3. A projectile is an object launched into space under the influence of gravity only.

4. A football is kicked at an angle of θ with respect to the horizontal. The acceleration is -9.8 m/s2 at all times.

5. At the highest point of the trajectory, the ball’s velocity is perpendicular to its acceleration.

6. A projectile fired horizontally will strike the ground in the same time as one dropped vertically from the same position if we neglect the effects of air resistance.

7. The range of a projectile is affected by its initial speed only.

9. (2 points) What is the initial vertical velocity, viy of the projectile?

10. (3 points) How far does the projectile travel horizontally (range), x before it hits the ground?

12. (2 points) What is horizontal velocity of the shell as it hits the ground?

13. (2 points) What is vertical velocity of the shell as it hits the ground?

15. (2 points) What is the initial vertical velocity, viy of the football?

16. (3 points) What is initial speed (magnitude of the velocity) of the football?

18. A projectile fired at any angle above the ground has a constant horizontal velocity in all points. – T

20. A projectile is an object launched into space under the influence of gravity only. – T

21. A football is kicked at an angle of θ with respect to the horizontal. The acceleration is -9.8 m/s2 at all times. – T

22. At the highest point of the trajectory, the ball’s velocity is perpendicular to its acceleration. – T

23. A projectile fired horizontally will strike the ground in the same time as one dropped vertically from the same position if we neglect the effects of air resistance. – T

24. The range of a projectile is affected by its initial speed only. – F

26. (2 points) What is the initial vertical velocity, viy of the projectile?

27. (3 points) How far does the projectile travel horizontally (range), x before it hits the ground?

29. Derivation:vix=vicosθ

30. Substitution:vix=30.0 m/scos60ovix=15.0 m/s

31. Find: viy=?- initial vertical velocity of the projectile

32. Derivation:viy=visinθ

33. Substitution:vix=30.0 m/ssin60ovix=26.0 m/s

34. Find: x=?- range

35. Derivation:x=vixt=(vicosθ)t where vy-viy=at vy=0 at the highest point

36. So the total time is: t=-2visinθa Therefore: x=-2vi2cosθsinθa

37. Substitution:x=-230.0 m/s2cos60osin60o-9.8 m/s2x=79.5 m

38. Find: t=?- time to reach the highest point of the trajectory

39. Derivation:vy-viy=at vy=0 at the highest point

40. So the time is: t=-visinθa

42. (2 points) What is horizontal velocity of the shell as it hits the ground?

43. (2 points) What is vertical velocity of the shell as it hits the ground?

45. Derivation:vi=vix=xt where y=viyt+12at2 initially, viy=0 m/s

46. So, t=2ya Therefore, vi=x2yavi=x a2y

47. Substitution:vi=1330 m-9.8 m/s22-80.0 mvi=329 m/s

48. Find: vx=?- horizontal velocity of the shell as it hits the ground

49. Derivation:horizontal velocity is constant. Therefore, vx=vix=vi

50. Substitution:vx=329 m/s

51. Find: vy=?- vertical velocity of the shell as it hits the ground

52. Derivation:vy2-viy2=2ay initially, viy=0 m/s so, vy=-2ay

53. Substitution:vy=-2-9.8 m/s2-80.0 mvy=-39.6 m/s

54. Find: v=?- velocity (magnitude and direction) as it hits the ground

55. Derivation:v=vx2+vy2 so from the equations from 05 and 07:v=x2a2y+4ay

56. Simplifying, v=ax2+8y22y tanθ=vyvx=-2ayx a2y=-4y2x=-2yx θ=tan-1-2yx

58. (2 points) What is the initial vertical velocity, viy of the football?

59. (3 points) What is initial speed (magnitude of the velocity) of the football?

61. Derivation:vix=xt

62. Substitution:vix=26 m3.4 svix=7.6 m/s

63. Find: viy=?- initial vertical velocity of the football

64. Derivation:y=viyt+12at2back to initial level, y=0 m so, 0=viyt+12at2

65. Therefore, viy=-12at

66. Substitution:viy=-12-9.8 m/s23.4 sviy=17 m/s

67. Find: vi=?- initial speed of the football

68. Derivation:vi=vix2+viy2 so from the equations from 09 and 10:vi=xt2+-12at2

69. Simplifying, vi=12t4x2+a2t4

70. Substitution:vi=123.4 s426 m2+9.8 m/s223.4 s4vi=18 m/s

71. Find: θ=?- angle of projection

72. Derivation:tanθ=viyvix=-12atxt=-at22x therefore, θ=tan-1-at22x

73. Substitution:θ=tan-1--9.8 m/s23.4 s2226 mθ=65o=65o above horizontal