x = v iy = 4.9m/s v fy = - 4.9m/s a y = -10 m/s 2 d y = ½ t = t =

48. v i = 40 m/s θ = 60º xy v ix = m/s v fx = m/s a x = 0 m/s d x = v iy = m/s v fy = - m/s a y = -10 m/s 2 d y = ½ t = t = 7.1 s

49. v i = m/s θ = º xy v ix = 49 m/s v fx = 49 m/s a x = 0 m/s d x = v iy = 98 m/s v fy = -98 m/s a y = -10 m/s 2 d y = ½ t = t = v i = m/s v ix v iy θ = º v ix = 49 m/s v iy = 98 m/s

51. v i = m/s θ = º xy v ix = 15 m/s v fx = 15 m/s a x = 0 m/s d x = v iy = 25 m/s v fy = -25 m/s a y = -10 m/s 2 d y = ½ t = t = 5 s

52. Which launch angle gives the greatest range? The cannonball launched at a 45-degree angle had the greatest range. The cannonball launched at a 60- degree angle had the highest peak height before falling. The cannonball launched at the 30-degree angle reached the ground first.

53. v i = θ = xy v ix = v fx = a x = 0 m/s 2 d x = v iy = v fy = a y = -10 m/s 2 d y = ½ t = t = 8s HW #4 v i = ? m/s Θ = ?° v ix = 5 m/s Must use protractor/ruler

54. 1 st thing to do is make an x-y axis using your protractor and ruler that is a 90° Set a scale- 1 cm = 2.5 m/s Draw in the vectors at the correct scaled length with the correct angle

55. v i = ° v ix = 5 m/s = 2cm v iy = 40 m/s= 16cm

56. v i = ° v ix = 5 m/s = 2cm v iy = 40 m/s= 16cm

57. v i = 16.1 cm = 40.25 m/s 83° v ix = 5 m/s = 2cm v iy = 40 m/s= 16cm

58. v i = θ = 70° xy v ix = v fx = a x = 0 m/s 2 d x = v iy = 100 m/s v fy = a y = -10 m/s 2 d y = ½ t = t = HW#5 v iy = 100 m/s Θ = 70°

59. 1 st thing to do is make an x-y axis using your protractor and ruler that is a 90° Set a scale- 1 cm = 10 m/s Draw in the vectors at the correct scaled length with the correct angle

60. 70° v iy = 100 m/s= 10cm

61. vi v iy = 100 m/s= 10cm

62. vivi v ix

63. v i = 106 m/s θ = 70° xy v ix = 35 m/s v fx = 35 m/s a x = 0 m/s 2 d x = v iy = 100 m/s v fy = -100m/s a y = -10 m/s 2 d y = ½ t = t = HW#5 v iy = 100 m/s Θ = 70° Find in order: 1.d y 2.1/2t 3.t 4.d x

64. d y =? “Drop”= v iy = 0m/s v fy 2 = v iy 2 + 2a y d y 2a y d y = vf 2 2a d y = (100 m/s) 2 2(-10 m/s 2 ) d y = (10000 m 2 /s 2 ) (-20 m/s 2 ) d y = 500 m 0

65. ½ t=? “Drop”= v iy = 0m/s t=? vf = vi + at vf = at 0 aa t = vf a t = -100 m/s -10 m/s 2 1/2t = 10 s t = 20 s

66. d x =? v ix = d x t tt d x = v ix t d x = 35 m/s (20s) d x = 700 m d x = v ix t + 1/2a x t 0 d x = v ix t d x = 35 m/s (20s) d x = 700 m

67. v i = 106 m/s θ = 70° xy v ix = 35 m/s v fx = 35 m/s a x = 0 m/s 2 d x = 700m v iy = 100 m/s v fy = -100m/s a y = -10 m/s 2 d y = 500 m ½ t = 10s t = 20s HW#5- Complete v iy = 100 m/s Θ = 70°

68. Practice Problem: A golf ball is hit with a velocity of 20 m/s at 45 above the horizon. Find the time it takes the golf ball to travel, the range of the golf ball, and the maximum height the golf ball reaches.

69. A player kicks a football from ground level at 27.0 m/s at an angle of 30.0 degrees above the horizontal. Find: a)its “hang time” (time that the ball is in the air), b)the distance the ball travels before it hits the ground, and c)its maximum height.

70. A cannon is fired. The cannonball reaches a maximum height of 125 m. The range of the cannonball is 900 m. At what angle was the cannon fired from? θ = ? Range = 900 m Height = 125 m