1. Vectors add direction to a magnitude.
Vectors can be used to represent a force, an acceleration, a velocity, etc.
Adding two direction vectors together gives a RESULTANT VECTOR.
The resultant gives a distance and direction to the target.

2. Adding vectors
To add vectors put the components together head to tail and draw the resultant from the start point to the end point. See the board for this.

3. Subtracting vectors
To subtract vectors, put their feet together and draw the resultant. OR you can reverse the direction of the second vector and use vector addition.

4. Suppose you have walked 5km at a 45˚ angle
Suppose you have walked 5km at a 45˚ angle. How far have you gone along both the x and y axis.

5. Home fun
Pg 87 q 1-5
Pg 91 q 1-5

6. Chapter 3 section 3 PROJECTILE MOTION

7. Let's say Wile E. Coyote runs off a cliff.
What happens in the cartoon?
What happens in real life?
As he leaves the cliff he has a horizontal velocity.
The only way to change that horizontal velocity is to exert an horizontal force (acceleration) on the coyote.
Since there is no horizontal acceleration, the coyote will travel horizontally at the same speed the whole time!

8. What is happening vertically
What is happening vertically? As soon as the coyote leaves the cliff he will experience a vertical force due to gravity. This force will cause him to start to accelerate in the vertical direction. As he falls he will be going faster and faster in the vertical direction.

9. The horizontal and vertical components of the motion of an object going off a cliff are separate from each other, and can not affect each other.

10. Example 1: A ball is thrown off the edge of a 15
Example 1: A ball is thrown off the edge of a 15.0m tall cliff horizontally at 8.0m/s. A. Determine how much time it takes to fall.

11. THINK VERTICAL -It would take the exact same amount of time for the object to hit the ground if it was just dropped straight down from the edge of the cliff , so just calculate the time to fall that way.
∆y = vit + 1/2 at2 Initial vertical velocity is zero so... ∆y = 1/2 at2 rearrange t = √(2 ∆y /a)
t = 1.75 s

12. b. Determine how far from the base of the cliff it hits the ground
b. Determine how far from the base of the cliff it hits the ground. THINK HORIZONTAL It was in the air for 1.75s (from the previous question), and it was moving at a constant speed of 8.0m/s in the x-direction the whole time.

13. v = ∆x /t rearrange ∆x = v t ∆x = 14m

14. c. Determine how fast it is moving vertically when it hits the ground
c. Determine how fast it is moving vertically when it hits the ground. THINK VERTICAL It has been accelerating down the whole time. We know that gravity is causing this acceleration, and that it wasn’t moving vertically at the start, so we can figure out how fast it is going (vertically) when it hits the ground.

15. vf2 = vi2 + 2ad vf = √(vi2 + 2ad) vf = 17 m/s

16. d. Determine what its total velocity is when it hits the ground
d. Determine what its total velocity is when it hits the ground. It’s total velocity is found by adding the horizontal and final vertical components of the velocity to find the resultant. c2 = a2 + b2 tanΘ = opp/adj

17. The object is moving at 19m/s at an angle of 25° below the horizontal when it hits the ground.

18. Some handy equations when a projectile falls from rest
vyf = -g ∆t
vyf2 = -2g ∆y
∆y = ½ g ∆t2
∆x = vx ∆t

19. Home fun pg 102 q 1-4 pg 114 q 30, 33, 34, 36, 37

20. Projectiles Launched at an Angle

21. There are two ways to do these types of problems, one based on the vertical velocity of the object, the other based on the vertical displacement.

22. The only big difference in these methods is how to calculate the time that the object spends in the air.

23. Vertical Velocity Method Imagine for a moment that you are watching an object as it rises into the air after you kick it upwards at an angle. When it left your foot, it was going the fastest that it can possibly move during its flight.

24. The instant it leaves your foot, a force begins to effect it
The instant it leaves your foot, a force begins to effect it. What is this force?
Gravity is pulling down on it, causing it to have less and less vertical velocity.
Remember that there will be no change in the horizontal component of its velocity.

25. When it reaches the highest point in its flight… it isn’t moving up… it isn’t moving down… for an instant of time… its vertical velocity is ZERO!

27. By the time it reaches the ground again, it will still be moving with its original horizontal velocity and will have the exact same vertical velocity as it left your foot with!

