Vectors and projectile motion and worked examples
- 1. Stage 2 Physics Section 1 Motion in Two Dimensions
- 2. What is motion? What types of motion are there? What causes motion? How to we describe motion in Physics?
- 3. What is a vector? Vector quantities have magnitude (size) and direction. Scalar quantities have magnitude only. Length represents the vectors magnitude.
- 4. Scalar Vector Distance Displacement Speed Velocity Mass Acceleration Force Time weight
- 5. Velocity vector of an angry bird
- 6. Resultant of Two Vectors The resultant is the sum or the combined addition of two vector quantities Vectors in the same direction: 6N 4N = 10 N Vectors in opposite directions: 6m 10 m =
- 7. Vectors • Vectors can be oriented to the gravitational field (up, down or some angle to the horizontal) or compass points (NESW). 5 ms-1 5 ms-1 30o above the horizontal
- 8. Velocity Vectors • Velocity can be resolved into its horizontal and vertical components at any instant. v vV vH
- 9. SOHCAHTOA Hypotenuse Opposite Adjacent
- 10. v vV = v sin vH = v cos
- 12. Example 1 Resolve the following velocity vector into its horizontal and vertical components 30o
- 13. Example 1 Resolve the following velocity vector into its horizontal and vertical components 30o
- 14. This problem can be solved in two ways (and you need to be able to do both) 1. Scale Diagram 2. Trigonometry
- 15. 1. Scale diagram By drawing a vector diagram (using a protractor and a ruler) to scale we can simply measure the size of the components ideally the vector should be 10 cm or larger (for accuracy)
- 16. 2. Trigonometry vvertical = v sin = 40 sin 30o = 20 m s-1 vhorizontal = v cos 30o = 40 cos 30o = 34.6 m s-1
- 17. Example 2 Determine the velocity vector with initial horizontal velocity component of 50 ms –1 and vertical 20 ms-1.
- 18. v=? vv= 20 m s-1 vh= 50 m s- 1
- 19. 1. Scale diagram Again we could accurately draw the figure and measure the resultant length and angle to find the direction of v. (Note: You need to have a clear Perspex ruler and a protractor for EVERY test and exam)
- 20. 2. Pythagoras & Trigonometry By Pythagoras theory: v2 = vV2 + vH2 v = v V 2 + vH 2 = 202 + 502 = 2900 = 53.9 m s-1 v=? vv= 20 m s-1 vh= 50 m s- 1
- 21. tan vV/vH tan = 20/50 = 21.8o ie. v = 53.9 m s-1 at 21.8o above the horizontal
- 22. Summary – Vectors in 2D • Given any vector quantity in 2D it can be resolved into horizontal and vertical components eg displacement, force, fields etc • Given the horizontal & vertical components you can determine magnitude and direction of the vector (formula)
- 23. Motion in a Uniform Gravitational Field
- 24. In the absence of gravity objects move with constant velocity in a straight line. An object will remain at rest, or continue to move at a constant velocity, unless a net force acts on it. Note: The following is all in the absence of air resistance.
- 25. When an object falls under the influence of gravity, the vertical force causes a constant acceleration
- 26. vH The resultant motion is a combination of both horizontal and vH vertical vV vV components
- 27. Horizontal Projection While the vertical component undergoes constant If an object is acceleration. projected horizontally, the horizontal component moves with constant velocity.
- 28. Three equations of motion Note that all the equations have “a” in them – they only apply under CONSTANT acceleration
- 29. Constant vertical acceleration Vertical formulae 2 v 2 v0 2as 1 2 Horizontal velocity is constant s vt at 2 Horizontal formula v vo at s vH t
- 30. Three equations of motion
- 31. Learning Symbols in Physics Quantity Quantity Symbol Units Unit symbol
- 33. Example A stone is dropped down a well and takes 3 seconds to hit the ground. a) How fast does it hit the bottom? b) How deep is the well?
- 34. A stone is dropped down a well at takes 3 seconds to hit the ground. a) How fast does it hit the bottom?
- 35. b) How deep is the well?
- 36. An arrow is fired upwards at 50ms-1. a) How high does the arrow fly?
- 37. An arrow is fired upwards at 50ms-1. b) How long does the arrow take to hit the ground?
- 40. Projectile Motion Problems Except for time, everything can be separated into horizontal and vertical components and treated separately. sV = height V0 s H = range t = time of flight
- 41. Projectile Motion Problems Horizontal projection: down is +ve Uni-level projection: Up is considered positive, and down is negative.(Acceleration due to gravity aV = -9.8ms-2) sV = height V0 s H = range t = time of flight
- 42. Projectile Motion Problems At the top of the parabolic path, vV= 0 ms-1 1 vV 0ms 2 aV 9.8ms sV height V0 sH range t = time of flight
- 43. Projectile Motion Problems Remember the time of flight is the time it takes to go up+ down. 1 vV 0ms 2 aV 9.8ms sV height V0 sH range t = time of flight
- 44. Example 1
- 55. Bi-level projection • An object is projected at a height
- 58. Maximum Range To get the maximum range sH max in a vacuum (no air resistance) the launch angle must be 45o sH max
- 59. For a projectile launched at ground level find by sample calculation the launch angle that results in a maximum range
- 60. Pairs of launch angles that yield the same range add up to 90o α + θ = ranges Projectile 90o for various angles of launch 500 450 400 350 300 height 250 200 150 100 50 0 0 200 400 600 800 1000 1200 range
- 61. α + θ = 90o Find the launch angle that yields the same range as 32o θ = 32 α=? α + θ = 90o
- 62. The Effect of Air Resistance Air resistance acts in the opposite direction to motion. vertical horizontal
- 63. The Effect of Air Resistance vertical horizontal This decreases the • height • range Slight decrease in time of flight of the projectile
- 65. The magnitude of Fair resistance Fair resistance
- 66. Speed
- 68. Size More surface area = more air resistance
- 70. Air density Low air density = less air resistance High air density = high air resistance
- 71. Projectiles in Sport Consider the effect of launch height on range
- 72. As the object has further to fall tflight is increased. As the object is in the air for longer it travels farther.
- 73. 45o 41o For objects at h=0 the optimal angle is 45o For heights › 0 θ max height is less than 45o