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Projectile motion can be described by vertical components and horizontal components of motion.

Unit 2 : Projectile Motion. Projectile motion can be described by vertical components and horizontal components of motion. Vectors and Scalars. Honors Physics. Scalar. A SCALAR is ANY quantity in physics that has MAGNITUDE , but has NO direction associated with it.

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Projectile motion can be described by vertical components and horizontal components of motion.

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

  2. Vectors and Scalars Honors Physics

  3. Scalar A SCALAR is ANY quantity in physics that has MAGNITUDE, but has NO direction associated with it. Magnitude – A numerical value with units.

  4. Vector A VECTOR is ANY quantity in physics that has BOTH MAGNITUDE and DIRECTION. Vectors are typically illustrated by drawing an ARROW above the symbol. The arrow is used to convey direction and magnitude.

  5. 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, E 30 m, E Notice that the SIZE of the arrow conveys MAGNITUDE and the way it was drawn conveys DIRECTION. 84.5 m, E

  6. 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

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

  8. 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. N W of N E of N N of E N of W E W N of E S of W S of E NOTE: When drawing a right triangle that conveys some type of motion, you MUST draw your components HEAD TO TOE. W of S E of S S

  9. 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. To find the value of the angle we use a Trig function called TANGENT. 109.8 km 55 km, N q N of E 95 km,E So the COMPLETE final answer is : 109.8 km, 30 degrees North of East

  10. 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? The goal: ALWAYS MAKE A RIGHT TRIANGLE! To solve for components, we often use the trig functions tan, sin and cosine. H.C. = ? V.C = ? 25 65 m

  11. 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. - = 23 m, E 12 m, W - = 14 m, N 6 m, S 20 m, N 14 m, N 35 m, E R q 23 m, E The Final Answer: 26.93 m, 31.3 degrees NORTH of EAST

  12. 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. 8.0 m/s, W 15 m/s, N Rv q The Final Answer : 17 m/s, @ 28.1 degrees West of North

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

  14. 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. 1500 km V.C. 40 5000 km, E H.C. 5000 km + 1149.1 km = 6149.1 km R 964.2 km q 6149.1 km The Final Answer: 6224.1 km @ 8.91 degrees, N of E

  15. MINUTE PHYSICS

  16. Unit 2B: Projectile Motion We’ve seen simple straight-line motion (linear ) Now, apply these ideas to curvedmotion (nonlinear) A combination of horizontalandvertical motion.

  17. 3.1Vector and Scalar Quantities ? acceleration (a) scalar quantity: speed 80 km/h size, length, ... Scalarquantityhas magnitude only Vectorquantityhas magnitude and direction vector quantity: velocity (v) 80 km/h north

  18. What if the plane flies against the wind? 3.2Velocity Vectors 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 100 km/h 20 km/h 60 km/h 80 km/h 80 km/h vector addition: same direction (ADD) opp. direction (SUB) 20 km/h

  19. 3.2Velocity Vectors Consider a plane flying 80 km/hnorth, but… caught in a strong crosswind of 60 km/h east. The two velocity vectors must be combined to find the 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. 80 km/h 100 km/h resultant HOW? 60 km/h

  20. 3.2Velocity Vectors Consider a plane flying 80 km/hnorth, but… caught in a strong crosswind of 60 km/h east. The two velocity vectors must be combined to find the resultant. vector addition: 100 km/h 80 km/h • draw vectors tail-to-head. • a2 + b2 = c2 (80)2 + (60)2 = c2 √(6400 + 3600) = c 60 km/h

  21. 3.2Velocity Vectors The 80 km/h and 60 km/h vectors produce a resultant vector of 100 km/h, but… in what direction? 100 km/h, 53o N of E (or 53o above + x-axis) 100 km/h tan(θ) = opp/adj 80 km/h opp θ = tan-1(opp/adj) θ = tan-1(80/60) θ adj θ = 53o N of E θ : “theta” 60 km/h

  22. 3.2Velocity Vectors 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 courseby a predictable amount.

  23. Quick Quiz! • Which of these expresses a vector quantity? • 10 kg • 10 kg to the north • 10 m/s • 10 m/s 23o N of E 3.1

  24. Quick Quiz. • An ultra-light aircraft traveling north at 40 km/h in a 30 km/h crosswind (at right angles) has a groundspeed of _____. • 30 km/h • 40 km/h • 50 km/h • 60 km/h a2 + b2 = c2 (30)2 + (40)2 = c2 √(900 + 1600) = c 40 km/h ??? km/h Check off the learning targets you can do after today. 30 km/h 3.2

  25. 3.3Components of Vectors You can resolve a single vector into two componentvectors at right angles to each other: Vectors X and Y are the horizontal and verticalcomponents of a vector V.

  26. 3.3Components of Vectors A ball’s velocity can be resolved into horizontal (x) and vertical (y) components.

