Caution! - Marcus Gollahonrgollahon.weebly.com/uploads/1/6/3/6/16361290/twodimensionalmo… ·...
Transcript of Caution! - Marcus Gollahonrgollahon.weebly.com/uploads/1/6/3/6/16361290/twodimensionalmo… ·...
Name: _____________________
Period: _____________________
Copyright © 2010, C. Stephen Murray cstephenmurray.com
Trigonometry Basics
Calculators and Trig You will have to find the sin, cos, or tan ratios OR you will have to find θ,
given the ratios. You must know how to use your calculator to do this.
Given the angle (θ), find sin, cos, or tan. Ex: sin 30° = ______
Scientific Calculator: Type “30”
Push “sin”
Answer: “0.5”
: sin 30° = 0.5Answer
Finding Unknowns The real power of trigonometry is that it relates the angles and sides of a right triangle.
If, for example, you know θ and a side, you can find all the other parts of the triangle.
Step 3: Solve
25º
60 cm Y
X
Problem: Find the
length of X.
Variables:
θ = 25º
opp. = Y
adj. = X
hyp.= 60 cm
Step 1:
Assign Variables
Step 2:
Choose a Formula
You know hyp and
need adj., so use cos.
adj.cosθ =
hyp.
adj.cosθ =
hyp.
Xcos25° =
60cm
X.9063 =
60cm
(60).9063 = X
X = 54.4cm
Sin (Sine) − ratio of the opposite side
to the hypotenuse.
Cos (Cosine) − ratio of the adjacent
side to the hypotenuse.
Tan (Tangent) − ratio of the opposite
side to the adjacent side.
Trigonometric Functions
oppSin θ =
hyp
adjCos θ =
hyp
oppTan θ =
adj
30º
60º 16 m
8 m
13.9 m
opp. 13.9mSin60° = = = 0.866
hyp. 16m
opp. 8mSin30° = = = 0.5
hyp. 16m
opp. 8mTan30° = = = .5774
adj. 13.9m
Trigonometric Ratios
for This Triangle
Sin, cos, and tan are ratios, telling you how big
(what percentage) one side is in relation to another.
Scientific Calculator: Press the DRG key
until degrees shows
in the display.
Recheck!
DRG MODE
Graphing Calculator: Push “sin”
Type “30”
Answer: “0.5”
Graphing Calculator: Press the MODE key. Find
where RADIANS is selected.
Select DEGREES instead.
Press ENTER. Recheck!
: θ = 25Answer °
Given sin, cos, or tan, find θ.
Scientific Calculator: Type “.4226”
Push “INV” then “sin”
Gives “25”
Graphing Calculator: Push “2nd” then “sin”
Type “.4226”
Gives “25”
-1
Ex: sinθ = .4226
θ = sin (.4226)
θ = ____
Use Degrees not Radians! When using sin, cos,
and tan your calculator MUST be in degrees or all
of your numbers will be wrong.
Quick Check: The sin 30º = 0.5!
If sin 30º ≠ 0.5, your calculator in in radians
and must be changed.
Caution!
Basic Terms
θ (theta) − variable for any angle.
Hypotenuse − longest side of a triangle.
Opposite − side opposite the angle (θ).
Adjacent − side next to the angle (θ).
θ1
Y
X
θ2 R
Y is opposite to θ1;
Y is adjacent to θ2.
X is adjacent to θ1; X is opposite to θ2.
R is the hypotenuse
for both θ1 and θ2. Which side is opposite?
It depends on the angle.
a
b
c
Pythagorean Theorem
Remember that
a2 + b2 = c2
Where a and b are
either of the sides and
c is the hypotenuse.
Name: _____________________
Period: _____________________
Copyright © 2010, C. Stephen Murray cstephenmurray.com
Using your calculator, give the following ratios.
Given the following ratios, use your calculator to find θ.
Cos θ = .8192;
θ = _________
Sin θ = 0.5
θ = _________
Tan θ = .8391
θ = _________
Sin θ = .866
θ = __________
Tan θ = 1.732
θ = __________
Cos θ = 0.5
θ = __________
Tan θ = 1
θ = __________
Cos θ = .866
θ = __________
Sin θ = .7071
θ = __________
Cos 30º = ___________
Tan 15º = ___________
Sin 60º = ___________
Tan 30º = ___________
Sin 45º = ___________
Tan 85º = __________
Cos 45º = __________
Cos 60º = __________
Sin 30º = ___________
Tan 45º = __________
40º
15 cm 9.6 cm
11.5 cm
θ = _________________
Opposite = ___________
Adjacent = ___________
Hypotenuse = ________
75º 8 in 2
in
7.7 in
θ = _________________
Opposite = __________
Adjacent = __________
Hypotenuse = ________
Opposite for 30º = ________
Hypotenuse for 60º = ______
Adjacent for 60º = ________
Hypotenuse for 30º = ______
30º
60º 20 m 10 m
17.3 cm
If opposite = 17.3 cm, then θ = __________ If adjacent = 10 m, then θ = _____________ If adjacent = 17.3 cm, then θ = __________
70º
20º
120 mm 10 mm
17.3 mm
Adjacent for 20º = ________
Hypotenuse for 70º = ______
Opposite for 70º = ________
Hypotenuse for 20º = ______
Adjacent for 70º = ________
Opposite for 20º = ________
If opposite = 17.3 cm, then θ = ________
30º
10 cm Y
Obviously Y ≠ 9.88 cm.
What went wrong?
Ysin30° =
10cm
Y.988 =
10cm
Y = 9.88cm
35º
Y 26 cm
Step 1: Assign Variables
26 cm = _____________
35º = ________________
Y = _________________
Step 2: Choose a Formula (Sin, Cos, or Tan?)
Step 3: Solve
Y = __________
45º
6 ft
X
Following the three steps at the left,
find the length of X.
How long is Y?
.8660 35º
What is the sin of 75º?
If tan θ = 0.868, solve for θ.
12 cm
? 13 cm
Two sides of this triangle are
given. Calculate the third side.
Trigonometry Basics-2
Name: _____________________
Period: _____________________
Copyright © 2010, C. Stephen Murray cstephenmurray.com
Name: _____________________
Period: _____________________
Copyright © 2010, C. Stephen Murray cstephenmurray.com
Vector Basics
We use arrows to represent
vectors because vectors
have both magnitude (size)
and direction (which way
it points).
When adding vectors graphi-
cally, put the arrows head to
tail. The resultant points
from start to finish. In this
example your total displace-
ment is the straight line
distance between your initial
and final position NOT the
distance you traveled.
