Transcript PM PPT

10/22
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I will not be available after school
today.
Yesterday was the vector test. If you were absent I am
available Thu and Fri am (6:30) and Thu pm (until 3:30) for
make ups
Tests will not be returned until Monday. I will try to post
grades this weekend.
Get calculator and both note sheets.
PM powerpoint and Bops are posted under the Projectile
Motion Tab on website
Notes to use for next test are the class notes you are picking
up today. Powerpoint will only be allowed if they are
printed and initialed before 10/27. No WS or labs will be
allowed.
Trig Test 2 tomorrow.
10/23
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I will be available after school today
& Friday am
Yesterday you picked up PM notes and we worked through Ex 1
Today you will take the Trig Test 2. No notes are permitted. No BOPS
(these were due the day of the original test). You will need a pencil,
scantron (I will give to you), scratch paper and a calculator. The score
on the blue side is NOT your last Trig test grade.
When completed place scantron and test in appropriate tray.
Pick up PM WS I. This will be due BOC Tue 10/28
Test corrections and retakes for the Vector Test will be next MonThur. Retakes pm only. You must bring your completed vector class
notes to be eligible for test corrections and retakes.
10/23
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TRIG TEST 2
Clear calulators: Scratch pad>Doc>b
On side 2 (green) of scantron
Name: record your name
Subject: Trig 2
Period: Record your period
Date: Today is 10/23
Under the Please Recycle Symbol: Record your
test form
Blue tests are Form A
Green tests are Form B
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10/24
Wednesday you picked up PM notes and went through Ex 1.
Thursday you were offered the Trig Test 2. Retakes am/pm
Monday and Tuesday of next week only. You must sign into
the test book to let me know when you will be here.
Thursday I assigend PM WS I. This will be due BOC Tue
10/28. Note: rounding ag to 10 m/s2 Correction on 3d: how
fast traveling horizontally when it strikes the water (vx)
Today is the last day to make up the Vector Test.
Test corrections and retakes for the Vector Test will be next
Mon- Thur. Retakes pm only You must bring your
completed vector class notes to be eligible for test
corrections and retakes. You must sign into the test book 1
day in advance to let me know when you will be here for test.
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10/27
Thursday you were offered the Trig Test 2. Retakes am/pm Monday
and Tuesday of this week only. You must sign into the test book to let
me know when you will be here.
Thursday I assigend PM WS I. This will be due BOC Tue 10/28.
Note: rounding ag to 10 m/s2 Correction on 3d: how fast traveling
horizontally when it strikes the water (vx)
Friday we did example 2 in notes and #1 on PM WS I
Get your notes out now.
Vector BOPS have not been graded yet
Test corrections and retakes for the Vector Test will be this Mon- Thur.
Retakes pm only You must bring your completed vector class notes
to be eligible for test corrections and retakes. You must sign into the
test book 1 day in advance to let me know when you will be here for
test.
10/29
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Pick up PM WS II. You will be working on this in class. It will be due
Tue 11/4. Expectations: Show table, then show all work below. Insert
answers into table.
You should review the riverboat (girl scout) type of problem
Tomorrow we will work on a lab
Monday we will review concepts (we may do this as a quiz) and any
extra time you will work on PM WS II
Yesterday we did the softball toss lab. The make up is in the mailbox
and will be due tomorrow.
Vector BOPS have not been graded yet
Test corrections and retakes for the Vector Test will be this Mon- Thur.
Retakes pm only You must bring your completed vector class notes
to be eligible for test corrections and retakes. You must sign into the
test book 1 day in advance to let me know when you will be here for
test.
First, let’s talk about The River Boat…
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If the boat has a speed of 10 m/s, crosses theriver
perpendicular to the current and the current is
5 m/s, what is the resultant velocity of the boat?
How long to travel across the river?
If the boat has a speed of 10 m/s and the current is
5 m/s, what is the resultant velocity of the boat?
VR
Θ
5 m/s downstream
10 m/s across
VR =
√10 2 + 52 =
11.18 m/s
Θ=
Tan-1 (5/10) =
26.57º downstream
How long to travel across the 120 m wide river?
The time to cross depends on the speed across the river.
t= d
v
= 120 m
10m/s
= 12 sec
How far downstream will the boat land on the far bank?
The distance downstream depends on the downstream
current speed and the time in the water.
d = vt
= (5 m/s)(12sec) = 60 m downstream
The perpendicular components of motion
are INDEPENDENT of each other
So… the velocity across the river is independent
of the velocity down the river.
We will use this rule again and again…
Projectile Motion
Thank you Physics Classroom: http://www.glenbrook.k12.il.us/gbssci/Phys/Class/usage.html
What forces are working on the arrow as
it flies horizontally through the air?
