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10/3
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 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 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)
Ball Toss Lab 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 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 MotionProjectile Motion
Thank you Physics Classroom: http://www.glenbrook.k12.il.us/gbssci/Phys/Class/usage.html
What forces are working on the arrow as What forces are working on the arrow as it flies horizontally through the air?it flies horizontally through the air?
15 mph
FORCEFORCE
A Push or PullA Push or Pull If velocity constant, the force of thrust is equal If velocity constant, the force of thrust is equal
but opposite the force of air frictionbut opposite the force of air friction Is the arrow falling? The downward force Is the arrow falling? The downward force
working on the arrow is GRAVITY. This is working on the arrow is GRAVITY. This is greater than the upward force of air resistance.greater than the upward force of air resistance.
Anything thrown or launched on this planet is Anything thrown or launched on this planet is under the influence of gravity.under the influence of gravity.
What keeps the arrow moving What keeps the arrow moving forward?forward?
InertiaInertia
a property of matter that opposes any change in its state of motion
Newton’s First LawNewton’s First Law
ProjectileProjectile
An object propelled through the air, especially An object propelled through the air, especially one thrown as a weaponone thrown as a weapon
Projectile MotionProjectile Motion
The process of movement horizontally and The process of movement horizontally and vertically simultaneously. vertically simultaneously.
The components are independent of one The components are independent of one anotheranother
Types of Projectile MotionTypes of Projectile Motion
.
http://www.glenbrook.k12.il.us/gbssci/Phys/Class/usage.html
Two Components of Projectile Two Components of Projectile MotionMotion
Horizontal MotionHorizontal Motion Vertical MotionVertical Motion THEY ARE INDEPENDENT OF ONE THEY ARE INDEPENDENT OF ONE
ANOTHER!!!!!!!!ANOTHER!!!!!!!!
How would you describe the How would you describe the trajectory?trajectory?
ParabolicParabolic
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.
Who lands first?
Projectiles. From Physclips: Mechanics with Projectiles. From Physclips: Mechanics with animations and film.animations and film.
View the independence of vertical and View the independence of vertical and horizontal motionhorizontal motion
Ballistics cart demoBallistics cart demo
Show Mythbusters gun video here
If time permits
EX 1EX 1 A cannon ball is shot from a cannon A cannon ball is shot from a cannon with a horizontal velocity of 20m/s. What is with a horizontal velocity of 20m/s. What is
the vertical and horizontal displacement after the vertical and horizontal displacement after 1 second? After 2 seconds ? 1 second? After 2 seconds ?
Vertical displacement:Vertical displacement:What do you know?What do you know?
d =d =
vvi = =
vvf = =
a =a =
t =t =
horizontal horizontal velocity velocity of 20m/sof 20m/s
Vertical displacement:Vertical displacement:What do you know?What do you know?
d = d =
vvi = 0 m/s = 0 m/s
vvf = =
a = 9.8 m/sa = 9.8 m/s22
t = 1sect = 1sec
Which formula would you use to solve for d?Which formula would you use to solve for d?
dy = vdy = viyiy t + ½ a t + ½ ayy t t22
horizontal horizontal velocity velocity of 20m/sof 20m/s
To calculate vertical displacementTo calculate vertical displacementONLY USE VERTICAL INFO !ONLY USE VERTICAL INFO !
ddyy = v = viyiy t + ½ a t + ½ ayy t t22
What is vWhat is viyiy t = to? t = to?
ddyy = ½ a = ½ ayy t t22
Where:Where:ddyy = vertical displacement (y axis) = vertical displacement (y axis)aayy= g = gravity (9.8m/s= g = gravity (9.8m/s22) ) (some texts use negative to indicate downward. We will (some texts use negative to indicate downward. We will
assume gravity to be positive.)assume gravity to be positive.)t = time in secondst = time in seconds
Horizontal displacement (aka range):Horizontal displacement (aka range):What do you know?What do you know?