The components are the portions of the vector in the x or y direction, like coordinates on a graph.
Components retain
the units of their vector (and
vice-versa).
vector =
6 m/s
30o
x – component
= 6m/s(cos 30o)
= 5.2 m/s
y – component
= 6m/s(sin 30o)
= 3 m/s
If the vector was a plane, think of the x-component as a race
car trying to stay beneath the plane on the ground. The
y-component could be how fast the plane gains altitude.
start
final
Resultant
Total
displacement
Vector 1
Displacement 1
Vector 2
Displacement 2
The result of adding together two
or more vectors is called a resultant.
Order doesn’t matter when adding
vectors. The resultant will be the same.
V1 V2
Different order: same resultant.
start
finish
V1
V2 R1+2
V1
V2 R1+2
start
finish
o
o
ysin 30 =
40 m
y = (40m)sin 30
(40m)0.5 =y = 20 m
The components tell
you that you would
have to move 34.6 m
in the x-direction and
20 m in the y-direction
to move 40 m at 30º
o
o
xcos 30 =
40 m
x = (40m)cos 30
x = (40m )(.866) = 34.6 m
x – component = 34.6 m
40 m
30o
V1 + V1 = 2V1
V1 –V1
Subtracting vectors: add its opposite
(the negative of the vector).
Multiplying vectors: multiply the size of the vector.
Opposite of V1
Twice the size of V1
Adding Graphically
Components
y –
component
= 20 m
Components can be negative or zero.
50o
75 m
-x component
+y c
om
po
nent
130o 90o
+y c
om
po
nent
No x comp.
100% of
this vector
is vertical
Think directions NOT
angles. Your calculator
will give you positives and
negatives automatically IF
you give the calculator
correct directions.
Units Math and Vectors
Use 130° and your calculator
will give you the correct
+ and − components.
An example of no x-component: a cat climbs up a tree.
The cat doesn’t move horizontally, just vertically.
These angles are
equal (20º), but
point in different
directions.
0º
270º or -90º
90º
180º 20º
20º
180 + 20
= 200º
For your
calculator
only this is 20º.
20º
For your
calculator this
is -20º.
20º
90 + 20
= 110º
X-comp = Hyp.(cosθ) Y-comp = Hyp.(sinθ)
X and Y Components:
Magnitude (length) of the vector
Direction of the vector
Setting up vectors.
Before you calculate
components always
find direction of θ,
as shown above.
30º
90º
180º
-90º
0º
15m The angle is 30º, but the
direction is 150º (180−30).
This vector is 15 m at 150º
7 m/s
7 m/s
45° Same magnitude;
different directions.
4 m
6 m
Same direction;
different magnitudes.
-30º
-30º
Mass is a scalar,
which only requires
magnitude. You
don’t need the direc-
tion of the mass. 2 kg
Name: _____________________
Period: _____________________
Copyright © 2010, C. Stephen Murray cstephenmurray.com
1. Resolve
2. Magnitude
3. Resultant
4. Component
5. Direction
6. Vector
A. The portion of the vector on the
x or y axis.
B. To find the x– or y-component of
a vector.
C. The size of a vector (“35” of “35 m”).
D. Tells where a vector is pointing or the
angle of the vector.
E. What you find by adding two vectors
together.
F. Something that has magnitude and
direction.
V1
V2
R A.
V2
V1
R C.
V2
V1
R D.
V2
V1
R B.
7. In figures A—D, which
vectors are added correctly?
If wrong, why?
A.
B.
C.
D.
A. 42 m/s
35o
B.
20 m/s2
45o C.
70o
72 m
30o
20 m/s
V1 V3 V2 V4 V5 Using the vectors at the right, draw the resultants for
the following operations.
8. V2 –V5 =
9. 2V2 + V4 =
10. V3 + 2V4 − V5 =
11. 2V1 − 2V4 =
12. Add 2V1 +V4 mathematically.
13. If each of the vectors is 10º from
the closest axis, determine the
directions of each of the vectors.
θA =
θB =
θC =
θD =
θE =
θF =
θG =
θH =
15. A person walks 12 m across a room.
A. What is their horizontal component?
B. What is their vertical component?
16. Resolve this vector into its components.
17. Resolve these vectors into their components.
Vector Basics -2
0º
270º or -90º
90º
180º
B C
A
G
H E
F
D
14. Find calculator directions for the following vectors.
25º 45º
A. θ = B. θ = C. θ =
Name: _____________________
Period: _____________________
Copyright © 2010, C. Stephen Murray cstephenmurray.com
Name: _____________________
Period: _____________________
Copyright © 2010, C. Stephen Murray cstephenmurray.com
start
V1 =
V2 =
finish
1. Find the following information for vector 1.
A. How far does vector 1 move horizontally?
(This is the X-component.) X1 =
B. What is the Y-component of vector 1? Y2 =
C. How long is vector 1? (Find the magnitude of
vector 1.)
2. Resolve vector 2 into its x and y components.
(Do the same as in #1)
A. X2 =
B. Y2 =
C. Magnitude of vector 2 =
Each grid square represents 1 m. In this example, you may count squares.
3. Draw the resultant from the start to the finish
(Label it “R”).
4. Add together X1 and X2 =
(this is Xtotal)
5. Ytotal =
6. Using Xtotal and Ytotal, calculate the length of R
(with Xtotal and Ytotal you have two sides of a right triangle).
Ex 1
1. Find the following information for vector 1.
A. How far does vector 1 move horizontally?
(This is the X-component.) X1 =
B. What is the Y-component of vector 1? Y2 =
C. How long is vector 1? (Find the magnitude of
vector 1.)
2. Resolve vector 2 into its x and y components.
(Do the same as in #1)
A. X2 =
B. Y2 =
C. Magnitude of vector 2 =
Each grid square represents 1 m. In this example, you may count squares.
3. Draw the resultant from the start to the finish
(Label it “R”).
4. Add together X1 and X2 =
(this is Xtotal)
5. Ytotal =
6. Using Xtotal and Ytotal, calculate the length of R
(with Xtotal and Ytotal you have two sides of a right triangle).
start
V1 =
V2 =
finish Ex 1
Name: _____________________
Period: _____________________
Copyright © 2010, C. Stephen Murray cstephenmurray.com
start V1 =
V2 =
finish
1. A. X1 = B. Y2 =
C. Magnitude of V1 =
2. A. X2 = B. Y2 =
C. Magnitude of vector 2 =
3. Draw the resultant from the start to the finish.
4. A. Xtotal = B. Ytotal =
5. Calculate the magnitude of R.
6. Using X total and Y total, calculate the direction of R.
(see “Adding Vector” notes.)
