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Ch. 3 Vectors & Projectile Motion

Ch. 3 Vectors & Projectile Motion. Scalar Quantity. Described by magnitude only Quantity Examples: time, amount, speed, pressure, temperature. Vector Quantity. Describe magnitude AND direction Examples: velocity, force, acceleration, resistance

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Ch. 3 Vectors & Projectile Motion

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  1. Ch. 3 Vectors & Projectile Motion

  2. Scalar Quantity • Described by magnitude only • Quantity • Examples: time, amount, speed, pressure, temperature

  3. Vector Quantity • Describe magnitude AND direction • Examples: velocity, force, acceleration, resistance • Vector quantities can be represented as “vectors” in physics diagrams • Arrow: points in direction of vector and specifies the magnitude on top of the arrow 10 m/s

  4. Calculations using vectors • Can be added or subtractedIF they are in the same plane • Example: A boat travelling 10 m/s west encounters a down stream current, 5 m/s west. What will be the boats velocity travelling with this current?

  5. Calculations using vectors • Can be added or subtracted if they are in the same plane • Example: A boat travelling 10 m/s west encounters a down stream current, 5 m/s west. What will be the boats velocity travelling with this current? 10 m/s + 5 m/s = 15 m/s

  6. Journal #1 A plain travelling 50 m/s north encounters a head wind 10 m/s south. What is the resulting velocity of the plane?

  7. Journal #1 A plain travelling 50 m/s north encounters a head wind 10 m/s south. What is the resulting velocity of the plane? 50 m/s – 10 m/s = 40 m/s

  8. Journal #2: With the person sitting next to you….. 1. List what you remember about solving for the sides of a right triangle. Write down as much as you know, be specific!!!

  9. Vectors in different planes!!!! • Consider a boattravelling North and a current in the water that is moving towards the east…. • We will solve this using Trigonometry!

  10. Vectors in different planes!!!! • Consider a boattravelling North and a current in the water that is moving towards the east…. • We will solve this using Trigonometry!

  11. Vectors in different planes!!!! • Consider a boattravelling North and a current in the water that is moving towards the east…. • We will solve this using Trigonometry! this is the resultant (the path that the boat will take)

  12. a2+b2 = c2 a c b

  13. Important Concepts SOH CAH TOA Sin θ = opp Cos θ = adj Tan θ = opp hyp hyp adj Opposite Adjacent Hypotenuse θ

  14. Lets try a problem!!! • A boat travelling with a velocity of 5 m/s North, encounters a current 2 m/s to the west. • What is the resulting velocity? • Draw a vector diagram • Fill in the knowns • Solve for the Resultant

  15. Lets try a problem!!! • A boat travelling with a velocity of 5 m/s north, encounters a current 2 m/s to the west. • What is the resulting velocity? 2 m/s • Draw a vector diagram 5 m/s

  16. Lets try a problem!!! • A boat travelling with a velocity of 5 m/s north, encounters a current 2 m/s to the west. • What is the resulting velocity? 2 m/s • Draw a vector diagram 5 m/s X

  17. Lets try a problem!!! • A boat travelling with a velocity of 5 m/s north, encounters a current 2 m/s to the west. • What is the resulting velocity? 2 m/s ( 2m/s)2 + (5 m/s)2 = X2 5 m/s X

  18. Lets try a problem!!! • A boat travelling with a velocity of 5 m/s north, encounters a current 2 m/s to the west. • What is the resulting velocity? 2 m/s ( 2m/s)2 + (5 m/s)2 = X2 4 + 25 = X2 5 m/s X √(29) = √ (X2) 5.39 = x

  19. Lets try a problem on our own…Journal # 3 • A boat travelling at 9m/s south, encounters a current 16 m/s to the east. • What is the resulting velocity and angle of displacement for the boat?

  20. Lets try a problem on our own…Journal # 3 • A boat travelling at 9m/s south, encounters a current 16 m/s to the east. • What is the resulting velocity and angle of displacement for the boat? 92 + 162 = X2  81 + 256 = X2 9 m/s 337 =  X2 16 m/s X = 18.35

  21. Lets try a problem on our own…Journal # 3 • A boat travelling at 9m/s south, encounters a current 16 m/s to the east. • What is the resulting velocity and angle of displacement for the boat? tan  = opp  adj 9 m/s tan  = 16 16 m/s 9  = tan-1 (16/9) = 60

  22. Components of Vectors • Any vector can be “resolved” into its components. • Its X and Y plane

  23. Components of Vectors • Any vector can be “resolved” into its components. • Its X and Y plane • How??? By creating a right triangle!!!

  24. Components of Vectors • Any vector can be “resolved” into its components. • Its X and Y plane • How??? By creating a right triangle!!!

  25. Example…. What are the horizontal and vertical components of a 6.4 unit vector that is oriented 37° above the horizontal?