d =d =
vvi = =
vvf = =
a = a =
t =t =
horizontal horizontal velocity velocity of 20m/sof 20m/s
Horizontal displacement:Horizontal displacement:What do you know?What do you know?
d =d =
vvi = 20 m/s = 20 m/s
vvf = 20 m/s = 20 m/s
a = 0 m/sa = 0 m/st = We will use 1s and 2 sect = We will use 1s and 2 sec
Which formula would you use to solve for d?Which formula would you use to solve for d?
ddxx = v = vixix t + ½ a t + ½ axx t t22
horizontal horizontal velocity velocity of 20m/sof 20m/s
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. To calculate horizontal displacement. ONLY USE HORIZONTAL INFO ! ONLY USE HORIZONTAL INFO !
Time determined vertically.Time determined vertically.
ddxx = v = vii t + ½ a t t + ½ a t22
Since Since a a is zero, then is zero, then ½ a t½ a t22 = zero = zero
ddxx = v = vixix ** t t
d = vtd = vt
Where:Where:
ddxx = horizontal displacement (x axis) = horizontal displacement (x axis)
The subscript x refers to horizontalThe subscript x refers to horizontal
VVix ix = = initial horizontal velocity initial horizontal velocity t = time in secondst = 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 2EX 2 A ball is thrown horizontally at 25 m/s off a roof 15 A ball is thrown horizontally at 25 m/s off a roof 15
m high. m high. A.A. How long is this ball in flight?How long is this ball in flight? B.B. How far does the ball travel vertically?How far does the ball travel vertically? C.C. How far does the ball travel horizontally How far does the ball travel horizontally
(range)?(range)? How would I calculate final velocity horizontal? How would I calculate final velocity horizontal?
Vertical?Vertical?
Vertical (Y)Vertical (Y) Horizontal (X)Horizontal (X)
d =d = d = d =
vvii== vvii==
vvff== vvff==
a =a = a =a =
t =t = t =t =
Vertical (Y)Vertical (Y) Horizontal (X)Horizontal (X)
d = 15 m d = 15 m d = d = Use d = vUse d = viit + .5att + .5at2 2
vvii= 0 m/s = 0 m/s vvii= 25 m/s = 25 m/s
vvff= = Use vUse vf f = v= vii + at + at vvff= 25 m/s = 25 m/s
a = 9.8 m/sa = 9.8 m/s22 a = 0 m/sa = 0 m/s22
t = t = Use d = vUse d = viit + .5att + .5at2 2 t = determine from vertical t = determine from vertical informationinformation
How long is it in the air?How long is it in the air?
d = vd = viit + .5att + .5at2 2 Since vSince vii= 0, this can be = 0, this can be
simplified to:simplified to:
d = .5atd = .5at22
To solve for t:To solve for t:
t = d/.5at = d/.5a
1.75 sec1.75 sec
Using time from vertical motion, can Using time from vertical motion, can calculate distance for horizontal calculate distance for horizontal
motionmotion
ddxx = v = vii t + ½ a t t + ½ a t22
Since Since a a is zero, then is zero, then ½ a t½ a t22 = zero = zero
ddxx = v = vixix ** t t
d = vtd = vt
43.8m43.8m
2 Objects are dropped from a height of 10 m. 2 Objects are dropped from a height of 10 m. Object A has a mass of 50 g. Object B has a Object A has a mass of 50 g. Object B has a mass of 100g. If there is no air friction, then:mass of 100g. If there is no air friction, then:
A. Object A should hit the ground before A. Object A should hit the ground before Object BObject B
B. Object B should hit the ground before B. Object B should hit the ground before Object AObject A
C. Object A and Object B should hit the C. Object A and Object B should hit the ground at the same time.ground at the same time.