Ex 2
start V1 =
V2 =
finish
1. A. X1 = B. Y1 =
C. Magnitude of V1 =
2. A. X2 = B. Y2 =
C. Magnitude of vector 2 =
3. Draw the resultant from the start to the finish.
4. A. Xtotal = B. Ytotal =
5. Calculate the magnitude of R.
6. Using X total and Y total, calculate the direction of R.
(see “Adding Vector” notes.)
Ex 2
Name: _____________________
Period: _____________________
Copyright © 2010, C. Stephen Murray cstephenmurray.com
Name: _____________________
Period: _____________________
Copyright © 2010, C. Stephen Murray cstephenmurray.com
Vx1 = 30.3 m
Vx2 = 25 m
Example: A person walks 35 m at 30° then 50 m
at 60°. Calculate the person’s total displacement.
Step 1. Resolve vectors into their components.
Step 3. Draw a resultant triangle
with xtotal and ytotal . Then, calculate the
resultant’s magnitude and direction.
Use the Pythagorean
Theorem to find the
resultant’s magnitude.
2 2 2
total total
2 2 2
R = x + y
R =55.3 + 60.8 = 6754.73
R = 6754.73 = 82.2 m
Use inverse tangent to find
the resultant’s direction.
total
total
-1
yopp.tan θ =
adj. x
60.8tan θ =
55.3
60.8θ = tan = 47.7
55.3
=
�
Step 2. Calculate xtotal and ytotal by adding up all x-components
and all y-components.
Be sure to keep
track of negatives!
xtotal = 55.3 m
yto
tal =
60
.8 m
= 8
2.2
m
θ = 47.7o
R can be found graphically
(by drawing and measuring)
or mathematically.
v 1 = 35 m
30o
Vy1 =
17.5 m
v 2 =
50
m
60o
Vy2 =
43.3 o
2
2
Vy = 50(sin60 )
Vy = 43.3 m
o
2
2
Vx = 50(cos60 )
Vx = 25 m
o
1
1
Vy = 35(sin30 )
Vy = 17.5 mo
1
1
Vx = 35(cos30 )
Vx = 30.3 m
o
o
ysin120 =
45 m
45(sin120 ) = y
45(.866) = y
y = 39 m
o
o
xcos120 =
45 m
45(cos120 ) = x
All angles
MUST start
at the
+ x axis
D =
45 m
120o
y =
39 m
x = -22.5 m
60o
From
+ x axis
+x
Setting Up Individual Vectors
In this example all components are positive. By taking
your angle from the +x-axis, sine and cosine will give
you positive and negative components automatically.
Answer: R = 82.2 m at 47.7º
Before you begin, be sure all of your
vectors have the same units and all
angles start at the +x-axis. Then, your
calculator will automatically calculate
any positive or negative components.
Direction =
Res
ulta
nt (t
otal
displ
acem
ent)
Res
ulta
nt (t
otal
displ
acem
ent)
= 30.3 + 25 = 55.3 m
x1
x2
xtotal = x1 + x2
yto
tal =
y1 +
y2
...
= 4
3.3
+ 1
7.5
= 6
0.8
m
y1
y2
2
2
tota
l
tota
l
x
+ y
Mag
nitu
de (R
) =
-1 total
total
yθ = tan
x
Where H is the vector,
θ is the angle from the + x axis,
x is the x component of H,
and y is the y-component of H.
These 2 equations
work for any vector
(even if the vector
is vertical
[θ = 90º or 270º]
or horizontal
[θ = 0º or 180º]) .
x = Hcosθ and y = Hsinθ
X and Y Components
45(-0.5) = x
x = -22.5 m
Using 120°
gave a –x
component!
Tan Can’t See –X Tan can’t see the difference
between a negative x and
a negative y. Tan only
gives angles between +90°
and –90.°, so it will never
give you an angle in
quadrants 2 or 3.
( )
( )
1 1
1 1
-4tan tan 2 63.4
2
Tan sees these as the same!
4tan tan 2 63.4
-2
− −
− −
= − = − °
= − = − °
If Xtotal is negative add 180 degrees
1 10.4tan 60
-6
− = °
Since Xtotal is negative:
θ = 60 + 180 = 120º
But we know that this
direction is greater than 90º.
Yto
tal =
10
.4 m
Xtotal = -6 m
θ
90º
0º Which we can
see is true.
Adding Vectors
Name: _____________________
Period: _____________________
Copyright © 2010, C. Stephen Murray cstephenmurray.com
3. Resolve the following vectors into their components.
A. C.
25º
80 m 32º
40º
12 m/s
B.
6.5 m
1. A drag racer moves
350 m down a race track.
2. A person walks
150 m north.
4. Find the resultant of the following two vectors.
5. Add these vectors together. Assume each square = 1 m.
6. A person walks 30 m north, then 50 m at 35º. Find their total displacement.
A. Draw the resultant from
the start of the first vector
to the end of the second.
Label it “R”.
B. Xtotal =
C. Ytotal =
40º
X1 = 9.2 m
Y1 =
7.7 m
X2 = −5.2 m
V2 = 6 m
30º
Y2 =
3 m
V 1 =
12 m
start
finish
D. Calculate the magnitude
(length) of R.
E. Calculate the direction
of R.
V2
V1
A. X1 =
Y1 =
B. X2 =
Y2 =
C. Xtotal =
Ytotal =
D. Draw the resultant (R).
E. Calculate R’s magnitude.
F. Calculate R’s direction.
X =
Y =
X =
Y =
A. Below draw R from the start of
V1 to the end of V2. B. Resolve v1 and v2 into their
components (Step 1 on the front)
X1 =
Y2 =
Xtotal = X2 =
Y1 = Ytotal =
Magnitude = Direction =
35º V2
= 50 m
V1 =
30
m
7. Add these vectors: V1 = 55 m at 38o and V2 = 10 m at 26 0o.
A. Draw V1, V2, and R below.
B. Resolve v1 and v2 into their
components (Step 1 on the front)
X1 =
Y2 =
X2 =
Y1 =
Xtotal =
Ytotal =
Magnitude =
Direction =
Adding Vectors -2
Name: _____________________
Period: _____________________
Copyright © 2010, C. Stephen Murray cstephenmurray.com
Name: _____________________
Period: _____________________
Copyright © 2010, C. Stephen Murray cstephenmurray.com
Ex. 1—Add these vectors: 35 m at 30° and 60 m at 160°.