15 mph
FORCE
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A Push or Pull
If velocity constant, the force of thrust is equal
but opposite the force of air friction
Is the arrow falling? The downward force
working on the arrow is GRAVITY. This is
greater than the upward force of air resistance.
Anything thrown or launched on this planet is
under the influence of gravity.
What keeps the arrow moving
forward?
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Inertia
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a property of matter that opposes any
change in its state of motion
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Newton’s First Law
Projectile
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An object propelled through the air, especially
one thrown as a weapon
Projectile Motion
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The process of movement horizontally and
vertically simultaneously.
Types of Projectile Motion
.
http://www.glenbrook.k12.il.us/gbssci/Phys/Class/usage.html
Projectiles Facts
1.
2.
3.
Projectiles maintain a constant horizontal
velocity (neglecting air resistance) due to 1st
law of motion Vi and Vf are equal. We will
refer to these as VX (horizontal velocity)
Projectiles always experience a constant
vertical acceleration of “g” or 9.80 m/s2
(neglecting air resistance) due to 2nd law of
motion
Horizontal and vertical motion are completely
INDEPENDENT of each other.
Two Components of Projectile
Motion
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Horizontal Motion
Vertical Motion
THEY ARE INDEPENDENT OF ONE
ANOTHER!!!!!!!!
How would you describe the
trajectory?
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Parabolic
http://www.glenbrook.k12.il.us/gbssci/Phys/Class/usage.html
Suppose you shoot a gun a drop a
spare bullet at the same time.
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Who lands first?
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Projectiles. From Physclips: Mechanics with
animations and film.
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View the independence of vertical and
horizontal motion
Ballistics cart demo
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Show Mythbusters gun video here
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If time permits
EX 1 A cannon ball is shot from a cannon
with a horizontal velocity of 20m/s. What is
the vertical and horizontal displacement after
1 second? After 2 seconds ?
horizontal
velocity
of 20m/s
d=
vi =
vf =
a=
t=
Vertical displacement:
What do you know?
horizontal
velocity
of 20m/s
Vertical displacement:
What do you know?
d=
vi = 0 m/s
vf =
a = 9.8 m/s2
t = 1sec
Which formula would you use to solve for d?
dy = viy t + ½ ay t2
To calculate vertical displacement
ONLY USE VERTICAL INFO !
dy = viy t + ½ ay t2
What is viy t = to?
dy = ½ a y t2
Where:
dy = vertical displacement (y axis)
ay= g = gravity (9.8m/s2)
(some texts use negative to indicate downward. We will
assume gravity to be positive.)
t = time in seconds
horizontal
velocity
of 20m/s
d=
vi =
vf =
a=
t=
Horizontal displacement:
What do you know?
= vx
horizontal
velocity
of 20m/s
Horizontal displacement:
What do you know?
d=
vi = 20 m/s
=
v
x
vf = 20 m/s
a = 0 m/s
t = We will use 1s and 2 sec
Which formula would you use to solve for d?
dx = vix t + ½ ax t2
Of these three equations, the top equation is the most commonly
used. The other two equations are seldom (if ever) used. An
application of projectile concepts to each of these equations
would also lead one to conclude that any term with ax in it would
cancel out of the equation since ax = 0 m/s/s.
To calculate horizontal displacement.
ONLY USE HORIZONTAL INFO !
Time determined vertically.
dx = vi t + ½ a t2
Since a is zero, then ½ a t2 = zero
dx = vix * t
d = vt
Where:
dx = horizontal displacement (x axis)
The subscript x refers to horizontal
Vix = initial horizontal velocity
t = time in seconds
Calculate the displacement at 2
seconds
How does vertical displacement change as time
increases?
How does horizontal displacement change as time
increases?
EX 2
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A ball is thrown horizontally at 25 m/s off a roof 15
m high.
A. How long is this ball in flight?
B. How far does the ball travel vertically?
C. How far does the ball travel horizontally?
How would I calculate final velocity horizontal?
Vertical?
Hint: Determine how long the ball is in the air using
vertical information, then use calculated time to
determine horizontal distance.
Vertical (Y)
Horizontal (X)
d=
d=
vi=
vi=
vf =
vf =
a=
a=
t=
t=
= vx
Vertical (Y)
Horizontal (X)
d = 15 m
d = Use d = vit + .5at2
vi= 0 m/s
vi=
vf= Use vf = vi + at
vf =
a = 9.8 m/s2
a = 0 m/s2
t = Use d = vit + .5at2
t = determine from vertical
information
25 m/s
= vx
25 m/s
How long is it in the air?
d = vit + .5at2 Since vi= 0, this can be
simplified to:
d = .5at2
To solve for t:
t = d/.5a
1.75 sec
Using time from vertical motion, can
calculate distance for horizontal
motion
dx = vi t + ½ a t2
Since a is zero, then ½ a t2 = zero
dx = vix * t
d = vt
43.8m
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2 Objects are dropped from a height of 10 m.
