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Projectile Motion
Prior Knowledge
Review and Reflect.
 Guided question: What concepts
might we need to solve for the
motion of a projectile launched at an
angle, on the Earth?
 UM, UAM, Vectors, Resolving Vectors,
Gravity, Free fall, Independence of
Perpendicular Motions!
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Misconceptions: Diagnostics
Two balls A and B fall freely, dropped
simultaneously, with the height of ball A
greater then the height of ball B.
(a) Compare the velocity of the two balls
when they reach the ground.
(b) Compare the acceleration of the two
balls.
(c) Which ball will hit the ground first? Why?
Misconceptions: Diagnostics
A baseball is thrown at an angle from one
person to another.
(a) What is the velocity of the object at the
peak of the flight?
(b) What is the acceleration of the object at
the peak of the flight?
Misconceptions
Ball A falls freely from a specific height and
at the same time ball B is thrown
horizontally from the same height.
(a) Compare the velocity of the two balls
when they reach the ground.
(b) Compare the acceleration of the two
balls.
(c) Which ball will hit the ground first? Why?
Misconceptions
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Applets or animated gifs
are useful tools for
visual learners.
Here we see that heavier
objects hit the ground at
the same time as lighter
ones.
How might air resistance
affect this?
Common Misconceptions
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The parabolic motion of a projectile is different
from one-dimensional motions.
An object at rest drops to the ground before a
similar object initially pushed horizontally falls to
the ground.
Heavier objects fall faster.
At the top (peak) of its trajectory, the velocity of
a projectile is always momentarily zero.
The acceleration of the object is also zero, at the
peak.
Correcting Misconceptions
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Showing the parabolic nature of projectiles
leads the student to believe they are quite
different from linear falling or horizontal
motion.
An easy way to deal with this is to plot the
motion of a freely falling object and a
horizontally moving object on the same graph.
When the intersection points at the same
time, are connected, the result is a parabola.
Correcting Misconceptions
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Have students look at the horizontal and vertical
velocity vectors throughout parabolic flight.
Demonstrate that the velocity is not zero as there
must be some horizontal motion still (it doesn’t stop
in midair).
If you throw an object up in the air, it stops briefly
(speed is zero) but acceleration is not zero (gravity
doesn’t “shut off”). What is its role?
To change the direction of the velocity vector.
Is the velocity zero?
Misconcpetions
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The green ball
illustrates UM
The red ball illustrates
free fall and UAM
The combination of
these two motions at
equal time intervals
yields the parabolic
path shown by the blue
ball!
Independence is clear.
http://www.physicsclassroom.com/mmedia/vectors/bds.gif
http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/vectors/mzs.gif
Monkey and Banana
Projectile Motion
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Treat 2 dimensional problems as a
combination of two one-dimensional
motions.
In one dimension, the object undergoes
UM.
In the other dimension, the object
undergoes UAM.
We ignore air resistance.
Problem Solving
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Subscripts are used to denote vertical and
horizontal motions to alter our formulae as
follows:
sx = ux t
UM
sy = uy t + (½) at2
UAM
vy = uy + at
UAM
Projectile Problems
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Each problem is solved as a system of
equations in two independent motions.
Both motions (UM and UAM) occur at the
same time. (Thus Δt is exchangeable)
Feel free to use the other UAM equations
for problem solving.
Become familiar with the quadratic
equation.
Problem #1

A purple dinosaur is dropped from a cliff
40. m high. How long does it take to
strike the ground?
As the motion is all vertical:
sy = uyt + ½at2
- 40m = 0 + ½ (-9.81 m/s2) t2
t = 2.9 s
Problem #2
A biology teacher is thrown at 10.0 m/s
[E] from the top of a building that is 90.0
m tall. Calculate the horizontal
displacement of the “test dummy” from
the building upon initial contact with the
ground.
Step 1: Identify variables
uy = 0, ux = +10.0 m/s, a = -9.81 m/s2,
sy = - 90.0 m, t = ?, sx = ?
Step 2:Choose formulae
sx = uxΔt
To solve for sx we need Δt.
Use Vertical motion to find Δt
sy = uyΔt + ½aΔt2 with uy = 0
- 90.0 m = ½ (-9.81 m/s2) t2
t = 4.2835 s
Now sx = (10.0 m/s)(4.2835 s)
= 42.8 m [E]
Therefore, the annoying simpleton lands
42.8 m [E] of the base of the building.
Links
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http://www.physicsclassroom.com/shwave
/projectile
http://phet.colorado.edu/sims/projectilemotion/projectile-motion_en.html
Text work
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Page
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63
67
69
C, D
F
#4 – 11
# 17 - 31
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