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10/22 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 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 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 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. 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 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… . 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 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? Inertia a property of matter that opposes any change in its state of motion Newton’s First Law Projectile An object propelled through the air, especially one thrown as a weapon Projectile Motion 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 Horizontal Motion Vertical Motion THEY ARE INDEPENDENT OF ONE ANOTHER!!!!!!!! How would you describe the trajectory? 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. Who lands first? Projectiles. From Physclips: Mechanics with animations and film. View the independence of vertical and horizontal motion Ballistics cart demo Show Mythbusters gun video here 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 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 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 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 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 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) 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 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 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 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 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? 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) dx = vix•t dx = (3.07 m/s)(2.54 s) dx = 7.80 m