Additional Adding Vector Examples
D1 = 35 m
30o
x1 = 30.3 m
y1 =
17.5 m
160o
x2 = -56.4 m
y2 =
20.5 m
D2 = 60 m
x1
y1
x2
y2
= 30.3 − 56.4 = −26.1 m
yto
tal =
y1 +
y2
...
= 2
0.5
+ 1
7.5
= 3
8 m
R
xtotal = x1 + x2
2 2 2
total total
2 2 2
R = x + y
R =26.1 + 38
R =46.1 m
total
total
-1
yopp.tan θ =
adj. x
38tan θ =
26.1
38θ = tan = 56.6
-26.1
56.6 180 124.4
=
−
−
− + =
�
R =
46.1 m
θ = 124.4°
xtotal = −26.1m
yto
tal =
38
m 2
2
total
total
x
+ y
-1 total
total
yθ = tan
x
Add 180° to
your angle be-
cause it is obvi-
ously not in the
4th quadrant.
Step 1. Resolve vectors into their components.
Step 3. Calculate R’s magnitude and direction.
Step 2. Calculate xtotal and ytotal
12
50
m/s
at
90
°
2600 m/s at 320°
R
y1 =
1250
m/s
x2 = 1992 m/s
y2 =
−1670
m/s
R
y1
x2
y2
ytotal = y1 + y2 ... = 1250 −1670
= −421 m
xtotal = x1 + x2
= 1992
total
total
-1
yopp.tan θ =
adj. x
421tan θ =
1992
421θ = tan = 12
1992
=
−
− −
�
xt = 1992m
R
2 2 2
total total
2 2 2
R = x + y
R = 1992 + 421
R =2036 m
yt = −421 m
Ex. 2—Add these vectors: 1250 m/s at 90° and 2600 m at 320°.
These additional examples are for
students to visualize the process of
adding vectors. It is assumed that
students can already calculate
components.
Step 1. Resolve vectors into their components.
Step 3. Calculate R’s magnitude and direction.
Step 2. Calculate xtotal and ytotal
θ2 = 320º
Name: _____________________
Period: _____________________
Copyright © 2010, C. Stephen Murray cstephenmurray.com
Remember that the resultant is drawn from the start of the first vector to the end of the last vector. Not all motion is possible
Relative motion describes an object’s
motion relative (in relation) to different
frames of reference.
In the picture at the right the tortoise has a
speed of 0.1 m/s relative to the truck, but has
a speed of 5.1 m/s relative to the road.
In describing relative motion you must
always give your frame of reference
(the truck or the road).
Vtortoise to truck = 0.1 m/s
Vtruck to road = 5 m/s
Vtortoise to road = 5.1 m/s
X and Y are Independent
Imagine a person walking 1m/s across a
2 m wide railroad car. If the car is at rest
the person will take 2 seconds to cross.
start
end Vperson
= 1 m/s
Vtrain = 0 m/s
2 m
If the railcar is moving, the person still takes 2 seconds to cross the car.
The velocity of the railcar is irrelevant because the velocity of the car
is in the x-direction and the person is moving in the y-direction.
The x and y directions are independent of each other!
start
end Vtrain
Vperson
Vtotal
To calculate Vtotal, use Pythagorean theorem:
Vtotal2 = Vperson
2 + Vtrain2
In the example below, a boat encounters a strong current
in a river. If the boat captain aims straight across the
river, the river will push the boat downstream.
landing River
7 m/s
Boat
5 m/s
VB
VR
VT
launch
landing
Boat 5 m/s
River 7 m/s
launch
But what if the captain wishes to land at a point
straight across the river? The boat will have to
be aimed up stream, but at what angle?
The total velocity of the boat is found by adding the
vectors, which here requires only Pythagorean theorem,
because these vectors are already horizontal and vertical .
A vector triangle shows that this is not possible
because the hypotenuse is smaller than one of the
sides. The boat is not able to go straight across.
VT
VB =
5 m/s
VR = 7 m/s
Desired
path
Relative motion can be solved graphically or mathematically, just like all other vectors. Vectors Still Add
VWind = 20 m/s
45º
V Plane =
40
m/s
A plane flying has a velocity of 40 m/s, relative
to the ground, experiences a west wind that
is moving 20 m/s, relative to the ground.
What is the velocity of
the plane relative to
the ground?
VWind = 20 m/s
V Plane =
40
m/s
V total
Mathematical Solution:
Vxwind = 20
Vxplane = 40(cos45º) = 28.3
Vxtotal = 20 + 28.3 = 48.3m/s
Vywind = 0 (all horizontal)
Vyplane = 40(sin45º) = 28.3
Vytotal = 0 + 28.3 = 28.3m/s
2 248.3 + 28.3
56 m/s
28.3θ = tan-1
48.3
θ = 30.4
total
total
V
V
=
=
°Graphically:
See notes: “Adding Vectors”
Relative Motion
Name: _____________________
Period: _____________________
Copyright © 2010, C. Stephen Murray cstephenmurray.com
1. An moving walkway at the airport has a velocity of 2 m/s to the right.
A person walks at a steady pace of 3 m/s.
A. If the person is walking to the right, what is their
velocity relative to the walkway?
B. What is their velocity relative to the ground?
C. How long would it take them to travel to the food
court, 100 m away?
D. How long would it take them to walk back if
they have to walk on the same walkway?
3 m/s
2 m/s
E. How long would it take them to walk to the food
court and back without using the walkway?
2. A toy plane’s is flying 55° going 8 m/s. If the wind is pushing with a velocity of 3 m/s at 30º,
find the total velocity and direction of plane.
3. A boat is traveling 6 m/s at an angle of −30o . The water has a current
flowing 3 m/s directly south . Find the boat’s total velocity and direction.
4. A person can swim 4 m/s. The river has a current flowing 6 m/s directly east.
A. What will be the direction and velocity of the person if they aim directly across the river (north)?
B. If the person swims at constant speed, how long does it take them to swim across the 40 m wide river?
C. How far downstream will the person drift?
D. At what direction will the person have to swim to reach a point directly across the river?
E. If the river’s current increases (is faster), will the person take more or less time to cross the river?
Relative Motion -2
Name: _____________________
Period: _____________________
Copyright © 2010, C. Stephen Murray cstephenmurray.com
Hypo. can’t be smaller than a side.
Even if they swam directly
upstream they would be
pushed downstream. They
have to be swimming faster
than the river to get straight
across. Then the hypo is big-
ger than both sides.
The x and y directions are independent of each other. The x-direction speed of
the river has no effect on a person swimming in the y-direction.