  26. What are the horizontal and vertical components of a 6.4 unit vector that is oriented 37° above the horizontal? 6.4 37 °

  27. What are the horizontal and vertical components of a 6.4 unit vector that is oriented 37° above the horizontal? Solve for X and Y Y X 6.4 37 °

  28. What are the horizontal and vertical components of a 6.4 unit vector that is oriented 37° above the horizontal? Solve for sin 37 = Y/6.4 cos 37 = X/6.4 X and Y 6.4 sin 37 = Y 6.4 cos 37 = X Y Y = 3.85 X = 5.11 X 6.4 37 °

  29. Projectile Motion

  30. What Forces are acting on a Projectile? • Initial Force that caused motion • Force of gravity

  31. Gravity causes the object to curve downward in a parabolic path (trajectory)

  32. An Object’s motion can be broken down into it’s horizontal and vertical component vectors. • x and y vectors

  33. Important Rule: Horizontal motion does NOT affect vertical motion!!!!

  34. Vx is constant and there is 0 acceleration! Vy is changing and acceleration is due to gravity.

  35. At the top of a path, • there is no y velocity component • Vx component only!!!

  36. Projectile Problem Solving • Problems in which an object was dropped with a force in the x- axis V0 dy dx

  37. Projectile Problem Solving • Problems in which an object was dropped with a force in the x- axis From Free Fall dy = ½ gt2 V0 dy dx

  38. Projectile Problem Solving • Problems in which an object was dropped with a force in the x- axis From Free Fall dy = ½ gt2 From Linear Motion dx= v0t and that v0= dx / t V0 dy dx

  39. Ex. 1 An object travelling at 50 m/s falls out of a plane. It hits the ground 10 s later. What is the horizontal distance travelled by the object?

  40. Ex. 1 An object travelling at 50 m/s falls out of a plane. It hits the ground 10 s later. What is the horizontal distance travelled by the object? v0 = 50 m/s t = 10 s dx= ? Solving for the x-axis vector component dx= v0t

  41. Ex. 1 An object travelling at 50 m/s falls out of a plane. It hits the ground 10 s later. What is the horizontal distance travelled by the object? Solving for the x-axis vector component dx= v0t dx= (50 m/s)(10s) dx= 500 m ** Ch. 3 problem 41 in HW

  42. Ex. 2 You want to launch an object from the balcony into a pool. You are 50 m high and the pool is 10 m from the bottom of the building. How fast must the object initially launch in order to reach the pool?

  43. Ex. 2 You want to launch an object from the balcony into a pool. You are 50 m high and the pool is 10 m from the bottom of the building. How fast must the object initially launch in order to reach the pool? dy = 50 m dx= 10 m v0 is unknown v0= dx / t But wait, t is unknown…..

  44. Ex. 2 You want to launch an object from the balcony into a pool. You are 50 m high and the pool is 10 m from the bottom of the building. How fast must the object initially launch in order to reach the pool? • dy = 50 m • dx= 10 m • v0 is unknown v0= dx / t But wait, t is unknown….. And we can solve for it using dy = ½ gt2

  45. Ex. 2 You want to launch an object from the balcony into a pool. You are 50 m high and the pool is 10 m from the bottom of the building. How fast must the object initially launch in order to reach the pool? • dy = 50 m • dx= 10 m • v0 is unknown v0= dx / t dy = ½ gt2 OKAY – Solve for time

  46. Ex. 2 You want to launch an object from the balcony into a pool. You are 50 m high and the pool is 10 m from the bottom of the building. How fast must the object initially launch in order to reach the pool? • dy = 50 m • dx= 10 m • v0 is unknown v0= dx / t dy = ½ gt2 OKAY – Solve for time 50 m = ½ (10 m/s2) (t2) 10 s2 = t2 take the square root of both sides t = 3.16 s

  47. Ex. 2 You want to launch an object from the balcony into a pool. You are 50 m high and the pool is 10 m from the bottom of the building. How fast must the object initially launch in order to reach the pool? • dy = 50 m • dx= 10 m • v0 is unknown v0= dx / t Knowing that t = 3.16 s, we can now solve for V0. v0= dx / t = 10 m / 3.16 s = 3.16 m/s ** problems 42 and 44 in the hw

  48. Ex.3 You aim an arrow directly at a target that is 0.2 seconds away. How far below the target does the arrow hit?

  49. Ex.3 You aim an arrow directly at a target that is 0.2 seconds away. How far below the target does the arrow hit? The only known given is time and we are determining the distance in the vertical direction dy. Which equation should we use????

  50. Ex.3 You aim an arrow directly at a target that is 0.2 seconds away. How far below the target does the arrow hit? • The only known given is time and we are determining the distance in the vertical direction dy. • Which equation should we use???? • dy = ½ gt2

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