Physics 2 LT3: Projectile Motion Solutions
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  • 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 />
    • True or False
    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 />
    • A projectile fired at any angle above the ground has a constant horizontal velocity in all points.
    • 2. The time in going up of a projectile is the same to its time in going down back to its initial level.
    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 />
    • The vertical motion of a projectile is a uniformly accelerated motion.
    • 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.
    • 8. PROBLEM SOLVING
    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 />
    • (2 points) What is the initial horizontal velocity, vix of the projectile?
    • 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?
    • 11. (3 points) How long (time) does it take the projectile to reach the highest point in its trajectory?
    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 />
    • (3 points) Determine the initial velocity of the shell.
    • 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?
    • 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
    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 />
    • (2 points) What is horizontal velocity, vix of the football?
    • 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?
    • 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?
    <br />GOD Bless!!! John 3:16<br />Tract: dtquinsaat_07/31/2010/1832h <br />Answers and Solutions<br />
    • True or False
    • 18. A projectile fired at any angle above the ground has a constant horizontal velocity in all points. – T
    • 19. The time in going up of a projectile is the same to its time in going down back to its initial level. – T
    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 />
    • The vertical motion of a projectile is a uniformly accelerated motion. – 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
    • 25. PROBLEM SOLVING
    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 />
    • (2 points) What is the initial horizontal velocity, vix of the projectile?
    • 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?
    • 28. (3 points) How long (time) does it take the projectile to reach the highest point in its trajectory?
    Illustration:<br />Given:vi=30.0 m/s- initial velocity magnitude of the projectile<br />θ=60o- angle of projection<br />
    • Find: vix=?- initial horizontal velocity of the projectile
    • 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
    • 41. Substitution:t=-30.0 m/ssin60o-9.8 m/s2t=2.65 s
    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 />
    • (3 points) Determine the initial velocity of the shell.
    • 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?
    • 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
    Illustration:<br />Given:y=80.0 m- height of the cliff<br />x=1330 m- range<br />
    • Find: vi=?- initial velocity of the shell
    • 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
    • 57. Substitution:v=-9.8 m/s21330 m2+8-80.0 m22-80.0 mv=334 m/s
    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 />
    • (2 points) What is horizontal velocity, vix of the football?
    • 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?
    • 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?
    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 />
    • Find: vix=?- initial horizontal 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