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Projectile motioncan be described byvertical componentsandhorizontal componentsof motion.
Unit 2: Projectile Motion
Vectors and Scalars
Honors Physics
Scalar
A SCALAR is ANY quantity in physics that has MAGNITUDE, but has NO direction associated with it.
Magnitude – A numerical value with units.
Scalar Example
Magnitude
Speed 20 m/s
Distance 10 m
Age 15 years
Heat1000
calories
Vector
A VECTOR is ANY quantity in physics that has BOTH MAGNITUDE and DIRECTION.
VectorMagnitude & Direction
Velocity 20 m/s, N
Acceleration 10 m/s/s, E
Force 5 N, West
Faxv
,,,Vectors are typically illustrated by drawing an ARROW above the symbol. The arrow is used to convey direction and magnitude.
Applications of Vectors
VECTOR ADDITION – If 2 similar vectors point in the SAME direction, add them.
Example: A person travels 54.5 meters east, then another 30 meters easterly. Calculate the displacement relative to starting point.
54.5 m, E 30 m, E+
84.5 m, E
Notice that the SIZE of the arrow conveys MAGNITUDE and the way it was drawn conveys DIRECTION.
Applications of Vectors
VECTOR SUBTRACTION - If 2 vectors are going in opposite directions, you SUBTRACT.
Example: A man walks 54.5 meters east, then 30 meters west. Calculate his displacement relative to where he started?
54.5 m, E
30 m, W-
24.5 m, E
Non-Collinear VectorsWhen 2 vectors are perpendicular, you must use
the Pythagorean theorem.
95 km,E
55 km, N
Start
Finish
A man walks 95 km, East then 55 km, north. Calculate his RESULTANT DISPLACEMENT.The hypotenuse in Physics
is called the RESULTANT.
The LEGS of the triangle are called the COMPONENTS
Horizontal Component
Vertical Component kmc
c
bacbac
8.10912050
5595Resultant 22
22222
BUT……what about the direction?In the previous example, DISPLACEMENT was asked for
and since it is a VECTOR we should include a DIRECTION on our final answer.
NOTE: When drawing a right triangle that conveys some type of motion, you MUST draw your components HEAD TO TOE.
N
S
EW
N of E
E of N
S of W
W of S
N of W
W of N
S of E
E of S
N of E
BUT...what about the ANGLE VALUE..?Just putting North of East on the answer is NOT specific enough for the direction. We MUST find the VALUE of the angle.
N of E
55 km, N
95 km,E
To find the value of the angle we use a Trig function called TANGENT.
30)5789.0(
5789.095
55
1
Tan
sideadjacent
sideoppositeTan
109.8 km
So the COMPLETE final answer is : 109.8 km, 30 degrees North of East
What if you are missing a component?Suppose a person walked 65 m, 25 degrees East of North. What
were his horizontal and vertical components?
65 m25
H.C. = ?
V.C = ?
The goal: ALWAYS MAKE A RIGHT TRIANGLE!
To solve for components, we often use the trig functions tan, sin and cosine.
EmCHopp
NmCVadj
hypopphypadj
hypotenuse
sideopposite
hypotenuse
sideadjacent
,47.2725sin65..
,91.5825cos65..
sincos
sincosine
ExampleA bear, searching for food wanders 35 meters east then 20 meters north. Frustrated, he wanders another 12 meters west then 6 meters south. Calculate the bear's displacement.
35 m, E
20 m, N
12 m, W
6 m, S
- =23 m, E
- =14 m, N
23 m, E
14 m, N
3.31)6087.0(
6087.23
14
93.262314
1
22
Tan
Tan
mR
The Final Answer: 26.93 m, 31.3 degrees NORTH of EAST
R
ExampleA boat moves with a velocity of 15 m/s, N in a river which flows with a velocity of 8.0 m/s, west. Calculate the boat's resultant velocity with respect to due north.