Name: _____________________
Period: _____________________
Copyright © 2010, C. Stephen Murray cstephenmurray.com
Vi
Vxi = Vicosθ
Vyi =
Visinθ Vx
Vy
In the x-direction the object is at constant speed. So ax = 0 m/s2 and S = D/T.
In the y-direction it is in
freefall. So ay = –9.8 m/s2.
θ
At the top: Vy = 0 m/s,
so V = Vx
Ground
Vyf =
–Vyi stops
The object stops in x-direction
when it hits the ground in
the y-direction, so: ty = tx
Vxi = Vicosθ
A projectile is any thrown, shot, or launched object: a rock; a bullet; a volleyball; a person jumping. All
projectiles follow a parabolic path. The distance a projectile travels in the x-direction is known as its range.
Just as with any vector, the x and y-components of
the initial velocity can be found using sine and cosine.
You know that if you throw a rock from up on a cliff, it
will go farther than if you were on the ground. This is
because the higher up you are, the longer the rock is in
the air. Flight time is dependent on the y-direction only!
3 m
6 m
An object launched
from higher up stays in the
air longer and goes farther.
Time Comes From the Y-direction
Projectile Motion
If you took several
quick pictures of a
projectile in flight you
would notice that in the
y-direction it looks just
like freefall—being
pulled back to the
earth by gravity.
In the x-direction
you would see that the
object is at constant
speed, moving the same
distance each second.
To find the projectile’s
velocity and angle at any
point: calculate Vx and Vy,
then use Pythagorean
Theorem and inverse tan.
Vyi =
V(sinθ)
Vxi = V(cos θ)
θ
Vi
(Initial horizontal velocity)
(Initial vertical
velocity)
Initial velocity
(velocity it was
thrown or shot)
Vyi =
V(sinθ)
If an object is launched
horizontally, the initial
vertical velocity is 0 m/s.
Vi = Vxi
Vyi = 0 m/s
y−direction
ay = −9.8 m/s2
Vi = 0 m/s
Vf = not used
∆y = −8 m
t = ?
Solving Projectile Motion Problems
Use your “Freefall”
notes to help you assign
the y-direction variables.
Just as in all freefall
problems, you will use
one of the kinematic
equations to calculate
unknowns.
Often you will
solve for time in
the y-direction.
The accelerations will
ALWAYS be –9.8 m/s2
and 0 m/s2.
Any projectile motion problem can be easily solved if you : 1) draw the situation;
2) write out what you know in the x and y-directions; 3) solve for unknowns.
9 m/s
8 m
∆x (range) = ?
2
2
2
2
1( ) ( )
2
18 (0 ) ( 9.8)( )
2
8 4.9
81.63
4.9
1.63 1.28 sec
iy v t a t
t t
t
t
t
∆ = +
− = + −
− = −
−= =
−
= =
DS =
T
D = ST
D = 9(1.28)
D = 11.5 m
x−direction
ax = 0 m/s2
Vi = 9 m/s
Vf = 9 m/s
∆x =
t = 1.28 sec
Because it’s
at constant
velocity
From the
y-direction
Since the x-
direction is at
constant speed,
you can ALWAYS
use S = D/T for
your x-direction
equation.
Since horizontal:
Vy = 0 m/s
Name: _____________________
Period: _____________________
Copyright © 2010, C. Stephen Murray cstephenmurray.com
Pro
ject
ile
Mo
tio
n E
xa
mp
le
Pro
ble
m:
A c
an
no
n f
ires
a c
an
no
nb
all
. I
f
the
can
no
n’s
an
gle
is
35
o to
th
e g
rou
nd
an
d
the
mu
zzle
vel
oci
ty i
s 5
0 m
/s,
wh
at
is t
he
ran
ge
of
the
can
no
nb
all
?
Ste
p 1
: R
eso
lve
V i
nto
Vx
an
d V
y
θ Vi
VX
i
VY
i V
Yi =
V(s
in θ
)=
28
.7m
/s
VX
i =
V(c
os θ)
= 4
1 m
/s
Vi = 5
0 m
/s
θ =
35
o
X-c
om
po
nen
t
VX
= V
(co
s θ)
VX
= 5
0m
/s(c
os
35
o)
VX
= 4
1m
/s
Y-c
om
po
nen
t
VY
= V
(sin
θ)
VY
= 5
0m
/s(s
in 3
5o)
VY
= 2
8.7
m/s
Vi = 5
0 m
/s
θ =
35
o
∆X
= ?
Ste
p 2
: F
ind
tim
e fr
om
th
e y-
dir
ecti
on
.
ay =
–9
.8 m
/s2
Vi
= V
y =
Vsi
nθ
= 2
8.7
m/s
(se
e st
ep 1
)
Vf
= -
Vi
= -
28
.7m
/s
∆Y
= 0
m (
fro
m g
rou
nd
to
gro
un
d)
∆t
= ?
Ch
oo
se o
ne
of
the
kin
em
ati
c eq
ua
tio
ns:
In t
he
y-d
irec
tio
n,
the
canno
nb
all
is i
n
free
fall
.
Wri
te t
he
va
ria
ble
s y
ou
kn
ow
:
() 2
1(
)2
1 2
if
fi
i
yv
vt
vv
at
yv
ta
t
∆=
+
=+
∆
=+
Ste
p 3
: F
ind
∆x
fro
m t
he
x-d
irec
tio
n.
In t
he
x-d
irec
tio
n,
the
canno
nb
all
trav
els
at c
onst
ant
vel
oci
ty:
Vx,
the
x-c
om
po
nent
of
the
vel
oci
ty.
The
tim
e o
f fl
igh
t w
e ju
st
fou
nd
in t
he
y-d
irec
tio
n.
Wri
te t
he
va
ria
ble
s y
ou
kn
ow
:
ax =
0 m
/s2
Vi =
Vf
Vx =
Vco
sθ =
41
m/s
(se
e st
ep 1
)
t =
5.8
sec
(fr
om
y [
see
step
2])
∆x =
?
∆x =
23
7.8
m
By
kn
ow
ing
on
ly t
he
lau
nch
vel
oci
ty a
nd
an
gle
, yo
u c
an
ca
lcu
late
d t
hat
the
can
no
n-
ba
ll t
rave
led
237
.8 m
eter
s in
5.8
sec
on
ds.
land
ing
Muz
zle
velo
city
()
2
22
1 2 2
f
fi
yv
ta
t
vv
ay
∆=
−
=+
∆
DS
T= 4
1(5
.8)
DS
T
DS
T
D
= =
=
Sin
ce a
x =
0 m
/s2
RA
NG
E
This
is
the
mo
st b
asic
of
all
pro
ject
ile
mo
tio
n p
rob
lem
s: f
rom
the
gro
und
-to
-the-
gro
und
(A
to
E).
