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# Physics 2 LT3: Projectile Motion Solutions

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### Physics 2 LT3: Projectile Motion Solutions

1. 1. 556260047625Test No.:_________00Test No.:_________1228725-9525PHILIPPINE SCIENCE HIGH SCHOOL – Main Campus<br />Physics 2 – Advanced Topics in Physics I<br />LONG TEST 3: PROJECTILE MOTION<br /><ul><li>True or False</li></ul>Direction:<br />Write T if the statement is true and F if it is false. Note that air resistance is neglected in all statements.<br /><ul><li>A projectile fired at any angle above the ground has a constant horizontal velocity in all points.
2. 2. The time in going up of a projectile is the same to its time in going down back to its initial level. </li></ul>The range of a projectile is maximum if it is fired at angle of 45o.<br />A projectile fired at an angle of 25o covers the same range if it is fired at an angle of 65o with greater initial speed.<br /><ul><li>The vertical motion of a projectile is a uniformly accelerated motion.
3. 3. A projectile is an object launched into space under the influence of gravity only.
4. 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. 5. At the highest point of the trajectory, the ball’s velocity is perpendicular to its acceleration.
6. 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. 7. The range of a projectile is affected by its initial speed only.
8. 8. PROBLEM SOLVING</li></ul>Direction:<br />Solve all the problems completely and correctly in your answer sheet. Always follow the format in problem solving: Illustration (if necessary) -> Given -> Find -> Derivation -> Solution -> Final answer -> Box your final equation and answer. Follow significant figures rules.<br />Problems 01 to 04 pertain to the statement below:<br />A projectile is fired at an angle of 60.0o above the horizontal/ground with an initial speed of 30.0 m/s.<br /><ul><li>(2 points) What is the initial horizontal velocity, vix of the projectile?
9. 9. (2 points) What is the initial vertical velocity, viy of the projectile?
10. 10. (3 points) How far does the projectile travel horizontally (range), x before it hits the ground?
11. 11. (3 points) How long (time) does it take the projectile to reach the highest point in its trajectory?</li></ul>Problems 05 to 08 pertain to the statement below:<br />A shell is fired horizontally in the positive x direction from the top of an 80.0 m high cliff. The shell strikes the ground 1330 m from the base of the cliff.<br /><ul><li>(3 points) Determine the initial velocity of the shell.
12. 12. (2 points) What is horizontal velocity of the shell as it hits the ground?
13. 13. (2 points) What is vertical velocity of the shell as it hits the ground?
14. 14. (3 points) What is the velocity (magnitude and direction) of the shell as it hits the ground? Hint: use your answers in 06 and 07 and apply vector components</li></ul>Problems 09 to 12 pertain to the statement below:<br />A football is kicked with an initial speed vi at an angle θ above the horizontal ground. It covers a horizontal distance/range of 26 m in 3.4 s.<br /><ul><li>(2 points) What is horizontal velocity, vix of the football?
15. 15. (2 points) What is the initial vertical velocity, viy of the football?
16. 16. (3 points) What is initial speed (magnitude of the velocity) of the football?
17. 17. (3 points) What is the angle of trajectory, θ? What is the other angle that can give the same range of the football if it is kicked at the same speed?</li></ul> <br />GOD Bless!!! John 3:16<br />Tract: dtquinsaat_07/31/2010/1832h <br />Answers and Solutions<br /><ul><li>True or False
18. 18. A projectile fired at any angle above the ground has a constant horizontal velocity in all points. – T
19. 19. The time in going up of a projectile is the same to its time in going down back to its initial level. – T </li></ul>The range of a projectile is maximum if it is fired at angle of 45o. – T <br />A projectile fired at an angle of 25o covers the same range if it is fired at an angle of 65o with greater initial speed. – F <br /><ul><li>The vertical motion of a projectile is a uniformly accelerated motion. – T
20. 20. A projectile is an object launched into space under the influence of gravity only. – T
21. 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. 22. At the highest point of the trajectory, the ball’s velocity is perpendicular to its acceleration. – T
23. 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. 24. The range of a projectile is affected by its initial speed only. – F
25. 25. PROBLEM SOLVING</li></ul>Problems 01 to 04 pertain to the statement below:<br />A projectile is fired at an angle of 60.0o above the horizontal/ground with an initial speed of 30.0 m/s.<br /><ul><li>(2 points) What is the initial horizontal velocity, vix of the projectile?
26. 26. (2 points) What is the initial vertical velocity, viy of the projectile?
27. 27. (3 points) How far does the projectile travel horizontally (range), x before it hits the ground?