15 m/s, N
8.0 m/s, W
Rv
1.28)5333.0(
5333.015
8
/17158
1
22
Tan
Tan
smRv
The Final Answer : 17 m/s, @ 28.1 degrees West of North
ExampleA plane moves with a velocity of 63.5 m/s at 32o South of East. Calculate the plane's horizontal and vertical velocity components.
63.5 m/s
32
H.C. =?
V.C. = ?
SsmCVopp
EsmCHadj
hypopphypadj
hypotenuse
sideopposite
hypotenuse
sideadjacent
,/64.3332sin5.63..
,/85.5332cos5.63..
sincos
sinecosine
ExampleA storm system moves 5000 km due east, then shifts course at 40 degrees North of East for 1500 km. Calculate the storm's resultant displacement.
NkmCVopp
EkmCHadj
hypopphypadj
hypotenuse
sideopposite
hypotenuse
sideadjacent
,2.96440sin1500..
,1.114940cos1500..
sincos
sinecosine
5000 km, E
40
1500 km
H.C.
V.C.
5000 km + 1149.1 km = 6149.1 km
6149.1 km
964.2 kmR
91.8)1568.0(
1568.01.6149
2.964
1.62242.9646149
1
22
Tan
Tan
kmR
The Final Answer: 6224.1 km @ 8.91 degrees, N of E
MINUTE PHYSICS
We’ve seen simple straight-line motion
(linear )
Now, apply these ideas to curved motion
(nonlinear)
A combination of horizontal and vertical
motion.
Unit 2B: Projectile Motion
vector quantity:
velocity (v)
3.1 Vector and Scalar Quantities
Scalar quantity has magnitude only
Vector quantity has magnitude and direction
size, length, ...
scalar quantity:
speed
80 km/h north
80 km/h
acceleration (a)?
A plane’s velocity is often the result of combining two or more other velocities. • a small plane flies north at 80 km/h• a tailwind blows north at 20 km/h
3.2 Velocity Vectors
80 km/h
20 km/h
100 km/h
20 km/h
80 km/h
60 km/h
What if the plane flies against the wind?
vector addition:
same direction (ADD)
opp. direction (SUB)
Consider a plane flying 80 km/h north, but… caught in a strong crosswind of 60 km/h east.
The two velocity vectors must be combined to find the resultant.
3.2 Velocity Vectors
80 km/h
60 km/h
resultant
An 80 km/h plane flyingin a 60 km/h crosswindhas a resultant speed of 100 km/h relative to the ground.
HOW?
100 km/h
Consider a plane flying 80 km/h north, but… caught in a strong crosswind of 60 km/h east.
The two velocity vectors must be combined to find the resultant.
3.2 Velocity Vectors
80 km/h
60 km/h
1) draw vectors tail-to-head.
2) a2 + b2 = c2
vector addition:
(80)2 + (60)2 = c2
√(6400 + 3600) = c
100 km/h
The 80 km/h and 60 km/h vectors produce aresultant vector of 100 km/h, but…in what direction?
3.2 Velocity Vectors
80 km/h
60 km/h
100 km/h
θ = tan-1(opp/adj)
tan(θ) = opp/adj
θθ : “theta”
θ = 53o N of E
100 km/h, 53o N of E(or 53o above + x-axis)
opp
adjθ = tan-1(80/60)
Suppose that an airplane normally flying at80 km/h encounters wind at a right angle toits forward motion—a crosswind.
Will the airplane fly faster or slower than80 km/h?
Answer:A crosswind would increase the speed of the airplane but blow it off course by a predictable amount.
3.2 Velocity Vectors
1. Which of these expresses a vector quantity?
A. 10 kg
B. 10 kg to the north
C. 10 m/s
D. 10 m/s 23o N of E
Quick Quiz!
3.1
2. An ultra-light aircraft traveling north at 40 km/h in a 30 km/h crosswind (at right angles) has a groundspeed of _____.
