#### Transcript Lesson 2: Projectile Motion

```10/30 do now – on a new sheet
• What is the difference between g and Fg?
Homework: castle learning
Due Today
• Vector packet correction
Due Monday
• Newton’s Laws packet correction
Projectile Motion objectives
1. What is a Projectile?
2. Characteristics of a Projectile's Trajectory
3. Describing Projectiles with Numbers
a. Horizontal and Vertical Components of Velocity
b. Horizontal and Vertical Components of Displacement
4. Initial Velocity Components
5. Horizontally Launched Projectiles - Problem-Solving
6. Non-Horizontally Launched Projectiles - ProblemSolving
What is a projectile?
• An object that is launched into the air with
some INITIAL VELOCITY
• Can be launched at ANY ANGLE
• In FREEFALL after launch (no outside forces
except force of gravity)
• The path of the projectile is a PARABOLA
Free body diagram of a projectile
Fg, a
Fg, a
Fg, a
Fg, a
Fg, a
Fg, a
Fg, a
Fg, a
Fg, a
Path of a projectile
The path of a projectile is parabolic, or arc. The object
moves vertically as well as horizontally.
Projectile Motion and Inertia
• Since a projectile is in free fall after it is launched, the only
force acting on it is gravity, which influence the vertical
motion of the projectile, causing a vertical acceleration.
• The horizontal motion of the projectile is the result of inertia.
There is no horizontal force.
• The combination of inertia and gravity causes the parabolic
trajectory that is characteristic of projectiles.
Effect of air resistance
Ideal projectile
Projectile with air
resistance
Characteristics of a Projectile's
Trajectory
• There are the two components of the projectile's motion –
horizontal and vertical motion. These two perpendicular
components of motion are independent of each other, which
means one component does not affect the other component.
Horizontal components
Fx  0
ax  0
v x  constant
d x v x t
Vertical components
Fy  m  g
a y  g
v y  viy  a y  t
1
2
d y  viy  t  a y  t
2
Projectile components
• There are two components in projectile motion:
horizontal and vertical. These two components are
independent of each other.
• Since the only force acting on the projectile is gravity,
which is in the vertical downward direction, it cause
vertical acceleration only. The projectile’s vertical motion
is the same as if it is in free fall with ay = -g, the vertical
velocity changes by -9.81 m/s every second.
• There is no force acting on the projectile in horizontal
direction. The horizontal acceleration is zero. The
horizontal motion is only affected by inertia, which means
its horizontal velocity is constant.
Horizontal
Motion
Vertical
Motion
Forces
(Present? - Yes or No)
(If present, what dir'n?)
Acceleration
(Present? - Yes or No)
(If present, what dir'n?)
Velocity
(Constant or
Changing?)
Horizontal and vertical components are independent of each other. Change of horizontal speed
does not affect vertical motion. Change of vertical speed does not affect horizontal motion.
Motion graphs of projectile
d vs. t graphs
Horizontal motion
Vertical motion
dx
dy
t
t
v vs. t graphs
vx
vy
t
t
What we know about projectile motion
•
•
•
•
•
A projectile is any object upon which the only force is gravity,
projectiles travel with a parabolic trajectory due to the
influence of gravity.
In horizontal direction: there are no horizontal forces acting
upon projectiles and thus no horizontal acceleration. The
horizontal velocity of a projectile is constant (a never
changing in value).
In vertical direction: there is a vertical acceleration caused by
gravity; its value is 9.81 m/s/s, down. The vertical velocity of a
projectile changes by -9.81 m/s each second
The horizontal motion of a projectile is independent of its
vertical motion.
When solving projectile problems, we must separate
horizontal and vertical parameters. Apply equations only in
horizontal or vertical direction.
Example
• A student throws a 5.0-newton ball straight
up. What is the net force on the ball at its
maximum height?
Lab – Determine Initial Velocity of a
launched ball
Name, Partners Names, Date
Purpose: what is the initial velocity of a launched ball?
Material: launcher, meter stick, metal ball, clamp
Data table: make your own data table
Data Analysis
Conclusion
Describing Projectiles with Numbers:
Horizontal and Vertical Velocity
Horizontal components:
Fx  0
ax  0
v x  v ix
vertical components: Object
thrown up and down.
