2. Projectile motion can be described by vertical components and horizontal components of motion. Unit 2: Projectile Motion

3. Vectors and Scalars Honors Physics

4. Scalar A SCALAR is ANY quantity in physics that has MAGNITUDE, but has NO direction associated with it. Magnitude – A numerical value with units. Scalar Example Magnitude Speed20 m/s Distance10 m Age15 years Heat 1000 calories

5. Vector A VECTOR is ANY quantity in physics that has BOTH MAGNITUDE and DIRECTION. Vector Magnitude & Direction Velocity20 m/s, N Acceleration10 m/s/s, E Force5 N, West Vectors are typically illustrated by drawing an ARROW above the symbol. The arrow is used to convey direction and magnitude.

6. Applications of Vectors VECTOR ADDITION – If 2 similar vectors point in the SAME direction, add them. Example: A person travels 54.5 meters east, then another 30 meters easterly. Calculate the displacement relative to starting point. 54.5 m, E30 m, E + 84.5 m, E Notice that the SIZE of the arrow conveys MAGNITUDE and the way it was drawn conveys DIRECTION.

7. Applications of Vectors VECTOR SUBTRACTION - If 2 vectors are going in opposite directions, you SUBTRACT. Example: A man walks 54.5 meters east, then 30 meters west. Calculate his displacement relative to where he started? 54.5 m, E 30 m, W - 24.5 m, E

8. Non-Collinear Vectors When 2 vectors are perpendicular, you must use the Pythagorean theorem. 95 km,E 55 km, N Start Finish A man walks 95 km, East then 55 km, north. Calculate his RESULTANT DISPLACEMENT. The hypotenuse in Physics is called the RESULTANT. The LEGS of the triangle are called the COMPONENTS Horizontal Component Vertical Component

9. BUT……what about the direction? In the previous example, DISPLACEMENT was asked for and since it is a VECTOR we should include a DIRECTION on our final answer. NOTE: When drawing a right triangle that conveys some type of motion, you MUST draw your components HEAD TO TOE. N S E W N of E E of N S of W W of S N of W W of N S of E E of S N of E

10. BUT...what about the ANGLE VALUE..? Just putting North of East on the answer is NOT specific enough for the direction. We MUST find the VALUE of the angle. N of E 55 km, N 95 km,E To find the value of the angle we use a Trig function called TANGENT.  109.8 km So the COMPLETE final answer is : 109.8 km, 30 degrees North of East

11. What if you are missing a component? Suppose a person walked 65 m, 25 degrees East of North. What were his horizontal and vertical components? 65 m 25 H.C. = ? V.C = ? The goal: ALWAYS MAKE A RIGHT TRIANGLE! To solve for components, we often use the trig functions tan, sin and cosine.

12. Example A bear, searching for food wanders 35 meters east then 20 meters north. Frustrated, he wanders another 12 meters west then 6 meters south. Calculate the bear's displacement. 35 m, E 20 m, N 12 m, W 6 m, S - = 23 m, E -= 14 m, N 23 m, E 14 m, N The Final Answer: 26.93 m, 31.3 degrees NORTH of EAST R 

13. Example A boat moves with a velocity of 15 m/s, N in a river which flows with a velocity of 8.0 m/s, west. Calculate the boat's resultant velocity with respect to due north. 15 m/s, N 8.0 m/s, W RvRv  The Final Answer : 17 m/s, @ 28.1 degrees West of North

14. Example A plane moves with a velocity of 63.5 m/s at 32 o South of East. Calculate the plane's horizontal and vertical velocity components. 63.5 m/s 32 H.C. =? V.C. = ?

15. Example A storm system moves 5000 km due east, then shifts course at 40 degrees North of East for 1500 km. Calculate the storm's resultant displacement. 5000 km, E 40 1500 km H.C. V.C. 5000 km + 1149.1 km = 6149.1 km 6149.1 km 964.2 km R  The Final Answer: 6224.1 km @ 8.91 degrees, N of E

16. MINUTE PHYSICS

17. We’ve seen simple straight-line motion (linear ) Now, apply these ideas to curved motion (nonlinear) A combination of horizontal and vertical motion. Unit 2B: Projectile Motion

18. vector quantity: velocity (v) 3.1 Vector and Scalar Quantities Scalar quantity has magnitude only Vector quantity has magnitude and direction size, length,... scalar quantity: speed 80 km/h north 80 km/h acceleration (a) ?