  27. 3.3Components of Vectors A jet flies 340 m/s (mach 1) at 60o N of E. What are the vertical and horizontal components of the jet’s velocity? vy = ? vx = ? vy = v sin(θ) (v) vx = v cos(θ) 340m/s sin(θ) = opp/hyp (hyp) vy opp cos(θ) = adj/hyp adj 60o vx

  28. 3.3Components of Vectors A jet flies 340 m/s (mach 1) at 60o N of E. What are the vertical and horizontal components of the jet’s velocity? vy = ? vx = ? vy = v sin(θ) (v) vx = v cos(θ) 340m/s vy = (340 m/s) • sin(60) vy = vy (hyp) opp 294 m/s 294 m/s vx = (340 m/s) • cos(60) vx = adj 60o vx 170 m/s 170 m/s

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

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

  31. 3.4Projectile Motion projectile: any objectmovingthrough a path, acted on only by gravity. (no friction/no air resistance) Ex: cannonball, ball/stone, spacecraft/satellite, etc. gravity-free path projectile motion gravity only

  32. 3.4Projectile Motion Projectile motion is separated into components. • Roll a ballhorizontally, v is constant, b/c no acceleration from g horizontally. • Drop a ball, it accelerates downward covering a greater distance each second. • x & ycomponents are completely independent of each other.

  33. 3.4Projectile Motion Projectile motion is separated into components. • Roll a ballhorizontally, v is constant, b/c no acceleration fromghorizontally. • Drop a ball, it accelerates downward covering a greater distance each second. • x & y components are completely independent of each other. • combined they cause curved paths.

  34. 3.4Projectile Motion • 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) vx vy

  35. 3.4Projectile Motion vx2 vx4 vx3 vy2 vy3 vy4

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

  37. 3.5Projectiles Launched at an Angle The Y distancefallen is the samevertical distance it would fall if dropped from rest. Remember d = ½gt2

  38. 3.5Projectiles Launched at an Angle vx is constant, but vy changes. At the max height, vy = 0.(only Vx) Height & Range

  39. 3.5Projectiles Launched at an Angle Height & Range launch angleaffectsheight (y) and range (x) more angle: -more initial vy, more height -less initial vx, less range height height 75o 60o range range

  40. 3.5Projectiles Launched at an Angle • angles that add to 90° have equal ranges • max range usually at 45° Height & Range

  41. 3.5Projectiles Launched at an Angle vup = –vdown Velocity & Time 0 m/s 12 m/s 10 m/s 12 m/s 12 m/s –10 m/s 20 m/s Is it safe to shoot a bullet in the air? 12 m/s 12 m/s –20 m/s

  42. 3.5Projectiles Launched at an Angle tup=tdown vup = –vdown Velocity & Time ttotal = (2)tup

  43. 3.5Projectiles Launched at an Angle Height & Range vxconstant, but vy changes At hmax, vy = 0 (only Vx horizontal) Velocity & Time more angle: -more initial vy, more height -less initial vx, less range vup = –vdown height tup=tdown ttotal = (2)tup range

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

  45. 3.5Projectiles Launched at an Angle Solvingprojectile calculation problems in 3 easy steps: • Direction: get Vix&Viy (pick Horiz.orVert.) • ListVariables d = vi = a = vf = t = • Pick equation, Plug numbers, and Solve.

  46. 3.5Projectiles 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 40o above horizontal. Solve for each of the following variables: vix = viy = tup = ttotal = (time of flight) dx = (range) dymax = (peak height) g = –10 m/s2 vx = v cos(θ) vy = v sin(θ) v = vi + at ttotal = (2)tup d = vit + ½at2

  47. 9.4 m/s Vi= 9.4 m/s at 40o above horizontal 40o vix = (9.4 m/s) • cos(40o) = viy = (9.4 m/s) • sin(40o) = tup = ttotal(up AND dn) = (2)(0.604 s) = 7.20 m/s 6.04 m/s vfy=viy+at 0 = 6.04 + –10t 0.604 s 0 – 6.04 = –10 1.21 s

  48. 9.4 m/s Vo= 9.4 m/s at 40o above x-axis 40o tup = ttotal = vix = viy = 0.604 s 7.20 m/s 6.04 m/s 1.21 s 0 dx = dymax = d=vixt+ ½at2 8.71 m d = (7.20 m/s)(1.21 s) d=viyt+ ½at2 1.82 m d = (6.04 m/s)(0.604 s) + ½(–10 m/s2)(0.604 s)2 =

  49. 0.604 s 6.04 m/s 1.21 s 1.82 m 8.71 m 7.20 m/s

  50. 3.5Projectiles Launched at an Angle Sample Calculation #2 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. vix = viy = 0 m/s t = dx = 35.0 m dymax = 22.0 m vix 22.0 m 35.0 m

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