10/610/6
Tests are not gradedTests are not graded We are continuing with PM todayWe are continuing with PM today
EX 3EX 3
d = .5 at2
1.01 s = t
d = vtd / t = v
20 m / 1.01 s = v19.8 m/s = v
EX 4EX 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 = ½ at 2
d = vtd / v = t
198 m / 39.5 m/s = t5.01 s = t
d = ½(9.8 m/s2)(5.01)2
d = 123 m
EX 4EX 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?
vyf = at
vxf = 39.5 m/s
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)Vertical (Y) Horizontal (X)Horizontal (X)
d = 267 m d = 267 m d = d = Use d = vtUse d = vt
vvii= 0 m/s = 0 m/s vvii= 14.3 m/s = 14.3 m/s
vvff= = Use vf = vi + at or
vf2 = vi
2 + 2ax
vvff= 14.3 m/s = 14.3 m/s
a = 9.8 m/sa = 9.8 m/s22 a = 0 m/sa = 0 m/s22
t = t = Use d = vUse d = viit + .5att + .5at2 2 t = determine from vertical t = determine from vertical informationinformation
vf horizontal is constant at 14.3 m/s vf
2 = vi2 + 2ax to determine vf vertically
vfy = 72.3 m/s overall velocity? This is just determining the resultant using
Pythagoreans vr
2 = (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 angleSteps 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 downRemember horizontal acceleration = 0; vertical acceleration is 9.8 m/s2 due to gravity
4. Calculate peak time using vertical down column vf = vi + at5. 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 What can you say about a trajectory path?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 knownsIndicate knowns HorizHoriz
(X)(X)
Vert UpVert Up
(Y)(Y)
Vert down Vert down (Y)(Y)
dd
vvii 0 m/s
vvff 0 m/s
aa 0 m/s2 9.8 m/s2 9.8 m/s2
tt
How do we determine the initial How do we determine the initial velocities?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
VerticalSin (83.1°) (13.22 m/s)
13.22 m
/s
83.1°
HorizontalCos (83.1°) (13.22 m/s)
THESE ARE
YOUR INITIAL
VELOCITIES!!
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 knownsIndicate knowns HorizHoriz
(X)(X)
Vert UpVert Up
(Y)(Y)
Vert down Vert down (Y) (Y)
dd
vvii 1.59 m/s -13.1 m/s 0 m/s
vvff 1.59 m/s 0 m/s 13.1 m/s
aa 0 m/s2 9.8 m/s2 9.8 m/s2
tt
Time at PeakTime at Peakt = vt = vfyfy - v - viyiy
a ayy
13.1 m/s – 0 m/s13.1 m/s – 0 m/s9.8m/s9.8m/s22
t = 1.34 st = 1.34 s
Horizontal Time would be 2.68 secHorizontal Time would be 2.68 sec
Peak HeightPeak Height
d = .5at2
(.5)(9.8 m/s(.5)(9.8 m/s22)(1.34 s))(1.34 s)22
8.80m8.80m
Horizontal DisplacementHorizontal Displacement(Remember to double time)(Remember to double time)
ddxx = v = vixix•t•t
ddxx = (1.59 m/s)(2.68 s) = (1.59 m/s)(2.68 s)
ddxx = 4.26 m = 4.26 m
10/710/7
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 knownsIndicate knowns HorizHoriz
(X)(X)
Vert UpVert Up
(Y)(Y)
Vert down Vert down (Y)(Y)
dd
vvii 0 m/s
vvff 0 m/s
aa 0 m/s2 9.8 m/s2 9.8 m/s2
tt
How do we determine the initial How do we determine the initial velocities?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
VerticalSin (76.1°) (12.8 m/s)
12.8 m
/s
76.1°
HorizontalCos (76.1°) (12.8 m/s)
THESE ARE
YOUR INITIAL
VELOCITIES!!
