Transcript Projectile Motion Chapter 3.3

Projectile Motion
Chapter 3.3
Objectives

Recognize examples of projectile motion

Describe the path of a projectile as a
parabola

Resolve vectors into their components and
apply the kinematic equations to solve
problems involving projectile motion
Projectile Motion

How can you know the displacement, velocity
and acceleration of a ball at any point in time
during its flight?

Use the kinematic equations of course! 
Vector Components p.98
(Running vs Jumping)
While running, the person is only moving in one dimension
Therefore, the velocity only has one component.
V
While jumping, the person is moving in two dimensions
Therefore, the velocity has two components.
Vy
Vx
Definition of Projectile Motion

Objects that are thrown or launched into the
air and are subject to gravity are called
projectiles

Examples?
–
Thrown Football, Thrown Basketball, Long
Jumper, etc
Path of a projectile

Neglecting air resistance, the path of a projectile is a
parabola

Projectile motion is free fall with an initial horizontal
velocity

At the top of the parabola, the velocity is not 0!!!!!!
Vertical and Horizontal Motion
Horizontal Motion
Vertical Motion
Velocity = Vx
Displacement = Δx
Velocity = Vy
Displacement = Δy
Because gravity does not
act in the horizontal
direction, Vx
is always constant!
Gravity acts vertically,
therefore a = -9.81 m/s2
Equations for projectiles launched
horizontally
Horizontal Motion Vertical Motion
Δx=Vxt
Vy,i=0 (initial velocity in y direction is 0)
Vx is constant! v 2  v 2  2ay  0  2ay  2ay
y, f
y ,i
a=0
v y , f  v y ,i  at  0  at  at
1 2
1 2 1 2
y  v y ,i t  at  0  at  at
2
2
2
Revised Kinematic Eqns for projectiles
launched horizontally
Horizontal
Motion
x  vxt
Vertical Motion
1 2
y  at
2
v
2
y, f
 2ay
v y , f  at
Finding the total velocity

Use the pythagorean theorem to find the
resultant velocity using the components (Vx
and Vy)
Use SOH CAH TOA to find the direction
V
Vy

Vx
Example p. 102 #2

A cat chases a mouse across a 1.0 m high
table. The mouse steps out of the way and
the cat slides off the table and strikes the
floor 2.2 m from the edge of the table. What
was the cat’s speed when it slid off the table?
What is the cat’s velocity just before it hits
the ground?
What do we know and what are we
looking for?
Δx= 2.2 m
Δy= -1.0m (bc the cat falls down)
Vx= ?????
1.0 m
What are we looking for??
2.2m
How do we find Vx?
x  vxt

Equation for horizontal motion:

We have x…so we need t.

How do we find how long it takes for the cat to hit
the ground?

Use the vertical motion kinematic equations.
Vertical Motion



Δy= -1.0m
a=-9.81 m/s^2
What equation should we use?
1 2
y  at
2

Rearrange the equation, to solve for t then plug in
values.
2y
2(1.0m)
t

 .45s
m
a
 9.81 2
s
Horizontal equation
x  vxt

Rearrange and solve
for Vx:
x 2.2m
m
vx  
 4.89
t .45s
s

Cat’s Speed is 4.89 m/s
Cliff example

A boulder rolls off of a cliff and lands 6.39
seconds later 68 m from the base of the cliff.
–
–
–
What is the height of the cliff?
What is the initial velocity of the boulder?
What is the velocity of the boulder just as it strikes
the ground?
How high is the cliff?



Δy= ?
t = 6.39 s
Vy,i = 0
a=-9.81 m/s2
Vx=?
Δx= 68 m
1 2
y  at
2
1 2 1
m
y  at  (9.81 2 )(6.39) 2  200m
2
2
s
The cliff is 200 m high
What is the initial velocity of the
boulder?