A. 30 km/h
B. 40 km/h
C. 50 km/h
D. 60 km/h
Quick Quiz.
3.2
Check off the learning targets you can do after today.
40 km/h
30 km/h
??? km/h
a2 + b2 = c2
(30)2 + (40)2 = c2
√(900 + 1600) = c
You can resolve a single vector into two component vectors at right angles to each other:
3.3 Components of Vectors
Vectors X and Y are the horizontal and vertical components of a vector V.
A ball’s velocity can be resolved into horizontal (x) and vertical (y) components.
3.3 Components of Vectors
3.3 Components of Vectors
A jet flies 340 m/s (mach 1) at 60o N of E.What are the vertical and horizontal components of the jet’s velocity?
vy = ?
vx = ?
60o
vy
vx
opp
adj
340 m/s
vy = v sin(θ)
sin(θ) = opp/hypcos(θ) = adj/hyp
vx = v cos(θ)
(hyp)
(v)
3.3 Components of Vectors
A jet flies 340 m/s (mach 1) at 60o N of E.What are the vertical and horizontal components of the jet’s velocity?
vy = ?
vx = ?
60o
vy
vx
opp
adj
340 m/s
vy = v sin(θ)
vx = v cos(θ)
(hyp)
(v)
vy = (340 m/s) • sin(60)
vy =
vx = (340 m/s) • cos(60)
vx =
294 m/s
170 m/s
294 m/s
170 m/s
1. A ball launched into the air at 45° to the horizontal initially has…
A. equal horizontal and vertical components.
B. components that do not change in flight.
C.components that affect each other throughout flight.
D.a greater component of velocity than the vertical component.
Quick Quiz!
3.3
30o
2. A jet flies 680 m/s (mach 2) at 30o N of E. What is the vertical component of the jet’s velocity (vy)?
A. 589 m/s
B. 340 m/s
C.230 m/s
D.180 m/s
Quick Quiz.
680 m/s
vy = v sin(θ) = opp / hyp
vy = (680 m/s) • sin(30)
340 m/s
projectile:
any object moving through a path, acted on only by gravity. (no friction/no air resistance)Ex: cannonball, ball/stone, spacecraft/satellite, etc.
3.4 Projectile Motion
projectile motion
gravity-free path
gravity only
Projectile motion is separated into components.
a. Roll a ball horizontally, v is constant, b/c no acceleration from g horizontally.
b. Drop a ball, it accelerates downward covering a greater distance each second.
c. x & y components are completely independent of each other.
3.4 Projectile Motion
Projectile motion is separated into components.
a. Roll a ball horizontally, v is constant, b/c no acceleration from g horizontally.
b. Drop a ball, it accelerates downward covering a greater distance each second.
c. x & y components are completely independent of each other.
d. combined they cause curved paths.
3.4 Projectile Motion
• x component is constant (a = 0)(g acts only in y direction)
• both fall the same y distance in same time.(x and y are completely unrelated)
3.4 Projectile Motion
vx vy
3.4 Projectile Motion
vx2 vx4
vy2
vy4
vx3
vy3
1. When no air resistance acts on a fast-moving baseball, its acceleration is …
A. downward only
B. in the forward x direction it was thrown
C. opposite to the force of gravity
D. both forward and downward
Quick Quiz!
The Y distance fallen is the same vertical distance it would fall if dropped from rest.
Remember d = ½gt2
3.5 Projectiles Launched at an Angle
Height & Rangevx is constant, but vy changes.
At the max height, vy = 0.(only Vx)
3.5 Projectiles Launched at an Angle
launch angle affects height (y) and range (x)
3.5 Projectiles Launched at an Angle
Height & Range
height
range
height
range
60o 75o
more angle:
-more initial vy, more height
-less initial vx, less range
• angles that add to 90° have equal ranges
• max range usually at 45°
3.5 Projectiles Launched at an Angle
Height & Range
vup = –vdown
3.5 Projectiles Launched at an Angle
20 m/s
–20 m/s
12 m/s
12 m/s
12 m/s
12 m/s
10 m/s
12 m/s
–10 m/s
Velocity & Time
Is it safe to shoota bullet in the air?