Fy  m  g
ay  g
v y  viy  g  t
vy decreases while up,
increase while down.
Example: cannon ball is launched with vix = 73.1 m/s and
viy = 19.62 m/s upward at time t = 0.0 s. Fill in the blanks.
Time
0s
1s
2s
3s
4s
5s
ax
ay
vx
vy
11/2 do now
1. Describe a projectile’s path. (1)
2. Describe force, acceleration and velocity in
the horizontal direction of a projectile. (3)
3. Describe force, acceleration and velocity in
the vertical direction of a projectile. (3)
4. What does it mean when we say the
horizontal motion of a projectile is
independent of its vertical motion? (1)
Homework: castle learning
Due today
Newton’s Laws packet correction
Due Friday
Newton’s Laws test correction
Newton’s Laws practice packet correction
The symmetrical nature of a projectile
launched at an angle
vy = 0 at top
vfy = -viy
tup = tdown
ttotal = 2∙tup
Describing Projectiles With Numbers:
(Horizontal and Vertical Displacement)
Horizontal components:
Fx  0
ax  0
v x  v ix
d x v ix t
vertical components:
Object thrown up then
come down
Fy  m  g
ay  g
v y  viy  g  t
1
d y v iy t  g  t 2
2
2
2
v fy  viy  2 gy
Example: cannon ball is launched with vix = 73.1 m/s and
viy = 19.62 m/s upward at time t = 0.0 s. Fill in the blanks.
Time
Time
a
axx
a
ayy
v
vxx
v
vyy
d
dxx
d
dyy
00 ss
11 ss
22 ss
33 ss
44 ss
55 ss
The symmetrical nature of a projectile's trajectory: the vertical displacement of a projectile
t seconds before reaching the peak is the same as the vertical displacement of a projectile t
seconds after reaching the peak.
The horizontal distance traveled by the projectile each second is a constant value.
Horizontal and vertical components are
independent of each other
• The horizontal and vertical motions of a projectile are
independent of each other. The horizontal velocity of a
projectile does not affect how far (or how fast) a projectile falls
vertically.
• Only vertical motion parameters (initial vertical velocity, final
vertical velocity, vertical acceleration) determine the vertical
displacement.
• Only horizontal motion parameters (initial horizontal velocity,
final horizontal velocity, horizontal acceleration). Determine
the horizontal displacement.
• One of the initial steps of a projectile motion problem is to
determine the components of the initial velocity.
Initial Velocity Components
• Since velocity is a vector quantity, vector resolution is
used to determine the components of velocity.
SOH CAH TOA
sinθ = viy / vi
vi
viy
viy = vi∙sinθ
cosθ = vix / vi
θ
vix = vi∙cosθ
vix
Special case: horizontally launched projectile:
θ = 0o
viy = visinθ = visin0o = 0;
vix = vicosθ = vicos0o = vi
In these
equations,
angle θ is
with x axis
Example
• The vector diagram below
represents the velocity of a car
traveling 24 meters per second
35° east of north. What is the
magnitude of the component
of the car’s velocity that is
directed eastward?
•
Ken Fused is resolving velocity vectors into
horizontal and vertical components. For each case,
evaluate whether Ken's diagrams are correct or
incorrect. If incorrect, explain the problem or make
the correction.
Practices – determine horizontal and vertical
components of velocity
1. A water balloon is launched with a speed of 40 m/s at an angle
of 60 degrees to the horizontal.
2. A motorcycle stunt person traveling 70 mi/hr jumps off a
ramp at an angle of 35 degrees to the horizontal.
3. A springboard diver jumps with a velocity of 10 m/s at an
angle of 80 degrees to the horizontal.
Why components?
• The point of resolving an initial velocity vector into
its two components is to use the values of these two
components to analyze a projectile's motion and
determine such parameters as dx, dy, vfy, ttotal, tup, etc.
• ALWAYS USE COMPONENTS IN YOUR
EQUATIONS!!!