Object A has a mass of 50 g. Object B has a
mass of 100g. If there is no air friction, then:
A. Object A should hit the ground before
Object B
B. Object B should hit the ground before
Object A
C. Object A and Object B should hit the
ground at the same time.
Work through PM WS I #1now
EX 3
d = .5 at2
1.01 s = t
d = vt
d/t=v
20 m / 1.01 s = v
19.8 m/s = v
EX 4 A projectile is shot with a speed of 39.5 m/s straight
off a roof and lands 198 m away. From what elevation was it
shot? ? With what speed does it impact the ground
vertically and horizontally? With what overall velocity does
it impact the ground?
d = vt
d/v=t
198 m / 39.5 m/s = t
5.01 s = t
d = ½ at
2
d = ½(9.8 m/s2)(5.01)2
d = 123 m
EX 4 A projectile is shot with a speed of 39.5 m/s straight
off a roof and lands 198 m away. From what elevation was it
shot? ? With what speed does it impact the ground
vertically and horizontally? With what overall velocity does
it impact the ground?
vxf = 39.5 m/s
vyf = at
vyf = (9.8 m/s2)(5.01)
vyf = 49.1 m/s
vr = √(49.1 m/s)2 + (39.5 m/s)2
vr = 63.0 m/s
EX 5 A projectile is shot horizontally off a
267-m tall building with a speed of 14.3 m/s.
A. With what speed does it impact the ground
vertically and horizontally?
B. With what overall velocity does it impact the
ground?
Vertical (Y)
Horizontal (X)
d = 267 m
d = Use d = vt
vi= 0 m/s
vi=
14.3 m/s
= vx
vf= Use vf = vi + at or
vf2 = vi2 + 2ax
a = 9.8 m/s2
vf =
14.3 m/s
t = Use d = vit + .5at2
t = determine from vertical
information
a = 0 m/s2
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vf horizontal is constant at 14.3 m/s
vf2 = vi2 + 2ax to determine vf vertically
vfy = 72.3 m/s
overall velocity?
This is just determining the resultant using
Pythagoreans
vr2 = (14.3 m/s)2 + (72.3m/s)2
vr = 73.7 m/s
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Supposing a snowmobile is equipped with a
flare launcher which is capable of launching a
sphere vertically (relative to the snowmobile).
If the snowmobile is in motion and launches
the flare and maintains a constant horizontal
velocity after the launch, then where will the
flare land (neglect air resistance)?
a. in front of the snowmobile
b. behind the snowmobile
c. in the snowmobile
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Many would insist that there is a horizontal force
acting upon the ball since it has a horizontal motion.
This is simply not the case. The horizontal motion of
the ball is the result of its own inertia. When
projected from the truck, the ball already possessed a
horizontal motion, and thus will maintain this state of
horizontal motion unless acted upon by a horizontal
force. An object in motion will continue in motion
with the same speed and in the same direction ...
(Newton's first law). Remind yourself continuously:
forces do not cause motion; rather, forces cause
accelerations
Ex. 6 A plane flying at 115 m/s drops a package from
600m. How far from the drop point will it land?
Objects dropped from a moving vehicle have the same
velocity as the moving vehicle.
Horizontal:
Vx = 115 m/s
dx = ?
Vertical:
Viy = 0
dy = 600. m
a = 9.8 m/s2
This is the same problem we’ve been working…
dy = ½ at2
600. m= ½ (9.8m/s2)t2
t = 11.1 s
dx = (115 m/s)(11.1s)
dx = 1280 m
Example 7: A projectile is thrown upward
at a rate of 13.22 m/s and at an angle of
83.1° with the horizontal.
A. How long is the projectile in the air?
B. Calculate the range.
C. What is the peak height?
What can you say about a trajectory
path?
Example 7: A projectile is thrown
upward at a rate of 13.22 m/s and at an
angle of 83.1° with the horizontal.
Indicate knowns
Horiz
(X)
Vert Up Vert down
(Y)
(Y)
d
vi
vf
a
t
0 m/s
=vx
0 m/s2
0 m/s
9.8 m/s2
9.8 m/s2
How do we determine the initial
velocities?
Given 13.22 m/s at an angle of 83.1°
This describes the resultant of the horizontal and
vertical velocity components.