S
tep
s 1
and
3 w
ill
AL
WA
YS
be
the
sam
e.
Ste
p 2
is
just
fre
efall
and
wil
l ch
an
ge
acco
rdin
g t
o t
he
situ
atio
n.
Sin
ce y
ou
ha
ve
all
of
the
va
ria
ble
s, y
ou
ca
n
use
an
y f
orm
ula
. S
o u
se t
he
easi
est
on
e:
vf =
vi +
(a
t)
Put
in w
hat
yo
u k
no
w:
-28
.7 =
(28
.7)
+ (
-9.8
t)
-57
.4
= -
9.8
t
÷ b
y –
9.8
t =
5.8
sec
–2
8.7
–
28
.7
Projectile Motion -2
Name: _____________________
Period: _____________________
Copyright © 2010, C. Stephen Murray cstephenmurray.com
6 m/s
Projectile Motion Concepts
Y-direction
Questions
“How high…?” “How much time…” “How long will it be in the air?” - Since the y-direction
is freefall, being pulled down by gravity, these questions are only answered in the y-direction.
Total Speed A projectile’s speed at any point is the vector sum of Vx and Vy at that point (use Pythagorean Theorem).
Only if you are asked for the vertical or horizontal speed do you use one of the components.
Vy =
28.7 m/s
Vx = 41.0 m/s
V = 50 m
/s
35°
Total Speed
Other words:
“Speed”
“Total velocity”
“Initial Velocity”
“Velocity”
Horizontal Speed
Vertical Speed
2 2Total Speed = Vx +Vy
Vyi = Vi =
5 m/s
Vy = V=
0 m/s Since this projectile is
thrown straight up (at
90°), its speed (V) is
equal to the vertical
speed (Vy) at all times.
Since they are at the same height, both balls
will hit the ground at the same time, but
Ball A will go twice as far in that time.
12 m/s B
Greatest Range
v = 35 m/s
60°
vy =
30.3 m/s
vx = 17.5 m/s
D 20°
v = 50 m/s vy =
17.1 m/s
vx = 47.0 m/s C
Object C has the greater initial speed, but object D
will go higher and take longer to get back to the
ground because D has the greater vertical speed.
If projectiles are shot with the same velocity,
but different angles, the range increases the
closer the launch angle gets to 45°. Notice
that 20° and 70° hit at the same point and
both are 25° away from 45°. Also, notice
that 70° goes the highest because it has
v = 3
5 m
/s
60°
vy =
30.3 m/s
vx = 17.5 m/s
20°
v = 50 m/s
vy =
17.1 m/s
vy = 47.0 m/s C
Even though 60° is closer to 45° than 20°, we cannot assume that object D
will have the greatest range, since the balls do not have the same initial
speed. Instead, we must calculate the range as with any ground-to-ground
projectile problem (calculate time in y-direction and range in the
x-direction). This reveals that object C actually has the greater range.
Different Initial Speeds Same Velocity, Different Angles
0
10
20
30
40
50
60
70
80
90
100
0 20 40 60 80 100 120 140 160 180 200
x position (m)
y p
osit
ion
(m
)
45° 35°
20°
65°
70°
Same Initial Speed, Different Angles
D
17.1 = 17.1 9.8
34.2 9.8
3.5 sec
f iv v at
t
t
t
= +
− −
− = −
=
47.0(3.5)
164.5m
DS
T
D ST
D
D
=
=
=
=
30.3 = 30.3 9.8
60.6 9.8
6.18sec
f iv v at
t
t
t
= +
− −
− = −
=
17.5(6.18)
108.2m
DS
T
D ST
D
D
=
=
=
=
How you determine which projectile has the greatest range depends on whether the projectiles
have the same initial velocity.
A
Name: _____________________
Period: _____________________
Copyright © 2010, C. Stephen Murray cstephenmurray.com
Projectile Motion Special Situations
All projectile motion problems work the same. First you resolve the initial velocity into Vxi and Vyi. Second, you write
everything you know in the x and y-directions. Third, remembering that ty = tx (times are the same in both directions),
you solve. This, of course, assumes that you know the basics, such as ay = –9.8 m/s2 and ax = 0 m/s2, etc.
Horizontal Launch
2 m/s
4 m
For any horizontally launched object, θ = 0º, Vxi = V and Vyi = 0 m/s.
You should already know:
ay = –9.8m/s2 ax = 0m/s2
∆y = –4 m Vx = 2 m/s
Vyi = 0 m/s
As always, find time in the y-direction:
2
2
2
2
1( ) ( )
2
14 (0 ) ( ( 9.8) )
2
4 4.9
.82
0.9 sec
iy v t at
t t
t
t
t
∆ = +
− = + −
− = −
=
=
Variables:
ay = –9.8m/s2
∆y = –4 m
Vyi = 0 m/s
t = _____
Vyf = not used
( )
2(0 .9)
1 .8 m
DS
T
D or x ST
x
x
=
∆ =
∆ =
∆ =
No acceleration
in the x-direction,
so use the easy
equation!
Maximum Height “Find the maximum height” or “how high?”: these are purely y-direction questions,
so the x-direction can be ignored.
40º
58 m/s
Example 1: A ball is shot 2 m/s horizontally from 4 m up. How far away will it land?
Example 2: A ball is shot 58 m/s at 40º. How high up will it go?
Step 1: Find Vyi.
40º
58 m/s Vyi =
58sin40º
= 37.3 m/s
Write what we know in
the y-direction and solve:
Variables:
ay = –9.8m/s2
∆y = _____
Vyi = 37.3 m/s
Vyf = 0 m/s (at the top)
t = not used
2 2
2
(2 )
0 (37.3 ) (2( 9.8) )
0 1391.29 ( 19.6 )
1391.29 19.9
1391.29
19.9
71 m =
f iV V a y
y
y
y
y
y
= + ∆
= + − ∆
= + − ∆
− = − ∆
−= ∆
−
∆
As with a ground-to-ground example, these two special situations work the same way every time.
More importantly, though, is for you to see the commonality of all projectile motion problems so that you can
solve new problems, when they arise.
How high?
