11-14-11
Agenda
HOMEWORK CHECK
Place on your desk:
*Calculator & PROTRACTOR
*Orange homework chart
*Projectiles at an Angle/Quiz
 1) What is the speed of a projectile at
the top of the parabolic pathway?
1)
Warm-Up 5 min
2)
Turn-in Blue “Vector” Folder!
3)
Vocab. Words 10 min
4)
Finish Projectile Motion fill-inblank Notes. 10 min
5)
New Formulas 5 min
6)
Example Problems 5 min
7)
Virtual Lab: Projectiles at an
angle 45 min
8)
Take-Home Quiz
0 m/s
1. Axis: the straight line about which rotation takes
place.
2. Rotation: also called spin; the spinning motion that
takes place when an object rotates about an axis
located within the object.
3. Revolution: motion of an object turning around an
axis outside the object.
Projectile fired at an Angle:
When a projectile is fired at an angle with the
horizontal, the principle of independent velocities still
holds. The initial velocity of the projectiles can be
“resolved” into 2 components. One component is
directed vertically and the second is directed
horizontally. These components are treated separately
when solving problems.
Horizontally: Look at all the horizontal vectors---They are all
equally spaced apart. There is still no acceleration in the
horizontal direction so the cannonball moves equal horizontal
distances in equal time intervals---constant velocity.
Vertically: Look at all the vertical vectors---They get shorter
and shorten then disappear at the top of the path, then get
longer and longer. This is because there is acceleration
vertically (in the direction of the earth’s gravity). The
vertical velocity and therefore distance gets bigger and
bigger each second the object is still moving. Notice that the
horizontal component vector is the same length at any point
along the cannonball’s path. The vertical component gets
smaller then, disappears then, gets bigger.
These vectors represent the horizontal and vertical
component of velocity. The ACTUAL velocity of the
object is represented by the diagonal of the
parallelogram formed by the components.
At the top of the path, the vertical component
vanishes, or becomes zero, so the ACTUAL velocity of
the cannonball at the top of the path is the exact
same as the horizontal velocity at all other points
(instead of a combination of the horizontal and
vertical).
The above picture has the same launching speed as
the last picture and is the same idea as the last
picture except the object is projected at a steeper
angle.
Notice the initial velocity vector has a greater vertical
component than in the last picture when the
projection angle was smaller. This bigger vertical
component results in a higher path, but the horizontal
component is less so the range is less.
DUE TODAY:
• Projectile fill-in-blanks (KEEP)
• Virtual Lab
• Blue “Vector” Unit Folder
“There is no better high than discovery.”
E. O. Wilson
DUE NEXT CLASS:
• Projectile Motion [Take-Home Quiz]
• Study for Vocab. Quiz
• Bring Calculator & Protractor!
 Projectile
motion under the
influence of near-Earth--gravity will produce a curved
path.
 The vertical direction is on
the
“y “axis. The only force
acting on any projectile
vertically will be gravity. The
acceleration will be the
constant g = 9.806 65 m/s2.
 KEY:
What happens to a
particle vertically does not
affect its horizontal motion.
The opposite is also true.
The axes are independent.
 NOTE
what axis is CHANGING!!
Those projectiles that
accelerate only in a vertical
direction while moving at a
constant horizontal velocity
have a path that forms a
PARABOLA.
 The
range is the horizontal
distance.
 Solve for the time needed for the
projectile to reach its maximum
height
(remember that gravity is acting)
by using the “vertical” distance.
 Once you know the time of flight,
solve for the range by using:
DH = Horizontal velocity X
Time
 Satellite
– an object that
falls around Earth or some
other body rather than
falling into it because of its
tremendous speed.
◦Page 127 (Hewitt book)

If no force acts horizontally, the motion
found is at constant velocity- due to
inertia.
 The
equation of motion is then DH =
VH t.
 BUT, to find VH when you have 2dimensional motion, resolve the
initial velocity into its horizontal
component by using the formula :
VH = Vi cos q.
 To
find Vv , resolve the
initial velocity into its
vertical component.
 The formula to do this is:
Vv = Vi sin Ø
 The
time an object
moves horizontally is
the same as the time
the object moves
vertically.
 So, Total time is the
“time in the air” for a
projectile at-an-angle.
To find time:
Remember that gravity is affecting the motion
and therefore, the amount of time the
projectile is in the air for “at an angle” will be
TOTAL time.
 So,
ttotal = Vv / g so DH = VH ( t t)

A vertical and a horizontal
component. With the trajectory
being the initial velocity*
 Acceleration
is constant for a
projectile.
 Speed and velocity change at
each point along the
parabolic pathway.
 What is the speed of the
projectile at the very top of
its pathway?
 What is the acceleration of
the projectile at the very top
of its pathway?