Transcript Projectile Motion

10/3
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Tests are not graded. If you were absent yesterday
you can take the test today after school
Solve now: A stone is dropped from a window 33.1m
above the ground. How long does it take the stone to
land? With what speed did it hit the ground?
2.60 s 25.5 m/s
10/8 Pick up lab sheet
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Tests are available for viewing before or after school.
Today we are doing a ball toss lab outside. This will
be due at BOC tomorrow.
Thursday we will discuss special situations with PMyou will then be able to complete the rest of PM I
Friday is Q & A
Tuesday is the PM test (PM I WS due)
Ball Toss Lab
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Measure distance between the two people in feet
Each person will throw the ball 3x. Record and
average your toss time. The 2nd person will then
repeat this. You will need a third person to time.
Everyone must have their own times!
If you know the total time and distance, what else can
you figure out?
This is due at the beginning of class tomorrow. Show
all work on back. Answer the questions on the
bottom. In table, for unknowns, enter formula used
rearranged to solve for variable if needed.
10/9
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Turn in lab to sorter
Report cards will be handed out tomorrow
Today will discuss special situations with PM-you will
then be able to complete the rest of PM I
Friday is Q & A
Tuesday is the PM test (PM I WS due)
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.
The components are independent of one
another
Types of Projectile Motion
.
http://www.glenbrook.k12.il.us/gbssci/Phys/Class/usage.html
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 (aka range):
What do you know?
horizontal
velocity
of 20m/s
Horizontal displacement:
What do you know?
d=
vi = 20 m/s
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
(range)?
How would I calculate final velocity horizontal?
Vertical?
Vertical (Y)
Horizontal (X)
d=
d=
vi=
vi=
vf =
vf =
a=
a=
t=
t=
Vertical (Y)
Horizontal (X)
d = 15 m
d = Use d = vit + .5at2
vi= 0 m/s
vi= 25 m/s
vf= Use vf = vi + at
vf= 25 m/s
a = 9.8 m/s2
a = 0 m/s2
t = Use d = vit + .5at2
t = determine from vertical
information
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.
10/6
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Tests are not graded
We are continuing with PM today
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
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
Example 6: 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?
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)
What can you say about a trajectory
path?
Example 6: 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
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 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 6: 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
1.59 m/s -13.1 m/s 0 m/s
vf
1.59 m/s 0 m/s
13.1 m/s
a
0 m/s2
9.8 m/s2
t
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
10/7
Yesterday we looked at Projectiles launched at an
angle from the horizontal. (Ex 7)
We will go over example 8 in class.
You will work on PM WS I problems 1-11 while I
pass back tests.
After looking over your test, return it to the blue
sorter.
AP I will not be offering test corrections for a 70 for
the remainder of the school year.
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.
A. How long is the arrow
in the air?
B. Calculate the range.
C. Determine the peak
height of the projectile
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
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
Compare Ranges
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Refer to problems 10 & 11 on PM WS I
Example 8:
Katniss is standing on a tree limb 5
meters above the ground. She 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
D. What is the arrow’s
velocity upon impact.
Include magnitude and
angle.
Example 8:
Katniss is standing on a tree limb 5
meters above the ground. She launches an arrow
upward at a rate of 12.8 m/s and at an angle of 76.1°
with the horizontal.
A. Determine vfy and vfx
B. Determine time for arch above the limb. Use
vertical down information. Use this info to
determine peak height but remember to add to
height off ground. Multiply by 2 to get time of
arch.
C. To determine time falling below, must first
determine vfy. Then determine time.
D. Determine range by adding all times. Use vfx
Example 8: Katniss is standing on a tree limb 5 meters above
the ground. She launches an arrow upward at a rate of 12.8
m/s and at an angle of 76.1° with the horizontal.
A.
B.
C.
D.
Determine vfy and vfx 12.4m/s 3.07m/s
Determine time for arch above the limb. Use
vertical down information. 1.27s Use this info to
determine peak height but remember to add to
height off ground. 12.9m Multiply by 2 to get time
of arch. 2.54s
To determine time falling below, must first
determine vfy. 15.9m/s Then determine time.
0.357s
Determine range by adding all times. Use vfx
8.90m
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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. 9 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:
Voy = 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 10
James Bond is standing on a bridge 15 meters
above the river below. He needs to escape his
pursuers. He sees a speed boat in the distance
coming toward him. The boat is moving at
constant velocity of 2.5 m/s. How far away should
the boat be when 007 jumps off the bridge if he
wants to land in the boat? Neglect air resistance.
Example 10
What do you know about 007?
d = 15 m
vi = 0
a = 9.8 m/s2
What do you know the boat?
v = 2.5 m/s
What do we want them to have in
common?
Time!!!!
What determines the time?
007 fall
Example 10
Determine time with data from 007.
What formula?
Δd = viΔt + ½aΔt2
t = 1.75 seconds
Use the time it takes 007 to fall to
determine distance of boat when he
jumps.
What formula?
vavg = Δd/Δt
d = 4.38 m
EX 11 A soccer ball is kicked horizontally off a 22.0-meter high
hill and lands a distance of 35.0 meters from the edge of the hill.
Determine the initial horizontal velocity of the soccer ball.
Horizontal
Vx = ?
dx = 35.0m
Vertical
Voy = 0
dy = 22.0m
a = 9.8 m/s2
d = Vot + ½at2
22.0 m = 0 + ½(9.8 m/s2)t2
t = 2.12 s
Vx = d
t
= 35.0 m
2.12s
Vx = 16.5 m/s
Horizontal
What is the vertical velocity just at impact? (Vyf)
Vertical
Voy = 0
dy = 22.0m
a = 9.8 m/s2
Vfy = ?
Vx = 16.5m/s
dx = 35.0m
t = 2.12 s
Vfy = Vo + at
Vfy = 0 + (9.8m/s2)(2.12s)
Vfy = 20.8 m/s
What is the resultant velocity of the ball at impact?
Vfy
VR2 = (16.5m/s)2 + (20.8 m/s)2
VR
θ
Vx
Tan θ = 20.8 m/s
16.5 m/s
= 26.5 m/s
= 51.6º to the ground
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Projectile Motion 2.03 Phet
Open up Projectile Motion Animation from
LMS
Place settings at
Adult Human
an initial speed of 15 m/s
Target diameter 0.5 m
Launch at 90, 60, 45, 30 without air ?
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Projectile Motion 2.03 Phet
Open up Projectile Motion Animation from
LMS
Place settings at
Tank shell
an initial speed of 15 m/s
Target diameter 0.15 m
What angle(s) must you launch at to score?
Projectile Motion
How does launch angle effect
trajectory?
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Google Image Result for
http://www.animations.physics.unsw.edu.au/i
mages/download_projectiles1.gif