Determination of the TIME OF FLIGHT for
projectile launched at an angle,
given initial speed and angle:
For a projectile with given launching angle and initial velocity, we
can determine the initial horizontal and vertical velocity: viy = visinθ;
vix = vicosθ
Its vertical motion is the same as free fall with initial vertical velocity:
vy = 0 at top;
vfy = -viy
vfy = viy + a∙t
0 = viy - g∙tup
viy
tup 
g
tup = tdown
ttotal = 2∙tup
ttotal  2tup 
2viy
g
If we know the initial vertical velocity, we can determine the time to
reach the highest the point and the total time of flight.
Time is the same for both vertical and horizontal components
Example
• A projectile is launched with 15 m/s -velocity
at an angle of 60.° above the horizontal. What
is the total flight time of projectile?
Example
• A projectile is launched at an angle above the
ground. The horizontal component of the
projectile’s velocity, vx, is initially 40. m/s. The
vertical component of the projectile’s velocity,
vy, is initially 30. m/s. What is the projectile’s
time of flight? [Neglect friction.]
Example
A football is thrown at an angle of 30.° above the horizontal. The
magnitude of the horizontal component of the ball’s initial velocity is
13.0 meters per second. The magnitude of the vertical component of
the ball’s initial velocity is 7.5 meters per second. [Neglect friction.]
1. Sketch a graph representing the relationship between the
horizontal displacement of the football and the time the football
is in the air. 
2. The football is caught at the same height from which it is thrown.
Calculate the total time the football was in the air. [Show all
work, including the equation and substitution with units.]
11/3 do now
• The vector diagram below
represents the velocity of a car
traveling 24 meters per second
30° east of north. What is the
magnitude of the component
of the car’s velocity that is
directed eastward?
Homework: castle learning
Due Friday
Newton’s Laws test correction
Newton’s Laws practice packet correction
30o
Determination of HORIZONTAL DISPLACEMENT
for projectile launched at an angle,
given initial speed and angle and time of flight
•
For the given projectile, we can determine the initial horizontal
and vertical velocity: viy = visinθ; vix = vicosθ
•
For a projectile launched at an angle, its horizontal motion is
constant. To determine its horizontal displacement we can use
x  vix  t
Range  vix  ttotal  vix 
2vi  cos 
Range  vi  sin  
g
2viy
g
vi  sin 2
Range 
g
Range is the longest when θ is 45o
2
Example
• A projectile is launched with 15 m/s -velocity at an angle of
60.° above the horizontal. What is the range of the projectile?
Example
• A projectile is launched at an angle above the
ground. The horizontal component of the projectile’s
velocity, vx, is initially 40. m/s. The vertical
component of the projectile’s velocity, vy, is initially
30. m/s. What is the projectile’s range? [Neglect
friction.]
Determination of the PEAK HEIGHT for
projectile launched at an angle,
given initial speed and angle
For the given projectile, we can determine the initial horizontal
and vertical velocity: viy = vi∙sinθ; vix = vi∙cosθ
For a projectile launched at an angle, its vertical motion is the
same as free fall with initial vertical velocity:
• At the very top, vy = 0
2
• vy2 = viy2 + 2ay∙y
iy
• 0 = viy2 – 2g∙ypeak (ay = -g)
y peak 
v
2g
Example
• A projectile is launched with 15 m/s -velocity
at an angle of 60.° above the horizontal. What
is the maximum height reached by the
projectile?
Example
• A projectile is launched at an angle above the
ground. The horizontal component of the projectile’s
velocity, vx, is initially 40. m/s. The vertical
component of the projectile’s velocity, vy, is initially
30. m/s. What is the maximum height reached by the
projectile? [Neglect friction.]
Total flight time, range, max height
Time to go up
t = visinθ / g
As θ increases, flight time
increase.
Max time: θ = 90o
Range
Range = vi2sin2θ /g
Maximum range: θ = 45o
Max height
hmax = (visinθ)2/2g
As θ increases, flight height
increase.
Max height: θ = 90o
Example
• A football is kicked with an initial velocity of 25 m/s at an
angle of 45-degrees with the horizontal. Determine the time
of flight, the horizontal displacement, and the peak height of
the football.
example
• A long jumper leaves the ground with an initial velocity of 12
m/s at an angle of 28-degrees above the horizontal.
Determine the time of flight, the horizontal distance, and the
peak height of the long-jumper.
