Post on 03-Jan-2016
Projectile Motion
CCHS Physics
Projectile Properties?
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
Projectile Motion
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
Projectile Motion
The ball is in free fall vertically and moves at constant speed horizontally!!!
Projectile Motion
Video
What’s wrong with this picture ?
Answer: It never happens ! Only whenthere is no gravity.
Projectile Motion
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
Horizontal Range
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).
Projectile Motion• We can see that complementary angles have
the same range because sin = sin2
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
Projectile Motion