#### Transcript CHAPTER 6

```CHAPTER 6
MOTION IN 2 DIMENSIONS
PROJECTILE MOTION
• Projectile- any object that is shot or thrown
through the air. (bullet, ball, drop of water,
etc.)
– After the initial thrust (force on the object) the
only force acting on the object is the force of
gravity, if air resistance is ignored.
– The path of all projectiles is going to be a
parabola; generally moves upward for a distance
reaches a maximum height and the travels
downward.
Trajectory- the path a projectile takes
through space; If the initial force of thrust
is known you can calculate the trajectory.
Important parts of an object’s trajectory.
• Maximum height
• Distance object travels- range- X component
of the motion- horizontal motion
• Flight time
Independence of motion
• The parabolic path of an object is the
combination of 2 separate motions:
 Vertical
 Horizontal
+
What if we viewed the 2 people from the previous slide
are viewed from behind the person throwing the ball?
What would the path of the ball look like?
It would appear as if the ball went straight up and then
down.
What if the 2 people were viewed from directly above?
What would the path of the ball look like?
It would appear to be moving from one player to the
other in a straight line at constant speed
A ball is dropped
and another is
given a horizontal
velocity of 2.0
m/s. What is
trajectories of the
balls? Why do the
balls behave this
way?
The horizontal motion of the
ball is uniform because the
velocity is constant. There is
no acceleration because
there is no horizontal force
acting on the ball; The
horizontal velocity is always
going to be equal to the
initial velocity.
The vertical motion is shown
to be accelerating because of
the force of gravity
Combining the 2 motions
created from the velocity
vectors gives the parabolic
pathway.
PROBLEM SOLVING
1. Separate the projectile motion problem into a vertical
motion problem and a horizontal motion problem
2. The vertical motion problem is exactly like an object being
dropped straight up or down.
1
2
df  di  vi (tf )  a (tf )
2
• a is equal to -9.8 m/s2
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.
1
df  di  vi (tf )  a (tf 2 )
2
.6= ½ (9.8)t2
t2= .122
t= .35 s
d= vt
d= (2.4) .35s
d= .84 m
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.
1
df  di  vi (tf )  a (tf 2 )
2
22m= 0 + (0)t + ½(9.8)t2
22 = -4.9 t2
t2= 4.49
t= 2.11 s
d= vt
35 m= v (2.11)
v = 16.6 m/s
HOMEWORK
pg. 150 1-3
Projectiles launched at an angle
• When a projectile is launched at an angle it
has a vertical component and a horizontal
component
• As the object is moving up it is slowing
down
• As the object comes down the object is
speeding up
• The velocities at each point in the vertical
direction are the same just in different
directions
• Maximum Height is when the velocity is
equal to 0 m/s
the horizontal component of an object launched at an angle
Equations for projectile motion problems
Time of Flight
Total distance travelled or Range
Maximum Height
A projectile is fired at 12.5 m/s at an angle of 53.1° with the horizontal from a
point 75.0 m above the ground.
a) How long does it take to reach the ground?
b) What maximum height does it reach?
c) What horizontal distance does it travel before striking the ground?
d) With what velocity does it strike the ground?
```