Kinematics- 2 Dimensions · path. This path is parabolic in nature. ... • Example: A plane...
Transcript of Kinematics- 2 Dimensions · path. This path is parabolic in nature. ... • Example: A plane...
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Kinematics- 2 Dimensions
Projectile Motion
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What is Projectile?
Projectile -Any object which projected by some means and continues to move due to its own inertia (mass) in a curved path
Projectiles are objects in free fall that have an initial horizontal velocity.
Free fall means the only force acting is gravity
Initial horizontal velocity results in motion in two dimensions
Examples of projectiles include
Bullets after they leave the barrel of the gun
Baseballs after they are thrown, unless they are thrown straight up
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Projectiles move in TWO dimensions
Since a projectile moves in 2 dimensions, it therefore has 2 components just like a resultant vector.
Horizontal
Vertical
Horizontal and vertical motions are independent.
The common factor is time.
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Horizontal “Velocity” Component
NEVER changes, covers equal displacements in equal time periods. This means the initial horizontal velocity equals the final horizontal velocity
In other words, the horizontal
velocity is CONSTANT. (Vx = Vox)
BUT WHY?
Gravity DOES NOT work
horizontally to increase or
decrease the velocity. (ax = 0 m/s2)
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Vertical “Velocity” Component
Changes (due to gravity), does NOT cover equal displacements in equal time periods.
Both the MAGNITUDE and DIRECTION change.
As the projectile moves up the MAGNITUDE
DECREASES & its direction is UPWARD. As it
moves down the MAGNITUDE INCREASES and
the direction is DOWNWARD.
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Combining the Horizontal & Vertical Components
Together, these components produce what is called a trajectory or flying path. This path is parabolic in nature.
Component Magnitude Direction
Horizontal Constant Constant
Vertical Changes Changes
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Horizontally Launched Projectiles
Projectiles which have NO upward trajectory and NO initial VERTICAL velocity.
0 /oyv m s
constantox xv v
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Horizontally Launched Projectiles
To analyze a projectile in 2 dimensions we need 2 equations. One for the “x” direction and one for the “y” direction. And we use the following kinematic equation: 21
2oxx v t at
oxx v t
Remember, the velocity
is CONSTANT
horizontally, so that
means the acceleration is
ZERO!
212
y gt
Remember that since the
projectile is launched
horizontally, the INITIAL
VERTICAL VELOCITY is
equal to ZERO.
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Horizontally Launched Projectiles
• Example: A plane traveling with a horiz-ontal velocity of 100 m/s is 500 m above the ground. At some point the pilot decides to drop some supplies to designated target below. (a) How long is the drop in the air? (b) How far away from point where it was launched will it land?
What do I
know?
What I want to
know?
vox=100 m/s t = ?
y = 500 m x = ?
voy= 0 m/s
g = -9.8 m/s2
2 2
2
1 1500 ( 9.8)2 2
102.04
y gt t
t t
10.1 seconds
(100)(10.1)oxx v t 1,010 m
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Vertically Launched Projectiles
Component Magnitude Direction
Horizontal Constant Constant
Vertical Decreases up, 0 @
top, Increases down Changes
Horizontal Velocity
is constant
Vertical
Velocity
decreases
on the way
upward
Vertical Velocity
increases on the
way down,
NO Vertical Velocity at the top of the trajectory.
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Vertically Launched Projectiles
Since the projectile was launched at a angle, the velocity MUST be broken into components!!!
cos
sin
ox o
oy o
v v
v v
vo
vox
voy
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Vertically Launched Projectiles
There are several things you must consider when doing these types of projectiles besides using components. If it begins and ends at ground level, the “y” displacement is ZERO: y = 0
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Vertically Launched Projectiles
You will still use kinematic equation, but YOU MUST use COMPONENTS in the equation.
cos
sin
ox o
oy o
v v
v v
vo
vox
voy
oxx v t 212oyy v t gt
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Example
A place kicker kicks a football with a velocity of 20.0 m/s and at an angle of 53 degrees.
(a) How long is the ball in the air?
(b) How far away does it land?
(c) How high does it travel?
cos
20cos53 12.04 /
sin
20sin 53 15.97 /
ox o
ox
oy o
oy
v v
v m s
v v
v m s
53
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Example
A place kicker kicks a football with a velocity of 20.0 m/s and at an angle of 53 degrees.