19. A plane’s velocity is often the result of combining two or more other velocities. a small plane flies north at 80 km/h a tailwind blows north at 20 km/h 3.2 Velocity Vectors 80 km / h 20 km / h 100 km / h 20 km / h 80 km / h 60 km / h What if the plane flies against the wind? vector addition : same direction (ADD) opp. direction (SUB)

20. Consider a plane flying 80 km/h north, but… caught in a strong crosswind of 60 km/h east. The two velocity vectors must be combined to find the resultant. 3.2 Velocity Vectors 80 km / h 60 km / h resultant An 80 km/h plane flying in a 60 km/h crosswind has a resultant speed of 100 km/h relative to the ground. HOW? 100 km / h

21. Consider a plane flying 80 km/h north, but… caught in a strong crosswind of 60 km/h east. The two velocity vectors must be combined to find the resultant. 3.2 Velocity Vectors 80 km / h 60 km / h 1) draw vectors tail-to-head. 2) a 2 + b 2 = c 2 vector addition : (80) 2 + (60) 2 = c 2 √(6400 + 3600) = c 100 km / h

22. The 80 km/h and 60 km/h vectors produce a resultant vector of 100 km/h, but… in what direction? 3.2 Velocity Vectors 80 km / h 60 km / h 100 km / h θ = tan -1 (opp/adj) tan(θ) = opp/adj θ θ : “theta” θ = 53 o N of E 100 km/h, 53 o N of E (or 53 o above + x-axis) opp adj θ = tan -1 (80/60)

23. Suppose that an airplane normally flying at 80 km/h encounters wind at a right angle to its forward motion—a crosswind. Will the airplane fly faster or slower than 80 km/h? Answer: A crosswind would increase the speed of the airplane but blow it off course by a predictable amount. 3.2 Velocity Vectors

24. 1.Which of these expresses a vector quantity? A.10 kg B.10 kg to the north C.10 m/s D.10 m/s 23 o N of E Quick Quiz! 3.1

25. 2.An ultra-light aircraft traveling north at 40 km/h in a 30 km/h crosswind (at right angles) has a groundspeed of _____. A.30 km/h B.40 km/h C.50 km/h D.60 km/h Quick Quiz. 3.2 Check off the learning targets you can do after today. 40 km / h 30 km / h ??? km / h a 2 + b 2 = c 2 (30) 2 + (40) 2 = c 2 √(900 + 1600) = c

26. You can resolve a single vector into two component vectors at right angles to each other: 3.3 Components of Vectors Vectors X and Y are the horizontal and vertical components of a vector V.

27. A ball’s velocity can be resolved into horizontal (x) and vertical (y) components. 3.3 Components of Vectors

28. A jet flies 340 m/s (mach 1) at 60 o N of E. What are the vertical and horizontal components of the jet’s velocity? v y = ? v x = ? 60 o vyvy vxvx opp adj 340 m / s v y = v sin(θ) sin(θ) = opp/hyp cos(θ) = adj/hyp v x = v cos(θ) (hyp) (v)(v)

29. 3.3 Components of Vectors A jet flies 340 m/s (mach 1) at 60 o N of E. What are the vertical and horizontal components of the jet’s velocity? v y = ? v x = ? 60 o vyvy vxvx opp adj 340 m / s v y = v sin(θ) v x = v cos(θ) (hyp) (v)(v) v y = (340 m/s) sin(60) v y = v x = (340 m/s) cos(60) v x = 294 m/s 170 m/s 294 m / s 170 m / s

30. 1.A ball launched into the air at 45° to the horizontal initially has… A.equal horizontal and vertical components. B.components that do not change in flight. C.components that affect each other throughout flight. D.a greater component of velocity than the vertical component. Quick Quiz! 3.3

31. 30 o 2.A jet flies 680 m/s (mach 2) at 30 o N of E. What is the vertical component of the jet’s velocity ( v y )? A.589 m/s B.340 m/s C.230 m/s D.180 m/s Quick Quiz. 680 m / s v y = v sin(θ) = opp / hyp v y = (680 m/s) sin(30) 340 m / s

32. projectile: any object moving through a path, acted on only by gravity. (no friction/no air resistance) Ex: cannonball, ball/stone, spacecraft/satellite, etc. 3.4 Projectile Motion projectile motion gravity- free path gravity only

33. Projectile motion is separated into components. a.Roll a ball horizontally, v is constant, b/c no acceleration from g horizontally. b.Drop a ball, it accelerates downward covering a greater distance each second. c.x & y components are completely independent of each other. 3.4 Projectile Motion

34. Projectile motion is separated into components. a.Roll a ball horizontally, v is constant, b/c no acceleration from g horizontally. b.Drop a ball, it accelerates downward covering a greater distance each second. c.x & y components are completely independent of each other. d.combined they cause curved paths. 3.4 Projectile Motion

35. x component is constant (a = 0) (g acts only in y direction) both fall the same y distance in same time. (x and y are completely unrelated) 3.4 Projectile Motion vxvx vyvy

36. vx2vx2 vx4vx4 vy2vy2 vy4vy4 vx3vx3 vy3vy3

37. 1.When no air resistance acts on a fast- moving baseball, its acceleration is … A.downward only B.in the forward x direction it was thrown C.opposite to the force of gravity D.both forward and downward Quick Quiz!