The boulder rolls off the cliff horizontally

Therefore, we are looking for Vx
x  vxt
x 68m
m
vx  
 10.6
t 6.39s
s
Important Concepts for Projectiles
Launched Horizontally
Horizontal Components
Horizontal Velocity is
constant throughout the
flight
Vertical Components
Initial vertical velocity is 0
but increases throughout
the flight
Horizontal acceleration is Vertical acceleration is
0
constant: -9.81 m/s2
Projectiles Launched at An Angle
Projectiles Launched
Horizontally
Projectiles Launched at
an Angle
•Vx
•Vx
is constant
•Initial Vy is 0
Vi = Vx
is constant
•Initial Vy is not 0
Vi
Vy,i
θ
Vx,i
Components of Initial Velocity for
Projectiles Launched at an angle
Use soh cah toa to find the Vx,i and Vy,i
Vi
Vy,i
θ
Vx,i
o v y ,i
sin    
h vi
a v x ,i
cos( )  
h vi
v y ,i  vi sin  
v x ,i  vi cos 
Revise the kinematic equations again
Horizontal Motion
Vertical Motion
v x  v x ,i  vi cos  vy, f  vy,i  at  vi sin    at
x  vx t  vi cos t
v y2, f  v y2,i  2ay  (vi sin(  )) 2  2ay
1
2
y  vi sin  t  a t 
2
Example p. 104 #3

A baseball is thrown at an angle of 25°
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?
What do we know?
Δx= 42.0 m
θ= 25°
Vi= 23.0 m/s
Vy at top = 0
Δt=?
Δy=?
25°
42.0 m
What can we use to solve the problem?
Find t using the horizontal eqn:
Δx=vxΔt = vicos(θ)t
t 

x
x
42.0m


 2.0 s
m
vx
vi cos  ( 23 )(cos 25
s
How to find Δy?
–
–
Vy,f = 0 at top of the ball’s path
2
2
What equation should we use? v y , f  v y ,i  2ay
2
y 
v y2, f  v y2,i
2a
m




0

(
23
)
sin
25


0  (vi sin  ) 2
s

  4.8m


m
2a

2  9.81 2 
s 

Cliff example

A girl throws a tennis ball at an angle of 60°North of
East from a height of 2.0 m. The ball’s range is 90 m
and it is in flight for 6 seconds.
–
–
–
–
–
What is the initial horizontal velocity of the ball?
What is the initial vertical velocity of the ball?
What is the total initial velocity of the ball?
How high above the initial position does the ball get?
What is the vertical velocity of the ball 2 seconds after it is
thrown?
What is the initial horizontal velocity of
the ball?




Δx= 90 m
Θ=60°
Total time= 6 s
Horizontal velocity is
constant: Vx
x  vxt
x 90
m
Vx  
 15 East
t
6
s
What is the initial vertical velocity of
the ball?
Vi
Vy,i
Vy, i
tan  
Vx, i
θ
Vx,i
V y ,i
m
m
 tan( 60)(15 )  25.98  26 North
s
s
What is the total initial velocity of the
ball?
Vi  V  V
2
Vi
Vy,i
2
x ,i
2
y ,i
θ
Vx,i
m
Vi  15  26  30 at 60 North of East
s
2
2
How high above the ground does the
ball get?

At the top of the parabola,
Vy is 0…so use the revised
kinematic equations
y 

v y2, f  v y2,i
2a
v
2
y, f
 v  2ay
2
y ,i
0  26

 34.45m
m

2  9.81 2 
s 

Add 2m to get the height
above the ground: 36.65 m
2
What is the vertical velocity of the ball
2 seconds after it is thrown?



Vy,i=+26 m/s
a= -9.81 m/s2
t = 2 seconds
v y , f  v y ,i
m
m
m
 at  26  (9.81 2 )( 2s )  6.4
s
s
s
Important Concepts for Projectiles
Launched at an Angle

At the top of the parabola, neither the
object’s velocity nor it’s acceleration is 0!!!!!
–
–
–
–
Only Vy is 0
Vx is constant throughout the flight
Horizontal acceleration is always 0
Vertical acceleration is always -9.81 m/s2