Transcript Projectile motion - netBlueprint.net

PROJECTILE MOTION
Projectile Examples
Hockey puck
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• Basketball
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• Volleyball
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• Arrow
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• Shot put
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• Javelin
These are all examples of things that are
projected, then go off under the
influence of gravity
•
Tennis ball
Golf ball
Football
Softball
Soccer ball
Bullet
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Not projectiles
•
•
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Jet plane
Rocket
Car (unless it looses contact with ground)
Understanding Projectiles
 The key to understanding
projectile motion is to realize
that gravity acts vertically
 it affects only the vertical
part of the motion, not the
horizontal part of the motion
Demonstration
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•
•
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We can see that the
horizontal and vertical
motions are independent
The red ball falls
vertically
The yellow ball was
given a kick to the right.
They track each other
vertically step for step
and hit the ground at the
same time
Projectile Paths
In the absence of gravity a bullet
would follow a straight line forever.
With gravity it FALLS AWAY from
that straight line!
Shoot the Monkey
Sample Problem
A zookeeper finds an escaped monkey hanging
from a light pole. Aiming her tranquilizer gun
at the monkey, she kneels 10.0 m from the
light pole,which is 5.00 m high.
The tip of her gun is 1.00 m above the ground.
At the same moment that the monkey drops a
banana, the zookeeper shoots.
If the dart travels at 50.0 m/s,will the dart hit the
monkey, the banana, or neither one?
Sample Problem
1 . Select a coordinate system.
The positive y-axis points up, and the positive x-axis
points along the ground toward the pole. Because
the dart leaves the gun at a height of 1.00 m, the
vertical distance is 4.00 m.
Sample Problem
2 . Use the inverse tangent function to find
the angle that the initial velocity makes
with the x-axis.
 y 
1  4.00 m 
  tan    tan 
  21.8
 x 
 10.0 m 
1
Sample Problem
3 . Choose a kinematic equation to
solve for time.
Rearrange the equation for motion along the x-axis to
isolate the unknown t, which is the time the dart takes to
travel the horizontal distance.
x  (vi cos  )t
x
10.0 m
t 

 0.215 s
vi cos  (50.0 m/s)(cos 21.8)
Sample Problem
4 . Find out how far each object will
fall during this time. Use the free-fall
kinematic equation in both cases.
For the banana, vi = 0. Thus:
yb = ½ay(t)2 = ½(–9.81 m/s2)(0.215 s)2 = –0.227 m
Sample Problem
The dart has an initial vertical component of velocity
equal to vi sin , so:
yd = (vi sin )(t) + ½ay(t)2
yd = (50.0 m/s)(sin 21.8)(0.215 s) +½(–9.81
m/s2)(0.215 s)2
yd = 3.99 m – 0.227 m = 3.76 m
Sample Problem
5 . Analyze the results.
Find the final height of both the banana and the dart.
ybanana, f = yb,i+ yb = 5.00 m + (–0.227 m)
ybanana, f = 4.77 m above the ground
Sample Problem
ydart, f = yd,i+ yd = 1.00 m + 3.76 m
ydart, f = 4.76 m above the ground
The dart hits the banana.
The slight difference is due to rounding.
No gravity is good for kickers
Newton’s First Law of Motion
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“Every object continues in its state of rest, or
of uniform motion in a straight line, unless it is
compelled to change that state of motion by
forces impressed upon it ”
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The tendency of matter to maintain its state
of motion is known as INERTIA.
Basketball – without gravity
Hitting the target – aim high, not
directly at the target
BULLSEYE!
Height
Path of the Projectile
falling
rising
g
v
v
projectile
Distance downfield
(range)
Vertical
velocity
Horizontal velocity
Horizontal Motion
vx  v0 x  axt
ax  0

vx  v0 cos 0
x  x0  v0 xt  axt
1
2
2
x  x0   v0 cos 0  t
Vertical Motion
y  y0  v0 yt  a yt
1
2
(2  15)
2
y  y0   v0 sin  0  t  gt
1
2
2
(4  22)
v y  v0 y  a yt
(2  11)
v y  v0 sin  0  gt
(4  23)
Projectile motion – key points
1)
2)
3)
4)
The projectile has both a vertical and
horizontal component of velocity
The only force acting on the projectile once
it is shot is gravity (neglecting air
resistance)
At all times the acceleration of the projectile
is g = 9.8 m/s2 downward
The horizontal velocity of the projectile
does not change throughout the path
Key points, continued
5)
6)
7)
On the rising portion of the path gravity
causes the vertical component of velocity to
get smaller and smaller
At the very top of the path the vertical
component of velocity is ZERO
On the falling portion of the path the vertical
velocity increases
More key points
8)
9)
10)
If the projectile lands at the same elevation as
its starting point it will have the same vertical
SPEED as it began with
The time it takes to get to the top of its path is
the same as the time to get from the top back to
the ground.
The range of the projectile (where it lands)
depends on its initial speed and angle of
elevation
Sample Problem
A 2.00 m tall basketball player wants to make a basket from
a distance of 10.0 m. If he shoots the ball at a 450 angle, at
what initial speed must he throw the ball so that it goes
through the hoop without striking the backboard?
y
y0
x
Equations to Choose from
y  y0  v0 yt  a yt
1
2
vx  v0 x  axt
(2  15)
2
y  y0   v0 sin  0  t  gt
1
2
v y  v0 y  a yt
v y  v0 sin  0  gt
2
ax  0
(4  22)

vx  v0 cos 0
(2  11)
x  x0  v0 xt  12 axt 2
(4  23)
x  x0   v0 cos 0  t
Maximum Range
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When an artillery shell is fired the initial speed of
the projectile depends on the explosive charge –
this cannot be changed
The only control you have is over the angle of
elevation.
You can control the range (where it lands) by
changing the angle of elevation
To get maximum range set the angle to 45°
Interactive
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http://galileo.phys.virginia.edu/classes/109N/
more_stuff/Applets/ProjectileMotion/jarapplet
.html
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http://jersey.uoregon.edu/vlab/Cannon/
The ultimate projectile: Orbit
• Imagine trying to
throw a rock around
the world.
• If you give it a large
horizontal velocity,
it will go into orbit
around the earth!