Unit 2: Projectile Motion (Chp 3) Projectile motion can be described by vertical components and horizontal components of motion.

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

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

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

Consider a plane flying 80 km/h north, but… 3.2 Velocity Vectors 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. 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

vector addition: Consider a plane flying 80 km/h north, but… 3.2 Velocity Vectors 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. vector addition: 80 km/h 100 km/h draw vectors tail-to-head. a2 + b2 = c2 (80)2 + (60)2 = c2 √(6400 + 3600) = c 60 km/h

The 80 km/h and 60 km/h vectors produce a 3.2 Velocity 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

Suppose that an airplane normally flying at 3.2 Velocity 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 course by a predictable amount.

Which of these expresses a vector quantity? 10 kg 10 kg to the north Quick Quiz! Which of these expresses a vector quantity? 10 kg 10 kg to the north 10 m/s 10 m/s to the north 3.1

Check off the learning targets you can do after today. 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

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

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

vy = ? vx = ? vy = v sin(θ) vx = v cos(θ) vy vx (v) 340 m/s 3.3 Components 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) 340 m/s sin(θ) = opp/hyp (hyp) vy opp cos(θ) = adj/hyp adj 60o vx

vy = ? vx = ? vy = v sin(θ) vx = v cos(θ) vy = (340 m/s) • sin(60) vy 3.3 Components 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) 340 m/s vy = (340 m/s) • sin(60) vy = (hyp) vy opp 294 m/s 294 m/s vx = (340 m/s) • cos(60) vx = adj 60o vx 170 m/s 170 m/s

A ball launched into the air at 45° to the horizontal initially has… 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

vy = v sin(θ) vy = (680 m/s) • sin(30) 680 m/s 340 m/s 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) vy = (680 m/s) • sin(30) 680 m/s 340 m/s 30o 3.3

3.4 Projectile Motion projectile: any object moving through 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

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

Projectile motion is separated into components. Roll a ball horizontally, v is constant, b/c no acceleration from g horizontally. 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.

x component is constant (a = 0) (g acts only in y direction) 3.4 Projectile 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

3.4 Projectile Motion vx2 vx3 vx4 vy2 vy3 vy4

in the forward x direction it was thrown 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

3.5 Projectiles Launched at an Angle The Y distance fallen is the same vertical distance it would fall if dropped from rest.

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

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

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

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

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

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

10 m/s for every second in the air. the same as the time going upward. 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.

Solving projectile calculation problems in 3 easy steps: 3.5 Projectiles Launched at an Angle Solving projectile calculation problems in 3 easy steps: Direction: get Vix & Viy (pick Horiz. or Vert.) List Variables d = vi = a = v = t = Pick equation, Plug numbers, and Solve.

vx = v cos(θ) vy = v sin(θ) vix = viy = tup = 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 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

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

tup = ttotal = vix = viy = dx = dymax = Vo = 9.4 m/s at 40o above x-axis 9.4 m/s 40o tup = ttotal = vix = viy = 7.20 m/s 0.604 s 6.04 m/s 1.21 s 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 =

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

vix = viy = 0 m/s t = vix dx = 35.0 m dymax = 22.0 m 3.5 Projectiles 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

√ vix viy = 0 m/s dx = 35.0 m dymax = 22.0 m t = vix = 3.5 Projectiles Launched at an Angle vix Sample Calculation #2 viy = 0 m/s dx = 35.0 m dymax = 22.0 m t = vix = 22.0 m 35.0 m 2.10 s d = viyt + ½at2 –22.0 = ½(–10)t2 √ 2(–22.0 m) = –10 d = vixt + ½at2 35.0 = vix(2.10) 16.7 m/s 35.0 m = 2.10 s

θ v v Horizontal Launch viy = 0 m/s vix = v 3.5 Projectiles Launched at an Angle v v θ Horizontal Launch viy = 0 m/s vix = v Angled Launch viy = v sin(θ) vix = v cos(θ) For ALL launches: a = g = –10 m/s2 for vertical motion a = 0 m/s2 for horizontal motion t is found vertically with: v = vi + gt or d = ½gt2