28. 28. (3 points) How long (time) does it take the projectile to reach the highest point in its trajectory?</li></ul>Illustration:<br />Given:vi=30.0 m/s- initial velocity magnitude of the projectile<br />θ=60o- angle of projection<br /><ul><li>Find: vix=?- initial horizontal velocity of the projectile
29. 29. Derivation:vix=vicosθ
30. 30. Substitution:vix=30.0 m/scos60ovix=15.0 m/s
31. 31. Find: viy=?- initial vertical velocity of the projectile
32. 32. Derivation:viy=visinθ
33. 33. Substitution:vix=30.0 m/ssin60ovix=26.0 m/s
34. 34. Find: x=?- range
35. 35. Derivation:x=vixt=(vicosθ)t where vy-viy=at vy=0 at the highest point
36. 36. So the total time is: t=-2visinθa Therefore: x=-2vi2cosθsinθa
37. 37. Substitution:x=-230.0 m/s2cos60osin60o-9.8 m/s2x=79.5 m
38. 38. Find: t=?- time to reach the highest point of the trajectory
39. 39. Derivation:vy-viy=at vy=0 at the highest point
40. 40. So the time is: t=-visinθa
41. 41. Substitution:t=-30.0 m/ssin60o-9.8 m/s2t=2.65 s</li></ul>Problems 05 to 08 pertain to the statement below:<br />A shell is fired horizontally in the positive x direction from the top of an 80.0 m high cliff. The shell strikes the ground 1330 m from the base of the cliff.<br /><ul><li>(3 points) Determine the initial velocity of the shell.
42. 42. (2 points) What is horizontal velocity of the shell as it hits the ground?
43. 43. (2 points) What is vertical velocity of the shell as it hits the ground?
44. 44. (3 points) What is the velocity (magnitude and direction) of the shell as it hits the ground? Hint: use your answers in 06 and 07 and apply vector components</li></ul>Illustration:<br />Given:y=80.0 m- height of the cliff<br />x=1330 m- range<br /><ul><li>Find: vi=?- initial velocity of the shell
45. 45. Derivation:vi=vix=xt where y=viyt+12at2 initially, viy=0 m/s
46. 46. So, t=2ya Therefore, vi=x2yavi=x a2y
47. 47. Substitution:vi=1330 m-9.8 m/s22-80.0 mvi=329 m/s
48. 48. Find: vx=?- horizontal velocity of the shell as it hits the ground
49. 49. Derivation:horizontal velocity is constant. Therefore, vx=vix=vi
50. 50. Substitution:vx=329 m/s
51. 51. Find: vy=?- vertical velocity of the shell as it hits the ground
52. 52. Derivation:vy2-viy2=2ay initially, viy=0 m/s so, vy=-2ay
53. 53. Substitution:vy=-2-9.8 m/s2-80.0 mvy=-39.6 m/s
54. 54. Find: v=?- velocity (magnitude and direction) as it hits the ground
55. 55. Derivation:v=vx2+vy2 so from the equations from 05 and 07:v=x2a2y+4ay
56. 56. Simplifying, v=ax2+8y22y tanθ=vyvx=-2ayx a2y=-4y2x=-2yx θ=tan-1-2yx
57. 57. Substitution:v=-9.8 m/s21330 m2+8-80.0 m22-80.0 mv=334 m/s</li></ul>θ=tan-1-280.0 m1330 mθ=-7o=7o below horizontal<br />Problems 09 to 12 pertain to the statement below:<br />A football is kicked with an initial speed vi at an angle θ above the horizontal ground. It covers a horizontal distance/range of 26 m in 3.4 s.<br /><ul><li>(2 points) What is horizontal velocity, vix of the football?
58. 58. (2 points) What is the initial vertical velocity, viy of the football?
59. 59. (3 points) What is initial speed (magnitude of the velocity) of the football?
60. 60. (3 points) What is the angle of trajectory, θ? What is the other angle that can give the same range of the football if it is kicked at the same speed?</li></ul>Illustration:<br />Given:x=26 m- range reached by the football<br />t=3.4 s- time to cover the range or time of flight<br /><ul><li>Find: vix=?- initial horizontal velocity of the football
61. 61. Derivation:vix=xt
62. 62. Substitution:vix=26 m3.4 svix=7.6 m/s
63. 63. Find: viy=?- initial vertical velocity of the football
64. 64. Derivation:y=viyt+12at2back to initial level, y=0 m so, 0=viyt+12at2
65. 65. Therefore, viy=-12at
66. 66. Substitution:viy=-12-9.8 m/s23.4 sviy=17 m/s
67. 67. Find: vi=?- initial speed of the football
68. 68. Derivation:vi=vix2+viy2 so from the equations from 09 and 10:vi=xt2+-12at2
69. 69. Simplifying, vi=12t4x2+a2t4
70. 70. Substitution:vi=123.4 s426 m2+9.8 m/s223.4 s4vi=18 m/s
71. 71. Find: θ=?- angle of projection
72. 72. Derivation:tanθ=viyvix=-12atxt=-at22x therefore, θ=tan-1-at22x
73. 73. Substitution:θ=tan-1--9.8 m/s23.4 s2226 mθ=65o=65o above horizontal