10/22 i will not be available after school today. yesterday was the vector test. if you were absent...
TRANSCRIPT
10/22 10/22 I will not be available after school I will not be available after school today.today. Yesterday was the vector test. If you were absent I am 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 available Thu and Fri am (6:30) and Thu pm (until 3:30) for make upsmake ups
Tests will not be returned until Monday. I will try to post Tests will not be returned until Monday. I will try to post grades this weekend.grades this weekend.
Get calculator and both note sheets.Get calculator and both note sheets. PM powerpoint and Bops are posted under the Projectile PM powerpoint and Bops are posted under the Projectile
Motion Tab on websiteMotion Tab on website Notes to use for next test are the class notes you are picking Notes to use for next test are the class notes you are picking
up today. Powerpoint will only be allowed if they are printed up today. Powerpoint will only be allowed if they are printed and initialed before 10/27. No WS or labs will be allowed.and initialed before 10/27. No WS or labs will be allowed.
Trig Test 2 tomorrow.Trig Test 2 tomorrow.
10/23 10/23 I will be available after school today I will be available after school today & Friday am& Friday am
Yesterday you picked up PM notes and we worked through Ex 1 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 Today you will take the Trig Test 2. No notes are permitted. No BOPS
(these were due the day of the original test). (these were due the day of the original test). You will need a pencil, You will need a pencil, scantron (I will give to you), scratch paper and a calculator. scantron (I will give to you), scratch paper and a calculator. The score The score on the blue side is NOT your last Trig test grade.on the blue side is NOT your last Trig test grade.
When completed place scantron and test in appropriate tray. When completed place scantron and test in appropriate tray. Pick up PM WS I. This will be due BOC Tue 10/28Pick up PM WS I. This will be due BOC Tue 10/28 Test corrections and retakes for the Vector Test will be next Mon- Test corrections and retakes for the Vector Test will be next Mon-
Thur. Retakes pm only. Thur. Retakes pm only. You must bring your completed vector class You must bring your completed vector class notes to be eligible for test corrections and retakes.notes to be eligible for test corrections and retakes.
10/23 10/23 TRIG TEST 2TRIG TEST 2
Clear calulators: Scratch pad>Doc>bClear calulators: Scratch pad>Doc>b On side 2 (green) of scantronOn side 2 (green) of scantron Name: record your nameName: record your name Subject: Trig 2Subject: Trig 2 Period: Record your periodPeriod: Record your period Date: Today is 10/23Date: Today is 10/23 Under the Please Recycle Symbol: Record your test Under the Please Recycle Symbol: Record your test
formform Blue tests are Form ABlue tests are Form A Green tests are Form BGreen tests are Form B
10/2410/24 Wednesday you picked up PM notes and went through Ex 1. Wednesday you picked up PM notes and went through Ex 1. Thursday you were offered the Trig Test 2. Retakes am/pm Thursday you were offered the Trig Test 2. Retakes am/pm
Monday and Tuesday of next week only. You must sign into the Monday and Tuesday of next week only. You must sign into the test book to let me know when you will be here.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. Thursday I assigend PM WS I. This will be due BOC Tue 10/28. Note: rounding aNote: rounding agg to 10 m/s to 10 m/s22 Correction on 3d: how fast traveling Correction on 3d: how fast traveling
horizontallyhorizontally when it strikes the water (v when it strikes the water (vxx)) Today is the last day to make up the Vector Test. Today is the last day to make up the Vector Test. Test corrections and retakes for the Vector Test will be next Mon- Test corrections and retakes for the Vector Test will be next Mon-
Thur. Retakes pm only Thur. Retakes pm only You must bring your completed vector You must bring your completed vector
class notes to be eligible for test corrections and retakes.class notes to be eligible for test corrections and retakes. You You must sign into the test book 1 day in advance to let me know must sign into the test book 1 day in advance to let me know when you will be here for test.when you will be here for test.
10/27 10/27 Thursday you were offered the Trig Test 2. Retakes am/pm Monday 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 and Tuesday of this week only. You must sign into the test book to let me know when you will be here.me know when you will be here.
Thursday I assigend PM WS I. Thursday I assigend PM WS I. This will be due BOC Tue 10/28This will be due BOC Tue 10/28. . Note: rounding aNote: rounding agg to 10 m/s to 10 m/s22 Correction on 3d: how fast traveling Correction on 3d: how fast traveling
horizontallyhorizontally when it strikes the water (v when it strikes the water (vxx)) Friday we did example 2 in notes and #1 on PM WS IFriday we did example 2 in notes and #1 on PM WS I Get your notes out now.Get your notes out now. Vector BOPS have not been graded yetVector BOPS have not been graded yet Test corrections and retakes for the Vector Test will be this Mon- Thur. Test corrections and retakes for the Vector Test will be this Mon- Thur.
