Transcript Forces & Motion in Two Dimensions

```Motion
in Two Dimensions
Chapter 7.2
Homework
P. 158

9,10,11
P.160.

12,13.
Projectile Motion
What is the path of a projectile as it moves
through the air?
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Parabolic?
Straight up and down?
 Yes, both are possible.
What forces act on projectiles?
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Only gravity, which acts only in the negative ydirection.
Air resistance is ignored in projectile motion.
Choosing Coordinates & Strategy
For projectile motion:
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Choose the y-axis for vertical motion where
gravity is a factor.
Choose the x-axis for horizontal motion. Since
there are no forces acting in this direction (of
course we will neglect friction due to air
resistance), the speed will be constant (a = 0).
Analyze motion along the y-axis separate from the
x-axis.
If you solve for time in one direction, you
automatically solve for time in the other direction.
The Trajectory of a Projectile
•What does the free-body diagram look like for force?
Fg
The Vectors of Projectile Motion
What vectors exist in projectile motion?
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Velocity in both the x and y directions.
Acceleration in the y direction only.
vx (constant)
ax = 0
vy (Increasing)
ay
Why does
velocity
is thethe
velocity
constant
increase
in the y-direction?
in the x-direction?
•Gravity.
•No force acting on it.
Ex. 1: Launching a Projectile
Horizontally
A cannonball is shot horizontally off a
cliff with an initial velocity of 30 m/s. If
the height of the cliff is 50 m:
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How far from the base of the cliff does the
cannonball hit the ground?
With what speed does the cannonball hit
the ground?
Diagram the problem
vi
50m
Fg = Fnet
a = -g
vx
x=?
vy
vf = ?
State the Known & Unknown
Known:
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xi = 0
vix = 30 m/s
yi = 0
viy = 0 m/s
a = -g
y = -50 m
Unknown:
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x at y = -50 m
vf = ?
Perform Calculations (y)
y-direction:
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t
vy = -gt
y = viyt – ½ gt2
2y
g
2  50m
t
 3.2s
2
 9.81m / s
Using the first formula above:
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vy = (-9.8 m/s2)(3.2 s) = 31 m/s
Perform Calculations (x)
x-Direction
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x = vixt
x = (30 m/s)(3.2 s) = 96 m from the base.
Using the Pythagorean Theorem:
 v = vx2 + vy2
 v =  (30 m/s)2 + (31 m/s)2 = 43 m/s
Ex. 2: Projectile Motion above the
Horizontal
A ball is thrown from the top of the Science Wing
with a velocity of 15 m/s at an angle of 50 degrees
above the horizontal.
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What are the x and y components of the initial velocity?
What is the ball’s maximum height?
If the height of the Science wing is 12 m, where will the ball
land?
Diagram the problem
Fg = Fnet
a = -g
y
viy
vi = 15 m/s
 = 50°
vi = 15 m/s
 = 50°
vix
x
12 m
Ground
x=?
State the Known & Unknown
Known:
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xi = 0
yi = 12 m
vi = 15 m/s
 = 50°
a = -g
Unknown:
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ymax = ?
t=?
x=?
viy = ?
vix = ?
Perform the Calculations
(ymax)
y-direction:
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Initial velocity: viy = visin
vyi
vi = 15 m/s
 viy = (15 m/s)(sin 50°)
 = 50°
 viy = 11.5 m/s
vxi
Time when vfy = 0 m/s: vfy = viy – gt
 t = viy / g
 t = (11.5 m/s)/(9.81 m/s2)
 t = 1.17 s
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Determine the maximum height: ymax = yi +viyt – ½ gt2
 ymax = 12 m + (11.5 m/s)(1.17 s) – ½ (9.81 m/s2)(1.17 s)2
 ymax = 18.7 m
Perform the Calculations (t)
Since the ball will accelerate due to gravity over the
distance it is falling back to the ground, the time for
this segment can be determined as follows
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Time when ball hits the ground: ymax = viyt – ½ gt2
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Since yi can be set to zero as can viy,
t = 2*ymax/g
t =  2(18.7 m)/ (9.81 m/s2)
t = 1.95 s
By adding the time it takes the ball to reach its maximum
height to the time it it takes to reach the ground will give
you the total time.
 ttotal = 1.17 s + 1.95 s = 3.12 s
Perform the Calculations (x)
x-direction:
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Initial velocity: vix = vicos
vyi
vi = 15 m/s
 vix = (15 m/s)(cos 50°)
 = 50°
 vix = 9.64 m/s
vxi
Determine the total distance: x = vixt
 x = (9.64 m/s)(3.12 s)
 x = 30.1 m
Analyzing Motion in the x and
y directions independently.
x-direction:
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dx = vix t = vfxt
vix = vicos
y-direction:
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dy = ½ (vi + vf) t = vavg t
vf = viy + gt
dy = viy t + ½ g(t)2
vfy2 = viy2 + 2gd
viy = visin
Key Ideas
Projectile Motion:
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Gravity is the only force acting on a
projectile.
Choose a coordinate axis that where the xdirection is along the horizontal and the ydirection is vertical.
Solve the x and y components separately.
If time is found for one dimension, it is
also known for the other dimension.
There are only so many types of
problems:
Vix = constant
Viy = 0
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Viy >0 (shooting upwards at an angle)
Viy < 0 (shotting down at an angle)
a=g
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a <>g
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