Name: _____________________
Period: _____________________
Copyright © 2010, C. Stephen Murray cstephenmurray.com
Assign variables for each of the following situations. Write “?” for any unknown.
y-dir.
ay =
Vyi =
Vyf =
∆y =
t = 3 sec
x-dir.
ax =
Vxi =
Vxf =
∆x =
t =
1. An object is shot 8 m/s horizontally from a 7 m tall cliff. How far away does it land?
2. An object is shot 40 m/s at an angle of 50º. How high does it go?
3. An object is shot 80 m/s at 60º. Calculate range.
y-dir.
ay =
Vyi =
Vyf =
∆y =
t = 6.9 sec
x-dir.
ax =
Vxi =
Vxf =
∆x =
t =
y-dir.
ay =
Vyi =
Vyf =
∆y =
t =
x-dir.
ax =
Vxi =
Vxf =
∆x =
t = 2.5 sec
Projectile Motion Concepts
Name: _____________________
Period: _____________________
Copyright © 2010, C. Stephen Murray cstephenmurray.com
y-dir.
ay =
Vyi =
Vyf =
∆y =
t = 2 sec
x-dir.
ax =
Vxi =
Vxf =
∆x =
t =
4. An object is shot 80 m/s at 60º. Find its maximum height.
5. An object is shot 40 m/s at an angle of 50º. How far away does it land?
6. An object is shot 8 m/s horizontally from a 7 m tall cliff. Find its range.
y-dir.
ay =
Vyi =
Vyf =
∆y =
t =
x-dir.
ax =
Vxi =
Vxf =
∆x =
t = 6 sec
y-dir.
ay =
Vyi =
Vyf =
∆y =
t = 3 sec
x-dir.
ax =
Vxi =
Vxf =
∆x =
t =
Name: _____________________
Period: _____________________
Copyright © 2010, C. Stephen Murray cstephenmurray.com
1. An object is thrown into the air going 80 m/s at an angle
of 60º. How high does it go?
A. Realizing that in the y-direction projectiles are just
freefall , fill in the y-direction variables.
B. Realizing that in the x-direction, projectiles are at
constant speed, fill in the x-direction variables.
C. In the y-direction, calculate how high the object goes.
2. An object is launched horizontally with a speed of 8 m/s.
A. Since it is launched horizontally, what is the
initial y-direction velocity?
B. What is its initial x-direction velocity?
C. Again, in the y-direction projectiles are just freefall,
fill in the y-direction variables.
D. In the x-direction, projectiles are at constant speed,
fill in the x-direction variables.
F. In the y-direction, calculate how much time it is in
the air before it hits the ground.
G. In the x-direction (at constant speed), what equation
will you use?
H. Calculate how far away it landed in the x-direction,
using the time you just found.
3. An object is shot 40 m/s at an angle of 50º from the
ground. How far away does it land?
A. Fill in the x and y variables for the object.
B. Calculate how long it was in the air, in the
y-direction.
C. In the x-direction (at constant speed), use the time
you just calculated to find how far away it landed.
Projectile Motion Concepts with Diagrams
Name: _____________________
Period: _____________________
Copyright © 2010, C. Stephen Murray cstephenmurray.com
4. An object is shot 50 m/s at an angle of 70º.
How high does it go?
A. Use trigonometry to calculate the initial
x and y velocities of the object.
B. Fill in the x and y variables.
C. Calculate how high the object rises.
5. An object is launched 15 m/s horizontally.
A. Fill in the variables for the object.
B. Solve for time in the y-direction.
C. Since the x-direction is constant speed,
solve for ∆x.
6. An object is launched from the ground at a speed of
30 m/s at an angle of 55º. If it lands back on the
ground, calculate how far it went horizontally.
A. Find the initial x and y velocities from the given
speed and direction.
B. Fill in the variables.
C. Calculate time in the y-direction.
D. Calculate ∆x.
Name: _____________________
Period: _____________________
Copyright © 2010, C. Stephen Murray cstephenmurray.com
Name: _____________________
Period: _____________________
Copyright © 2010, C. Stephen Murray cstephenmurray.com
Name: _____________________
Period: _____________________
Copyright © 2010, C. Stephen Murray cstephenmurray.com
2. Which has the greater
maximum height?
17. Which has the
greater acceleration?
18. Which has the greater range?
16. Which one has the greater range?
9. Which has the greater
maximum height? 7. Which has the greater
maximum height? 8. Which one hits the ground last?
10. Which has the greater
speed at the top of its path?
11. Which has the lesser
maximum height? 12. Which one hits the ground first?
3. Which one has the greatest range
(lands farthest from the table)?
4. Which one has the least range
(land closest to the table)?
5. Which as the greater
maximum height?
6. Which one has the greatest
acceleration when it leaves the table?
1.Which one hits the ground first?
14. Which one has the smaller
maximum height?
15. Which has the greater
vertical acceleration? 13. Which one has the greater
vertical displacement?
Projectile Motion Comparisons
Name: _____________________
Period: _____________________
Copyright © 2010, C. Stephen Murray cstephenmurray.com
Projectile Motion Comparisons -2
28. Which one has the
greatest range?
25. Which is in the air for the
most amount of time?
33. Which one has the greater range?
29. Which as the greater
maximum height?
31. Which one has the
greatest maximum height?
26. Which one has the
greater maximum height?
21. Which has the greater
horizontal acceleration? 19. Which has the greater
horizontal velocity? 20. Which one hits the ground first?
24. Which has the
smaller range? 23. Which one has the greatest vertical
acceleration at the top of its path?
22. Which one is in the
air for less time?
27. Which has the
greater range?
30. Which one has the greater
velocity at the top of it’s path?
32. Which one is in the air
for the least amount of time?
Name: _____________________
Period: _____________________
Copyright © 2010, C. Stephen Murray cstephenmurray.com
Satellites and Review
Imagine a ball
thrown at the
top of a hill.
Because the
ground falls
as the ball
falls, the ball
goes farther.
For dropped or thrown objects
Vyi = 0 m/s and ay = -9.8 m/s2.
So, in 1 second, it will drop:
21(0 ) ( 9.8)(1)
2
4.9(1) 4.9
y t
y m
∆ = + −
∆ = − = −
1 s
ec =
4.9
m
If you were able to throw the ball
fast enough, the earth would curve
away from the ball as fast as the
ball falls, causing it to be in orbit
around the earth. It would be a
special projectile: a satellite!
Satellites are still falling toward
the earth, but they are going fast
enough to stay in orbit.
The earth’s curvature takes about
8,000 m to falls 4.9 m. To be a satellite,
an object must go 8,000 m in the 1 second
it falls or 8,000 m/sec (8 km/s) or 18,000 mph.