0 m/s
3.5 Projectiles Launched at an Angle
Velocity & Time
tup = tdownvup = –vdown
ttotal = (2)tup
3.5 Projectiles Launched at an Angle
Height & Range
vx constant, but vy changes
At hmax, vy = 0 (only Vx horizontal)
Velocity & Time
tup = tdown
vup = –vdown
ttotal = (2)tup
more angle:
-more initial vy, more height
-less initial vx, less range
height
range
1. Without air resistance, the time for a vertically tossed ball to return to where it was thrown is …
A. 10 m/s for every second in the air.
B. the same as the time going upward.
C. less than the time going upward.
D. more than the time going upward.
Quick Quiz!
3.5 Projectiles Launched at an Angle
Solving projectile calculation problems in 3 easy steps:
1)Direction: get Vix & Viy (pick Horiz. or Vert.)
2)List Variables
d =
vi =
a =
vf =
t =
3. Pick equation, Plug numbers, and Solve.
3.5 Projectiles Launched at an Angle
Sample Calculation #1Bob Beamon’s record breaking long jump (8.9 m) at the 1968 Olympics resulted from an initial velocity of 9.4 m/s at an angle of 40o above horizontal.Solve for each of the following variables:
vix =
viy =
tup =
ttotal = (time of flight)
dx = (range)
dymax = (peak height)
vy = v sin(θ)
vx = v cos(θ)g = –10 m/s2
v = vi + at
d = vit + ½at2
ttotal = (2)tup
vix = (9.4 m/s) • cos(40o) =
viy = (9.4 m/s) • sin(40o) =
tup =
ttotal (up AND dn) = (2)(0.604 s) =
0 – 6.04 = –10
7.20 m/s
6.04 m/s
vfy = viy + at
0 = 6.04 + –10t0.604 s
1.21 s
Vi = 9.4 m/s
at 40o above horizontal 40o
9.4 m/s
dx =
dymax =
d = (7.20 m/s)(1.21 s)
d = vixt + ½at28.71 m0
d = (6.04 m/s)(0.604 s) + ½(–10 m/s2)(0.604 s)2 =
d = viyt + ½at21.82 m
Vo = 9.4 m/s
at 40o above x-axis 40o
9.4 m/s
vix =
viy =
7.20 m/s
6.04 m/s
tup =
ttotal =
0.604 s
1.21 s
7.20 m/s
6.04 m/s
0.604 s
1.21 s
8.71 m
1.82 m
3.5 Projectiles Launched at an Angle
A soccer ball is kicked horizontally off a 22.0 m high hill and lands a distance of 35.0 m from the edge of the hill. Determine the initial horizontal velocity of the soccer ball.
vix =
viy = 0 m/s
t =
dx = 35.0 m
dymax = 22.0 m
22.0 m
35.0 m
vix
Sample Calculation #2
3.5 Projectiles Launched at an Angle
viy = 0 m/s
dx = 35.0 m
dymax = 22.0 m
t =
vix =
22.0 m
35.0 m
vix
2(–22.0 m) = –10
d = viyt + ½at2
–22.0 = ½(–10)t2
0
√2.10 s
d = vixt + ½at2
35.0 = vix(2.10) 35.0 m = 2.10 s
016.7 m/s
Sample Calculation #2
3.5 Projectiles Launched at an Angle
θHorizontal Launchviy = 0 m/s
vix = v
Angled Launchviy = v sin(θ)
vix = v cos(θ)
vv
For ALL launches:
a = g = –10 m/s2 for vertical motion
a = 0 m/s2 for horizontal motion
t is found vertically with:
v = vi + gt or d = ½gt2