38. The Y distance fallen is the same vertical distance it would fall if dropped from rest. Remember d = ½gt 2 3.5 Projectiles Launched at an Angle

39. Height & Range v x is constant, but v y changes. At the max height, v y = 0.(only V x ) 3.5 Projectiles Launched at an Angle

40. launch angle affects height (y) and range (x) 3.5 Projectiles Launched at an Angle Height & Range height range height range 60 o 75 o more angle: -more initial v y, more height -less initial v x, less range

41. angles that add to 90° have equal ranges max range usually at 45° 3.5 Projectiles Launched at an Angle Height & Range

42. v up = – v down 3.5 Projectiles Launched at an Angle 20 m/s –20 m/s 12 m/s 10 m/s 12 m/s –10 m/s Velocity & Time Is it safe to shoot a bullet in the air? 0 m/s

43. 3.5 Projectiles Launched at an Angle Velocity & Time t up = t down v up = – v down t total = (2) t up

44. 3.5 Projectiles Launched at an Angle Height & Range v x constant, but v y changes At h max, v y = 0 (only V x horizontal ) Velocity & Time t up = t down v up = – v down t total = (2) t up more angle: -more initial v y, more height -less initial v x, less range height range

45. 1.Without air resistance, the time for a vertically tossed ball to return to where it was thrown is … A.10 m/s for every second in the air. B. the same as the time going upward. C. less than the time going upward. D. more than the time going upward. Quick Quiz!

46. 3.5 Projectiles Launched at an Angle Solving projectile calculation problems in 3 easy steps: 1)Direction: get V i x & V i y (pick Horiz. or Vert.) 2)List Variables d = v i = a = v f = t = 3.Pick equation, Plug numbers, and Solve.

47. 3.5 Projectiles Launched at an Angle Sample Calculation #1 Bob Beamon’s record breaking long jump (8.9 m) at the 1968 Olympics resulted from an initial velocity of 9.4 m/s at an angle of 40 o above horizontal. Solve for each of the following variables: v i x = v i y = t up = t total = (time of flight) d x = (range) d y max = (peak height) v y = v sin(θ) v x = v cos(θ) g = –10 m/s 2 v = v i + at d = v i t + ½at 2 t total = (2) t up

48. v i x = (9.4 m / s ) cos(40 o ) = v i y = (9.4 m / s ) sin(40 o ) = t up = t total (up AND dn) = (2)(0.604 s) = 0 – 6.04 = –10 7.20 m / s 6.04 m / s v f y = v i y + at 0 = 6.04 + –10t 0.604 s 1.21 s V i = 9.4 m/s at 40 o above horizontal 40 o 9.4 m/s

49. d x = d y max = d = (7.20 m / s )(1.21 s) d = v i x t + ½at 2 8.71 m 0 d = (6.04 m / s )(0.604 s) + ½(–10 m / s 2 )(0.604 s) 2 = d = v i y t + ½at 2 1.82 m V o = 9.4 m/s at 40 o above x-axis 40 o 9.4 m/s v i x = v i y = 7.20 m / s 6.04 m / s t up = t total = 0.604 s 1.21 s

50. 7.20 m / s 6.04 m / s 0.604 s 1.21 s 8.71 m 1.82 m

51. 3.5 Projectiles Launched at an Angle A soccer ball is kicked horizontally off a 22.0 m high hill and lands a distance of 35.0 m from the edge of the hill. Determine the initial horizontal velocity of the soccer ball. v i x = v i y = 0 m/s t = d x = 35.0 m d y max = 22.0 m 22.0 m 35.0 m vixvix Sample Calculation #2

52. 3.5 Projectiles Launched at an Angle v i y = 0 m/s d x = 35.0 m d y max = 22.0 m t = v i x = 22.0 m 35.0 m vixvix 2(–22.0 m ) = –10 d = v i y t + ½at 2 –22.0 = ½(–10)t 2 0 √ 2.10 s d = v i x t + ½at 2 35.0 = v i x (2.10) 35.0 m = 2.10 s 0 16.7 m / s Sample Calculation #2

53. 3.5 Projectiles Launched at an Angle θ Horizontal Launch v i y = 0 m/s v i x = v Angled Launch v i y = v sin(θ) v i x = v cos(θ) v v For ALL launches: a = g = –10 m/s 2 for vertical motion a = 0 m/s 2 for horizontal motion t is found vertically with: v = v i + gt ord = ½gt 2