4.9 m
8,000 m (not to scale)
Review
1. Resolve
2. Magnitude
3. Resultant
4. Component
5. Direction
6. Vector
A. “30°” of “35 m at 30°”.
B. “35m” of “35 m at 30°”.
C. How much of the vector is in the
x or y direction.
D. Using trigonometry to finding the x– or
y-components of a vector.
E. The total from start to finish after
adding vectors together.
F. Any physical quantity that requires both
magnitude and direction.
7. Scalar or vector?
A.____A person drives 35 m/s.
B. ____Velocity
C. ____Mass of a person.
D.____acceleration
A C
B D
E
F
G
H
8. A cat climbs 3 m up a tree. Resolve this into its
x and y components.
9. A = – ____
10. B + ___ = 0
11. A + D = ___
12. D – ____ = 2D
13. Which vector/s have +X and –Y components?
14. Which vectors have no x components?
15. Which vectors have equal x and y components?
16. Mathematically, what is A + H—D + 2E?
17. Draw 2E –G +F –C. 18. On the diagram below give two ways to make R.
R A
C
D
B
Name: _____________________
Period: _____________________
Copyright © 2010, C. Stephen Murray cstephenmurray.com
19. What is ax?
20. If Vxi = 12 m/s, what is Vxf?
21. So, how do Vxi and Vxf compare?
22. What is ay?
23. If a projectile is launched from the top of
an 9 m tall building, what is ∆y?
24. How do you calculate Vyi?
25. How do you calculate Vxi?
26. In which direction do you calculate time?
27. In which direction is a projectile freefall?
28. In which direction is a projectile in constant motion?
29. On the diagram at the right draw
Vx and Vy at each point on the
projectile’s path.
30. Which one will be in the air the longest?
31. Which one will have the greatest range?
B
A
C
D
E Vx
Vy
A
B
5 m/s
5 m/s 32. Which one goes higher into the air?
33. Which one has the greatest acceleration?
34. Which one has the greatest range?
A B 5 m/s 4 m/s
5 m/s 5 m/s
35. For a projectile shot from the ground to the ground
which angle gives the longest range?
36. Which angle gives a greater range: 20° or 30°?
37. Which angle gives a greater range: 40° or 50°?
38. Which angle gives a greater range: 20° or 70°?
39. A projectile is shot 25 m/s at an angle of 34°. How high up into the air does the projectile go?
A. Find Vy. C. Solve for “how high”.
B. Write the y-direction variables.
40. A person can swim 2 m/s in still water. The river they are swimming in has a current of 1 m/s.
A. How fast are they swimming relative to the water if they swim with the current?
B. How fast are they swimming relative to the shore if they swim with the current?
C. How fast are they swimming relative to the shore if they swim against the current?
D. If they swim straight across the
river how fast are they swimming
relative to the shore? Vriver
= 1 m/s
Vperson = 1 m/s
Vtotal = ?
Satellites and Review -2
Name: _____________________
Period: _____________________
Copyright © 2010, C. Stephen Murray cstephenmurray.com
Name: _____________________
Period: _____________________
Copyright © 2010, C. Stephen Murray cstephenmurray.com
Name: _____________________
Period: _____________________
Copyright © 2010, C. Stephen Murray cstephenmurray.com
Tw
o d
imen
sio
na
l m
oti
on
is
not
as
dif
ficu
lt
as
it m
ay
seem
if
you
un
der
stan
d t
he
log
ic.
Th
is f
low
ch
art
hel
ps
you
ask
th
e
rig
ht
qu
esti
on
s a
nd r
efer
s yo
u t
o t
he
ap
pro
pri
ate
ref
eren
ce m
ate
ria
l
(wo
rksh
eets
) to
un
der
sta
nd
any
two
dim
ensi
on
al
pro
ble
m.
No
te:
Th
is c
ha
rt a
ssu
mes
th
at
all
sit
ua
tio
ns
in w
hic
h V
an
d
a a
re i
n d
iffe
ren
t d
irec
tio
ns
are
pro
ject
ile
mo
tio
n.
Oth
er
po
ssib
ilit
ies
exis
t (a
fo
rce
pu
shin
g a
n o
bje
ct o
ff c
ou
rse,
fo
r
exa
mp
le),
bu
t th
ey a
re l
ess
com
mo
n.
Are
the
unit
s th
e sa
me
for
all
quan
titi
es (
all
m/s
, fo
r ex
am
ple
)?
Is t
he
acce
lera
tio
n (
forc
e) i
n t
he
sam
e d
irec
tio
n a
s th
e vel
oci
ty?
ST
AR
T
No
Yes
Yes
See
wo
rksh
eet:
“A
dd
ing
Vec
tors
”
No
R
eso
lve
vel
oci
ty
into
co
mp
onents
Pla
ce V
x a
nd
Vy i
nto
x a
nd
y d
irec
tio
ns
See
wo
rksh
eet:
“Pro
ject
ile
Mo
tio
n”
Fil
l in
x-d
irec
tio
n
var
iab
les,
rem
em
ber
ing
that
ax =
0 m
/s2
Use
the
kin
em
ati
c
equat
ions
at t
hat
ang
le
(this
is
the
hyp
ote
nu
se)
So
lve
for
un
kno
wn q
uanti
ties
Res
olv
e in
to c
om
po
nents
if
nec
essa
ry (
to f
ind
alt
itud
e [y
],
ho
w f
ast
eas
t [V
x],
etc
)
Is t
he
vec
tor
ver
tica
l o
r ho
rizo
nta
l?
No
Yes
Res
olv
e vec
tor
into
x a
nd
y c
om
po
nen
ts
Lab
el x
or
y
(rem
em
ber
+ o
r -)
Ad
d t
he
all
x’s
and
all
y’s
into
xto
tal a
nd
yto
tal
Fin
d m
ag
nit
ud
e
wit
h P
yth
ago
rean
theo
rem
Fin
d d
irec
tio
n
θ =
tan
-1(y
t/x
t)
Use
sp
ecia
l si
tuat
ions
on
“Fre
e-f
all”
wo
rksh
eet
to
fill
in y
-dir
ecti
on v
aria
ble
s
So
lve
for
any
un
kno
wn q
uanti
ties
y d
irec
tio
n
x d
irec
tio
n
BIG
HIN
T:
Tim
e is
th
e
con
nec
tio
n
bet
wee
n t
he
x a
nd
y
dir
ecti
on
s
So
lve
for
any
un
kno
wn q
uanti
ties
Tw
o D
imen
sio
na
l M
oti
on