Projectile Motion
-
Upload
galvin-williams -
Category
Documents
-
view
38 -
download
3
description
Transcript of Projectile Motion
Projectile Motion
• Describe the motion of an object in TWO dimensions
• Keep it simple by considering motion close to the surface of the earth for the time being (g = -9.8 m/s2 = constant in y direction)
• Neglect air resistance to make it simpler• Assume the rotation of the Earth has no effect
Launch speed = Return Speed. Speed is minimum at apex of parabolic trajectory.
Horizonal component
Net velocity
vx
vertical component
vy v
Above: Vectors areadded in geometricFashion.
Velocity Components at various points of the Trajectory
A History of Projectile Motion
Aristotle:The canon ball travels in a straight line until it lost its ‘impetus’.
Galileo: a result of Free Fall Motionalong y-axis and Uniform Motion along x-axis.
What’s the similarity between a freely-falling ball and a projectile ?
A dropped ball falls in the same time as a ball shot horizontally.Along the vertical, their motions are identical (uniformly accelerated motion (free-fall).
Along the horizontal, notice the ball fired horizontally covers the Same distance in the same unit time intervals (uniform motion along x)
x
yuniform motion
verticalmotion
Projectilemotion
Projectile Motion EquationsHorizontal (x)
x =vxt
Vertical (y)
y=y0 + v0yt−
12
gt2 or dy =12
gt2 + viyt
vy =v0y −gt or g=vfy
−viy
t
1. Along x, the projectile travels with constant velocity. vx=vxo x = vxot
2. Along y, the projectile travels in free-fall fashion.vy = vyo – gt y = vyot – (1/2) gt2 , g= 9.8 m/s2
Projectile motion = a combination of uniform motion along x and uniformly accelerated motion (free fall) along y.
Projectile Motion = Sum of 2 Independent Motions
Everyday Examples of Projectile Motion
1.Baseball being thrown2.Water fountains3.Fireworks Displays4.Soccer ball being kicked5.Ballistics Testing
QuickTime™ and aTIFF (Uncompressed) decompressor
are needed to see this picture.Type IA ball is kicked off a 125 m cliff with a horizontal velocity of 50 m/s. What is the range of the ball?
V = 50 m/s
125
m
Horizontal Verticald = d = vi = vi = a = a = t = t =
-125 m
0 m/s50 m/s
0 m/s2 -9.8 m/s2
5.1 s5.1 s
255 m
d =12
at2 +vit
−125 =1
2−9.8( ) t 2 + 0
d =12
at2 +vit
d =0+50(5.1)d =255 m
t =5.1s
Type IIA ball is kicked of a with a velocity of 50 m/s at an angle of 37° from the horizontal. What is the range of the ball?
Horizontal Verticald = d = vi = vi = a = a = t = t =
0 m
30 m/s40 m/s
0 m/s2 -9.8 m/s2
6.1 s6.1 s
244 m
d =12
at2 +vit
0 =12
−9.8( )t2 + 30t
d =12
at2 +vit
d =0+40(6.1)d =244 m
t =0 or 6.1s
v = 50 m/s37°
v = 50 m/svy = 50sin37 = 30 m/s
37°
vx = 50cos37 = 40 m/s
QuickTime™ and aTIFF (Uncompressed) decompressor
are needed to see this picture.
Type IIIA ball is kicked off a 125 m cliff with a velocity of 50 m/s at an angle of 37° from the horizontal. What is the range of the ball?
Horizontal Verticald = d = vi = vi = a = a = t = t =
-125 m
30 m/s40 m/s
0 m/s2 -9.8 m/s2
9.0 s9.0 s
360 m
d =12
at2 +vit
−125 =1
2−9.8( ) t 2 + 30t
d =12
at2 +vit
d =0+40(9.0)d = 360 m
t =9.0 or -2.8s
v = 50 m/s37°
125
mNote, this type
requires the use of the quadratic
equation
Maximum Range
• How to maximize horizontal range:– keep the object off the ground for as long
as possible.– This allows the horizontal motion to be a
maximum since x = vxt
– Make range longer by having a greater initial velocity velocity
QuickTime™ and aTIFF (Uncompressed) decompressor
are needed to see this picture.
vi
R
x =R=vxt
R=vi cos ( )t
Range Equation
d =12
at2 + vit
0 =12
gt2 + viyt ⇒ t=
2viy
g=2vi sin
g
What is total t? To solve, set vertical displacement = 0.
R =vi cos ( )2vi sin
g⎛⎝⎜
⎞⎠⎟=
vi2
g2sin cos( )
Trig Identity: 2sincos = sin(2)
R =vi2
gsin 2( ) REMEMBER: ONLY VALID WHEN VERTICAL
DISPLACEMENT IS ZERO (Type II problems).
At what angle do I launch for Maximum Range ?
Need to stay in air for the longest time, and with the fastest horizontal velocity componentAnswer: 45°
Projectile Motion
• What happens when we add air resistance?• Adds a new force on the ball• The force is in the opposite direction to the
ball’s velocity vector and is proportional to the velocity at relatively low speeds
• Need calculus to sort out the resulting motion• Lowers the angle for maximum range