You need to determine the horizontal and vertical
components
Vertical
Sin (83.1°) (13.22 m/s)
83.1°
Horizontal
Cos (83.1°) (13.22 m/s)
Example 7: A projectile is thrown
upward at a rate of 13.22 m/s and at an
angle of 83.1° with the horizontal.
Indicate knowns
Horiz
(X)
Vert Up
(Y)
Vert down
(Y)
d
vi
vf
a
t
= vx
1.59 m/s -13.1 m/s
0 m/s
1.59 m/s 0 m/s
13.1 m/s
0 m/s2
9.8 m/s2
9.8 m/s2
Time at Peak
t = vfy - viy
ay
13.1 m/s – 0 m/s
9.8m/s2
t = 1.34 s
Horizontal Time would be 2.68 sec
Peak Height
d = .5at2
(.5)(9.8 m/s2)(1.34 s)2
8.80m
Horizontal Displacement
(Remember to double time)
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dx = vix•t
dx = (1.59 m/s)(2.68 s)
dx = 4.26 m
Projectiles at a known velocity and angle
Steps to determine time, height , and range
1. Determine X component (C=A/H)
This yields the horizontal vi and vf
2. Determine Y component (S=O/H)
This yields the vertical up vi and vertical down vf
3. Make 3 column table of knowns: Horizontal, Vertical Up, and
Vertical down
Remember horizontal acceleration = 0; vertical acceleration is 9.8
m/s2 due to gravity
4. Calculate peak time using vertical down column vf = vi + at
5. Total time in air (horizontal) is 2 x peak time
6. Calculate peak height using vertical information x = .5at2
(vi t = 0 in vertical down column)
7. Calculate range using horizontal information x = vi t (.5at2 = 0)
Concept Review
Neglect Air
Resistance
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Describe velocity for the horizontal and
vertical components of the arrow
Describe the acceleration for the horizontal
and vertical components of the arrow
Describe the time for the horizontal and
vertical components of the arrow
Concept Review
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A ball is thrown straight into the air with an
upward velocity of 5 m/s. What will be the
velocity when it is caught? (same height from
ground)
Concept Review
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In the absence of air, what would hit the
ground first, an elephant or a feather if
dropped from the same height at the same
time?
Which projectile
lands first?
Which projectile
lands first?
30º, 45º, 60º, or 75º
In the absence of air resistance
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What angle do you think gives you the greatest
horizontal distance?
What angle do you think gives you the greatest
vertical height?
You walk at a constant speed.
While doing so you drop a ball.
Where are you in relation to the
ball when it hits the ground?
You are sitting in a vehicle moving
at a constant speed. While doing
so you toss a quarter directly into
the air. Where does the quarter
land?
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A stream flows due south at 16.4 m/s. A toy
boat crosses the stream at a rate of 0.28 m/s.
A. If the stream is 4.65 m wide, how long does
it take the toy boat to cross the stream?
B. What is the the resultant velocity of the toy
boat in respect to the river bank?
C. How far downstream is the boat when it
reaches the other side?
Example 8: Katniss launches an arrow
upward at a rate of 12.8 m/s and at an angle
of 76.1° with the horizontal.
A. How long is the arrow
in the air?
B. Calculate the range.
C. Determine the peak
height of the projectile
Example 8:
Katniss launches an arrow
upward at a rate of 12.8 m/s and at an angle
of 76.1° with the horizontal.
Indicate knowns
Horiz
(X)
Vert Up Vert down
(Y)
(Y)
d
vi
0 m/s
vf
a
t
0 m/s
0 m/s2
9.8 m/s2
9.8 m/s2
How do we determine the initial
velocities?
Given 12.8 m/s at an angle of 76.1°
This describes the resultant of the horizontal and
vertical velocity components.
You need to determine the horizontal and vertical
components
Vertical
Sin (76.1°) (12.8 m/s)
76.1°
Horizontal
Cos (76.1°) (12.8 m/s)
Example 7:
Katniss launches an arrow
upward at a rate of 12.8 m/s and at an angle
of 76.1° with the horizontal.
Indicate knowns
Horiz
(X)
Vert Up Vert down
(Y)
(Y)
d
vi
3.07 m/s -12.4 m/s 0 m/s
vf
3.07m/s 0 m/s
12.4 m/s
a
0 m/s2
9.8 m/s2
t
9.8 m/s2
Time at Peak
t = vfy - viy
ay
12.4 m/s – 0 m/s
9.8m/s2
t = 1.27 s
Horizontal Time would be 2.54 sec
Peak Height
d = .5at2
(.5)(9.8 m/s2)(1.27 s)2
7.90 m
Horizontal Displacement
(Remember to double time)
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dx = vix•t
dx = (3.07 m/s)(2.54 s)
dx = 7.80 m