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 knownsIndicate knowns HorizHoriz
(X)(X)
Vert UpVert Up
(Y)(Y)
Vert down Vert down (Y) (Y)
dd
vvii 3.07 m/s -12.4 m/s 0 m/s
vvff 3.07m/s 0 m/s 12.4 m/s
aa 0 m/s2 9.8 m/s2 9.8 m/s2
tt
Time at PeakTime at Peakt = vt = vfyfy - v - viyiy
a ayy
12.4 m/s – 0 m/s12.4 m/s – 0 m/s9.8m/s9.8m/s22
t = 1.27 st = 1.27 s
Horizontal Time would be 2.54 secHorizontal Time would be 2.54 sec
Peak HeightPeak Height
d = .5at2
(.5)(9.8 m/s(.5)(9.8 m/s22)(1.27 s))(1.27 s)22
7.90 m7.90 m
Horizontal DisplacementHorizontal Displacement(Remember to double time)(Remember to double time)
ddxx = v = vixix•t•t
ddxx = (3.07 m/s)(2.54 s) = (3.07 m/s)(2.54 s)
ddxx = 7.80 m = 7.80 m
Compare RangesCompare Ranges
Refer to problems 10 & 11 on PM WS IRefer 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. Determine vfy and vfx 12.4m/s 3.07m/s
B. 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
C. To determine time falling below, must first determine vfy. 15.9m/s Then determine time. 0.357s
D. Determine range by adding all times. Use vfx 8.90m
Supposing a snowmobile is equipped with a Supposing a snowmobile is equipped with a flare launcher which is capable of launching a flare launcher which is capable of launching a sphere vertically (relative to the snowmobile). sphere vertically (relative to the snowmobile). If the snowmobile is in motion and launches If the snowmobile is in motion and launches the flare and maintains a constant horizontal the flare and maintains a constant horizontal velocity after the launch, then where will the velocity after the launch, then where will the flare land (neglect air resistance)?flare land (neglect air resistance)?
a. in front of the snowmobile a. in front of the snowmobile b. behind the snowmobileb. behind the snowmobile c. in the snowmobilec. in the snowmobile
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/sdx = ?
Vertical:Voy = 0dy = 600. ma = 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 10What 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 10Determine time with data from 007.
What formula?
Δd = vΔd = viiΔΔt + ½at + ½aΔΔtt22
t = 1.75 seconds
Use the time it takes 007 to fall to
determine distance of boat when he
jumps.
What formula?
vvavgavg = Δd/ = Δd/ΔΔtt
d = 4.38 md = 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 Vertical
Vx = ?dx = 35.0m
Voy = 0dy = 22.0ma = 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 Vertical
Vx = 16.5m/sdx = 35.0mt = 2.12 s
Voy = 0dy = 22.0ma = 9.8 m/s2
Vfy = ?
What is the vertical velocity just at impact? (Vyf)
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?
VR
Vx
Vfy
VR2 = (16.5m/s)2 + (20.8 m/s)2 = 26.5 m/s
θTan θ = 20.8 m/s 16.5 m/s
= 51.6º to the ground
Projectile Motion 2.03 PhetProjectile Motion 2.03 Phet Open up Projectile Motion Animation from Open up Projectile Motion Animation from
LMSLMS Place settings at Place settings at Adult HumanAdult Human an initial speed of 15 m/san initial speed of 15 m/s Target diameter 0.5 mTarget diameter 0.5 m Launch at 90, 60, 45, 30 without air ?Launch at 90, 60, 45, 30 without air ?
Projectile Motion 2.03 PhetProjectile Motion 2.03 Phet Open up Projectile Motion Animation from Open up Projectile Motion Animation from
LMSLMS Place settings at Place settings at Tank shellTank shell an initial speed of 15 m/san initial speed of 15 m/s Target diameter 0.15 mTarget diameter 0.15 m What angle(s) must you launch at to score?What angle(s) must you launch at to score?
Projectile MotionProjectile MotionHow does launch angle effect How does launch angle effect
trajectory?trajectory? Google Image Result for Google Image Result for
http://www.animations.physics.unsw.edu.au/ihttp://www.animations.physics.unsw.edu.au/images/download_projectiles1.gifmages/download_projectiles1.gif
