Projectile Motion!
Physics 12
Mr. Reeves
•Velocity Components Review
•
•
A projectiles total
velocity is composed of
how many
components?
What are these
components?
Vertical (Y)
Velocity
Total Velocity
Horizontal (X)
Velocity
•Angry Birds!!
They’re angry. And following the laws of Physics!
•Today’s Objectives:
1
Review components of an objects velocity
2
Introduce the formula for projectile motion
3
Using the formula, students will be able to solve a projectile motion
problem when given initial conditions.
•The motion of an object can be explained
using this formula:
PROJECTILE MOTION:
•
D = Vi * t + (1/2) *a * t^2
•
D = displacement
Vi = Initial velocity
t = Time of travel
a = acceleration
•
•
•
•
*Remember, this formula can be applied to both
horizontal and vertical components!
•Steps to solving a projectile motion
problem
Step 1
•Make
sure to
label units for
each value.
• Record all known values from the
problem
Step 2
•The
better your
booking keeping
is, the easier your
calculations will
be!
• Use the vertical components to calculate
the time the ball will be in the air
Step 3
• Use the time and the horizontal
components to calculate the distance the
ball will travel
•Lets review projectile motion!
•
The boy on the tower throws a ball a distance of 20m. At
what speed is the ball thrown?
•Solution:
First let’s list our known values.
Horizontal Information
• Dx = 20m
• vix = ???
• ax = 0 m/s/s
•
Vertical Information
Dy = 4.9m
viy = 0 m/s
ay = 9.8 m/s/s
•We can use the vertical components of
the ball’s motion to determine the time it
will travel.
Dy = Viy*t+(1/2)Ay*t^2
4.9m = (0 m/s)•t + 0.5•(9.8 m/s/s)•t2
4.9m = (4.9 m/s/s)•t2
1 s2 = t 2
t=1s
We now use the time which we calculated
and our horizontal components to
calculate the initial velocity.
Dx = Vix*t+(1/2)Ax*t^2
20m = (Vix)•(1s) + 0.5•(0 m/s/s)•(1s)2
20m = (Vix)•(1s)
Vix= 20m/1s = 20m/s
•Lets try another!
•
•
A pool ball leaves a 0.60meter 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.
•Solution:
First let’s list our known values.
Horizontal Information
• Dx = ???
• vix = 2.4 m/s
• ax = 0 m/s/s
Vertical Information
Dy = 0.60 m
viy = 0 m/s
ay = 9.8 m/s/s
•
2.4 m/s
|
.6m
|
<---???--->
9.8m/s/s
•We can use our vertical components to
find time!
Dy = Viy*t+(1/2)Ay*t^2
0.60 m = (0 m/s)•t + 0.5•(9.8 m/s/s)•t2
0.60 m = (4.9 m/s/s)•t2
0.122 s2 = t2
t = 0.350 s
Now we can use our horizontal
components to determine where the ball
will land.
Dx = Vix*t+(1/2)Ax*t^2
Dx = (2.4 m/s)•(0.3499 s) + 0.5•(0 m/s/s)•(0.3499 s)2
Dx = (2.4 m/s)•(0.3499 s)
Dx = 0.84 m
•Theory
At the instant a horizontally held rifle is fired, a bullet
held at the rifle’s side is released and drops to the
ground. Which bullet strikes the ground first?
http://www.youtube.com/watch?v=D9wQVIEdKh8
•Before you leave!
On a separate paper answer these questions:
•Does
the time of travel for a projectile depend on the
horizontal components or vertical components of its
motion? How about the distance it travels?
•What
is one thing about today’s lesson that you still
have questions about? If there is nothing, what is one
thing you learned today?
Have a good day!