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

2. 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.

3. ? 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

4. 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

5. 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

6. 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
√( ) = c
60 km/h

7. 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

8. 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.

9. 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

10. 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
√( ) = c
40 km/h
??? km/h
Check off the learning targets you can do after today.
30 km/h
3.2

11. 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.

12. 3.3 Components of Vectors
A ball’s velocity can be resolved into horizontal (x) and vertical (y) components.

13. 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

14. 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

15. 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

16. 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

17. 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

18. 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.

19. 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.

20. 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

21. 3.4 Projectile Motion
vx2
vx3
vx4
vy2
vy3
vy4

22. 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

23. 3.5 Projectiles Launched at an Angle
The Y distance fallen is the same vertical distance it would fall if dropped from rest.

24. 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)

25. 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

26. 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°

27. 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

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

29. 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

30. 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.

31. 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.

32. 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

33. 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 = –10t
0.604 s
0 – 6.04 =
–10
1.21 s

34. 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 =

35. 0.604 s
6.04 m/s
1.21 s
1.82 m
8.71 m
7.20 m/s

36. 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

37. √ 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

38. 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