Example: Fill in the blanks
11/4 do now - Fill in the blanks
Homework: castle learning
Due Friday
Newton’s Laws test correction
Newton’s Laws practice packet correction
Example - Fill in the blanks
14.9
2.93
19.2
80
19.7
5.85
16.1
1.64
3.28
19.7
2
4.01
0.36
0.71
3.5
42
164
240
Horizontally Launched Projectile
special case: vix = vi, viy = 0
A projectile is launched with an initial horizontal velocity from an
elevated position and follows a parabolic path to the ground.
Horizontal direction
• ax = 0
• vix = vi
• dx=vi∙t
vertical direction
• ay = -g
• viy = 0, vy = -g∙t
• dy = - ½ g∙t2
• Predictable unknowns
include the initial speed of
the projectile, the initial
height of the projectile, the
time of flight, and the
horizontal distance of the
projectile.
Example
• A pool ball leaves a 0.60-meter high table with an initial horizontal
velocity of 2.4 m/s. Predict the time required for the pool ball to fall
to the ground and the horizontal distance between the table's edge
and the ball's landing location.
Example
• A soccer ball is kicked horizontally off a 22.0-meter high hill
and lands a distance of 35.0 meters from the edge of the hill.
Determine the initial horizontal velocity of the soccer ball.
example
•
1.
A cannon elevated at an angle of 35° to the horizontal fires a cannonball,
which travels the path shown in the diagram. [Neglect air resistance and
assume the ball lands at the same height above the ground from which it
was launched.] If the ball lands 7.0 × 102 meters from the cannon 14.0
seconds after it was fired,
what is the horizontal component of its initial velocity?
2.
what is the vertical component of its initial velocity?
11/5 Do now
• A student throws a 5.0-newton ball straight
up. What is the net force on the ball at its
maximum height?
A. 0.0 N
B. 9.8 N
C. 5.0 N
D. 4.9 N
objectives
• Projectile review
• Due Friday
– projectile motion (physics classroom)
• Due Monday
– Regents Physics – Projectile Practice
• Due Tuesday
– Worksheet 1.2.3-ground launched projectile
• Projectile test is on Tuesday
11/6 Do now
A student throws a 15.0newton ball straight up.
1. What is the net force on
the ball at its maximum
height?
2. What is the acceleration
of the ball at its
maximum height?
3. What is the velocity of
the ball at its maximum
height?
objectives
• Projectile review
• Due today
– projectile motion (physics classroom)
– Newton’s Laws test correction
– Newton’s Laws practice packet correction
• Due Monday
– Regents Physics Projectile Practice
• Due Tuesday
– Worksheet 1.2.3-ground launched projectile
• Projectile test is on Tuesday
11/9 do now
1. A cannonball is fired from a cliff that is 50 meters
above the ground. The cannonball is fired horizontally
with a speed of 120 meters per second. Calculate the
horizontal distance that the cannonball will travel.
objectives
• Projectile review
• Due today
– Regents Physics Projectile Practice
• Due tomorrow
– Worksheet 1.2.3-ground launched projectile
• Projectile test is on Tuesday
• Due Thursday
– projectile test extra credit (kinematic graphing)
• Due Monday
– Projectile project:
http://www.physicsclassroom.com/shwave/projectile
11/10 objectives
• Projectile test
• Due today
– Worksheet 1.2.3-ground launched projectile
• Due Thursday
– projectile test extra credit (kinematic graphing)
• Due Monday
– Projectile project:
http://www.physicsclassroom.com/shwave/projectile
11/10 do now
• A ball rolls toward the edge of a table that is 110
centimeters high and lands 2.0 meters away from its
edge. Determine the speed with which the ball rolls
off the edge.
11/10 objectives
• Projectile test
• Due today
– Worksheet 1.2.3-ground launched projectile
• Due Thursday
– projectile test extra credit (kinematic graphing)
• Due Monday
– Projectile project:
http://www.physicsclassroom.com/shwave/projectile
NAMES, DATE, TITLE
PURPOSE
How far will a horizontally launched ball land?
MATERIALS launcher, meter stick, target paper
PROCEDURE
• Write a paragraph to describe how the lab is done so that