Retakes pm only Retakes pm only You must bring your completed vector class notes You must bring your completed vector class notes to be eligible for test corrections and retakes.to be eligible for test corrections and retakes. You must sign into the You must sign into the test book 1 day in advance to let me know when you will be here for test book 1 day in advance to let me know when you will be here for test.test.
10/29 10/29 Pick up PM WS II. You will be working on this in class. It will be due 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 Tue 11/4. Expectations: Show table, then show all work below. Insert answers into table.answers into table.
You should review the riverboat (girl scout) type of problemYou should review the riverboat (girl scout) type of problem Tomorrow we will work on a labTomorrow we will work on a lab Monday we will review concepts (we may do this as a quiz) and any Monday we will review concepts (we may do this as a quiz) and any
extra time you will work on PM WS IIextra time you will work on PM WS II Yesterday we did the softball toss lab. The make up is in the mailbox Yesterday we did the softball toss lab. The make up is in the mailbox
and will be due tomorrow.and will be due tomorrow. Vector BOPS have not been graded yetVector BOPS have not been graded yet Test corrections and retakes for the Vector Test will be this Mon- Thur. Test corrections and retakes for the Vector Test will be this Mon- Thur.
Retakes pm only Retakes pm only You must bring your completed vector class notes to You must bring your completed vector class notes to be eligible for test corrections and retakes.be eligible for test corrections and retakes. You must sign into the test You must sign into the test book 1 day in advance to let me know when you will be here for 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…
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?
10 m/s across
5 m/s downstream
VR
VR =
√10 2 + 52 = 11.18 m/s
Θ
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 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.
Types of Projectile MotionTypes of Projectile Motion
.
http://www.glenbrook.k12.il.us/gbssci/Phys/Class/usage.html
Projectiles Facts1. 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)
2. Projectiles always experience a constant vertical acceleration of “g” or 9.80 m/s2
(neglecting air resistance) due to 2nd law of motion
3. Horizontal and vertical motion are completely INDEPENDENT of each other.
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 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:Horizontal 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
= vx
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
= vx
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 m A ball is thrown horizontally at 25 m/s off a roof 15 m
high. 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? How would I calculate final velocity horizontal? Vertical?How would I calculate final velocity horizontal? Vertical? Hint: Determine how long the ball is in the air using Hint: Determine how long the ball is in the air using
vertical information, then use calculated time to determine vertical information, then use calculated time to determine horizontal distance.horizontal distance.
Vertical (Y)Vertical (Y) Horizontal (X)Horizontal (X)
d =d = d = d =
vvii== vvii==
vvff== vvff==
a =a = a =a =
t =t = t =t =
= vx
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
= vx
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.
Work through PM WS I #1nowWork through PM WS I #1now
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
= vx
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
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. 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/sdx = ?
Vertical:Viy = 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 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 What can you say about a trajectory path?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 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
=vx
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 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 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
= vx
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
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)
Concept ReviewConcept ReviewNeglect Air Neglect Air ResistanceResistance
Describe velocity for the horizontal and Describe velocity for the horizontal and vertical components of the arrowvertical components of the arrow
Describe the acceleration for the horizontal Describe the acceleration for the horizontal and vertical components of the arrowand vertical components of the arrow
Describe the time for the horizontal and Describe the time for the horizontal and vertical components of the arrowvertical components of the arrow
Concept ReviewConcept Review A ball is thrown straight into the air with an A ball is thrown straight into the air with an
upward velocity of 5 m/s. What will be the upward velocity of 5 m/s. What will be the velocity when it is caught? (same height from velocity when it is caught? (same height from ground) ground)
Concept ReviewConcept Review In the absence of air, what would hit the In the absence of air, what would hit the
ground first, an elephant or a feather if ground first, an elephant or a feather if dropped from the same height at the same dropped from the same height at the same time? time?
Which projectile Which projectile lands first?lands first?
Which projectile Which projectile lands first?lands first?
30º, 45º, 60º, or 75º30º, 45º, 60º, or 75ºIn the absence of air resistanceIn the absence of air resistance
What angle do you think gives you the greatest What angle do you think gives you the greatest horizontal distance?horizontal distance?
What angle do you think gives you the greatest What angle do you think gives you the greatest vertical height?vertical height?
You walk at a constant speed. You walk at a constant speed. While doing so you drop a ball. While doing so you drop a ball. Where are you in relation to the Where are you in relation to the
ball when it hits the ground?ball when it hits the ground?
You are sitting in a vehicle moving You are sitting in a vehicle moving at a constant speed. While doing at a constant speed. While doing so you toss a quarter directly into so you toss a quarter directly into the air. Where does the quarter the air. Where does the quarter
land?land?
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 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
