Transcript Projectile motion - netBlueprint.net

PROJECTILE
MOTION
Projectile Examples
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Tennis ball
Golf ball
Football
Softball
Soccer ball
Bullet
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Hockey puck
Basketball
Volleyball
Arrow
Shot put
Javelin
These are all examples of things that are
projected, then go off under the
influence of gravity
Not projectiles
• Jet plane
• Rocket
• Car (unless it looses contact with ground)
 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
• 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
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?
1 . Select a coordinate system.
The positive y-axis points up, and the positive xaxis 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.
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
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)
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
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
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
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.
Football without gravity
No gravity is good for kickers
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) The projectile has both a vertical and
horizontal component of velocity
2) The only force acting on the projectile
once it is shot is gravity (neglecting air
resistance)
3) At all times the acceleration of the
projectile is g = 9.8 m/s2 downward
4) The horizontal velocity of the projectile
does not change throughout the path
Key points, continued
5) On the rising portion of the path gravity
causes the vertical component of velocity
to get smaller and smaller
6) At the very top of the path the vertical
component of velocity is ZERO
7) On the falling portion of the path the
vertical velocity increases
More key points
8) If the projectile lands at the same
elevation as its starting point it will have
the same vertical SPEED as it began with
9) 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.
10)The range of the projectile (where it
lands) depends on its initial speed and
angle of elevation
Example:
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
vx  v0 x  axt
x  x0  v0 xt  12 axt 2
ax  0
x  x0   v0 cos 0  t

vx  v0 cos 0
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)
Maximum Range
• 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
• http://galileo.phys.virginia.edu/classes/109
N/more_stuff/Applets/ProjectileMotion/jara
pplet.html
• http://jersey.uoregon.edu/vlab/Cannon/
The ultimate projectile:
Putting an object into 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!
THE END…