28. We can use this information about its vertical movement to make some calculations.

29. We know...
gravity (-9.81m/s2) causes the acceleration on the object vertically.
the initial vertical velocity of the object.
the final vertical velocity of the object.
We can even use this two ways, since we can say that the final vertical velocity happens at the halfway point (zero m/s), or when it gets back to the ground (same as it left the ground at).

30. This gives us enough information to calculate the maximum height of the flight, and the time it spends in the air. After that, we can calculate just about anything…

31. Example 1: You kick a soccer ball at an angle of 40° above the ground with a velocity of 20m/s. Now we can answer these questions a. How high will it go? b. How much time does it spend in the air? c. How far away from you will it hit the ground (what is the range of its flight)? d. What is the ball’s velocity when it hits the ground?

32. Before we can calculate anything we need to break the original velocity into components.
We do this so we have a vertical component to do the first couple calculations with. The horizontal component will be used later.

33. How high will it go? sinΘ = opp/hyp rearrange opp = sinΘ (hyp) = sin 40° (20m/s) opp = vy = 13m/s

34. At its maximum height, halfway through its flight, the object won't be going up or down, so its final velocity at that point is zero.
vf2 = vi2 + 2ad d = (vf2 - vi2) / 2a = ( ) / 2(-9.81) d = 8.6 m
The ball will reach a maximum height of 8.6 m.

35. b. How much time does it spend in the air?
THINK VERTICAL!
a = (vf - vi) / t t = (vf - vi) / a = (0 - 13) / -9.81
t = 1.33 s
But this is only the time to the halfway point, so the final answer is 2.66s. (2.7s with s.f.)

36. c. How far away from you will it hit the ground (what is the range of its flight)?
THINK HORIZONTAL
It is moving at a constant velocity horizontally during the whole time.
cosΘ = adj/hyp adj = cos Θ (hyp) = cos 40° (20m/s) adj = vx = 15m/s

37. Now we can use the equation for velocity
v = ∆x /t rearrange
∆x = vt = 15m/s (2.7s) ∆x = 41 m

38. d. What is the ball’s velocity when it hits the ground?
Simple enough…The ball’s velocity when it hits the ground is exactly the same as when it was originally launched… 20 m/s at 40° up from the horizontal.

39. Vertical Displacement Method
To do the problem this way, we must assume that the object strikes the ground at exactly the same height that it was launched from.
This is the case for most questions.

40. When you look at how the object has moved, it has a total vertical displacement of zero.

41. Redo part (b) of the previous example
Redo part (b) of the previous example. The other parts should still be done the same way.

42. b. How much time does it spend in the air?
Use the vertical velocity we calculated (13m). The vertical displacement is zero (from start to end), so…
d = vit + 1/2 at2 0 = vit + 1/2 at2 -vit = 1/2 at2 ….- divide both sides by "t" -vi = 1/2 at ….- solve for "t" t = -2vi / a = -2(13) / t = s (2.7s)

43. Some handy equations for projectiles launched at an angle
vx = vi (cos Θ)
∆x = vi (cos Θ) ∆t
vyf = vi (sin Θ) –g ∆t
vyf2 = vi2 (sin Θ)2 – 2g ∆y
∆y = vi (sin Θ) ∆t – ½ g ∆t2
vi2 = g ∆x/ (sin Θ)(cos Θ)

44. Home fun
Pg 104 q 1-5
Pg 115 q 35, 38, 39, 41

45. Chapter 3 Section 4 Relative Motion
Relative motion is just a way of saying that sometimes different people will say different things about the motion of the same object.

46. Motion is relative- All motion is relative to the observer or to some fixed object.
When you see a car drive by at 25m/s, it is moving with respect to you.
If you are in a car that is going at the same speed, in the same direction as the other car, then the other car will not be moving with respect to you.
If you are standing on the side of the road the car will be moving at 25 m/s.
If you are going 25m/s in the opposite direction…how fast does it appear that the other car is moving?

47. Point of reference
In talking about motion, it is important to indicate your point of reference. In the case of moving automobiles, it is usually assumed the speed is with respect to the ground. But there are situations where the speed or velocity may be with respect to another object or an observer.
For example, suppose a car was traveling at 60 miles per hour (mph) and hit another car. The second car is traveling in the same direction at 59 mph. What will the result be?

48. Velocity measurements differ in different frames of reference