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
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
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.
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)
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
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?
38.
39.
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
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