(a) How long is the ball in the air?
What I know What I want to
know
vox=12.04 m/s t = ?
voy=15.97 m/s x = ?
y = 0 ymax=?
g = - 9.8 m/s2
2 2
2
1 0 (15.97) 4.92
15.97 4.9 15.97 4.9
oyy v t gt t t
t t t
t
3.26 s
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Example
A place kicker kicks a football with a velocity of 20.0 m/s and at an angle of 53 degrees.
(b) How far away does it land?
What I know What I want
to know
vox=12.04 m/s t = 3.26 s
voy=15.97 m/s x = ?
y = 0 ymax=?
g = - 9.8 m/s2
(12.04)(3.26)oxx v t 39.24 m
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Example
A place kicker kicks a football with a velocity of 20.0 m/s and at an angle of 53 degrees.
(c) How high does it travel?
CUT YOUR TIME IN HALF!
What I know What I want
to know
vox=12.04 m/s t = 3.26 s
voy=15.97 m/s x = 39.24 m
y = 0 ymax=?
g = - 9.8 m/s2
2
2
12
(15.97)(1.63) 4.9(1.63)
oyy v t gt
y
y
13.01 m
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X Motion-
Constant velocity
Y-Motion-
Constant Acceleration
tvx x
2
21 tatvy yoy
Equations of Motion for projectiles
2m/s 81.9ya
constant xfxox vvv
tavv yoyfy
0xa
yavv yoyfy 222
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A scared kangaroo once cleared a fence by jumping with a speed of 8.42 m/s at an angle of 55.2o from the ground. If the jump lasted 1.40 s, how high was the fence? What was the kangaroo’s horizontal displacement?
Hint: The time for the maximum height was half the total time.
Kangaroo Problem
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Step 1: Write down givens, unknowns and model.
x = ?
55.2o
t = 1.40s for
whole jump.
t= 0.70 s for
maximum
height.
Projectile launched at an angle
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Step 2: Decide on a plan or equation to use.
tvx x
For the horizontal
displacement:
For the vertical displacement:
2
21 tatvy yiy
2s
m81.9yaWe will use trigonometry to find the initial
velocities.
Note: The only triangle here is initial velocity and its components!
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Step 3: Carry out the plan.
)cos(i
voxv
)cos(i
voxv
sm91.6
)2.55sin()sm42.8(
)sin(
)sin(
oyv
oyv
ivoyv
iviy
v
o
)2.55cos( ) 42.8(sm ooxv
vox
voy
sm81.4oxv
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Step 3: Carry out the plan.
m 6.7s) 4.1)( 81.4(sm x
tvx xs 1.4 t 81.4 s
m xox vv
2
21 attvy oy
2s
m m/s 81.9 s 0.70 t 91.6 avoy
m 43.2403.2837.4
)s 7.0)(81.9(- s) 7.0)(91.6( 2
s
m21
sm
2
y
y
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Step 4: Check your answer.
x = 6.7 m
55.2o
t = 1.40s
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Baseball Problem
A baseball is thrown at an angle of 25o relative to the ground at a speed of 23.0 m/s. If the ball was caught 42.0 m from the thrower, how long was it in the air? How high was the tallest spot in the ball’s path?
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Baseball Problem
m/s 23.0 m 042 i v. x We will split the initial
velocity into x and y components.
We can use the x-equation to find the time it was in the air.
We can then use the y-equations to find the maximum height.
θ=25o
This is a projectile. The
x motion is constant
velocity. The y motion is
constant acceleration.
twhole flight =?
ΔyMAX = ?
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Baseball Problem
m/s 23.0 m 042 i v. x
t vx x tv
x
x
m/s 20.8
)25)(cos(023()25cos(sm
oxx
iox
vv
. v v oo
s 0.2 20.8
m 42
sm
t
vox
m/s 9.72
)25)(sin(023()25sin(sm
oo . v v ioy
vox
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Baseball Problem
The highest point will be half-way through the flight. t = ½ x 2.0 s = 1.0 s
m 8.4
s) 0.1)(m/s 81.9( - s) m/s)(1.0 72.9( 22
21
y
y
2
21 att vΔy oy