scrambled eggs
up to no good
pipe down or downpipe
see eye to eye
3.2 Equations of Kinematics in Two Dimensions
• Equations of Kinematics
v  vo  at
x  12 vo  v  t
v  v  2ax
2
2
o
x  vot  at
1
2
2
3.2 Equations of Kinematics in Two Dimensions
– the horizontal component
vx  vox  axt
x  voxt  a x t
1
2
x
2
1
2
vox  vx  t
v  v  2a x x
2
x
2
ox
3.2 Equations of Kinematics in Two Dimensions
– the vertical component
v y  voy  a y t
y  voyt  a yt
1
2
y
1
2
v
oy
2
 vy  t
v  v  2a y y
2
y
2
oy
3.2 Equations of Kinematics in Two Dimensions
• The x part of the motion occurs exactly as it
would if the y part did not occur at all, and
vice versa.
Projectile Motion
Chapter 3
Section 3
What is a projectile?
• When a long jumper approaches his jump, he
runs along a straight line which can be called the
x-axis.
• When he jumps, as shown in the diagram below,
his velocity has both horizontal and vertical
components
What is a projectile?
• Projectiles are objects in free fall that have an
initial horizontal velocity.
• Free fall implies that the only force acting on the object
is gravity.
• Initial horizontal velocity results in motion in two
dimensions.
• Examples of projectiles include
• An arrow flying towards a target
• A baseball thrown in the air at an angle other than 90o,
unless they are thrown straight up.
Projectile motion
• Projectiles follow parabolic trajectories.
• Trajectory is the path of a flying object.
• The horizontal and vertical motions are
independent.
• We can write separate equations of motion for each
direction.
• Horizontal motion on the x-axis and
• the vertical motion on the y-axis
What is a projectile?
• If a ball is thrown horizontally, (yellow) and
another ball is dropped (red)-----they will both
• HIT THE GROUND AT THE SAME TIME!!!!!!!!
• We shall do a demo of this
Solving Projectile Motion Problems
• You need to separate the motion down into 2
parts!
• The first part is the horizontal movement
• The second part is the vertical movement
• Keep the two sets of information separate
using a table when solving such problems
Horizontal motion equations and
constants
x  vox t  12 a x (t ) 2
a x  0  gravity only affects the vertical component of a projectile
Δx  v ox Δt
vx  vox  a x t
vx  vox  vx
vx  constant
Vertical motion equations and
constants
y  voy t  a y ( t )
1
2
2
a y  g  9.81 sm2
y  voy t  (  g )(t )
1
2
v y  voy  gt
g
v y  voy
Δt
2
Calculating time of fall
• If a projectile is launched horizontally, this tells
us that the object has an initial horizontal
velocity (ie: in the x-axis)
• For such projectiles, the initial vertical velocity is
zero (ie. in the y-axis) (voy=0)
y  voy t  ( g )(t )
1
2
y  ( g )(t )
1
2
t 
2y
g
2
2
Example 1
• People in movies often jump from
buildings into pools. If a person jumps
from the 10th floor (30.0 m) to a pool that
is 5 m away from the building, with what
initial horizontal velocity must the person
jump?
Step 1: Write down Givens and unknowns.
• You will have to separate your y-components from
your x-components using a table
X- Component
Y - Component
x = 5.0 m (from building to pool)
y = -30.0 m (negative due to
downward direction of fall)
vox = ?
voy = 0 (horizontal projectile)
ax = 0
ay = -9.81 m/s2
Step 2: Write down equations that might
solve it.
• We can use the y equation to solve for the time of the jump.
t 
2y
g
• We then use this calculated time in the x equation to solve for
the initial velocity.
x  vox t
• These are the two common equations you will use in this
section!!!
Step 3: Carry out the solution.
Time for the jump
• Using the y equation to
solve for the time of the
jump.
2y
t 
g
2(-30 )
t 
- 9.81
t  2.47 s
Step 4: Carry out the solution.
Using our timeto calculate the initial horizontal velocity
• Using the x equation to solve for the initial
horizontal velocity.
x  vox t
x
 vox
t
5.0 m
 2.0 m/s
2.47 s
Practice Problem 1
• An autographed baseball rolls off of a
0.70m high desk. And strikes the floor
0.25m away from the desk. How fast did it
roll off the table?
Practice Problem # 1
Step 1: Write down what’s Given and your unknowns. (remember to
separate your X and Y components using a table)
X- Component
Y - Component
x = 0.25 m (from base of desk)
y = -0.7 m (negative due to downward
direction of fall)
vox = ?
voy = 0 (horizontal projectile)
ax = 0
ay = -9.81 m/s2
Step 2: Write down equations that might
solve it.
•
We can use the y equation to solve for the
time of the jump.
2y
t 
g
• We then use the time in the x equation to
solve for the initial velocity.
x  vox t
Step 3: Carry out the solution.
Time for the jump
• Using the y equation to
solve for the time of the
jump.
2y
t 
g
2(-0.7 )
t 
- 9.81
t  0.378 s
Step 3: Carry out the solution.
Using our timeto calculate the initial horizontal velocity
• Using the x equation to solve for the initial
velocity.
x  vox t
x
 vox
t
0.25 m
 0.66 m/s
0.378 s
Let us review the concepts for
Horizontally launched
projectiles
• Acceleration due to gravity does not affect
horizontal motion 𝑎𝑥 = 0 𝑚/𝑠 2
• Initial horizontal velocity = final horizontal
velocity
• 𝑣𝑜𝑥 = 𝑣𝑥
• Initial vertical velocity = 0m/s
• 𝑣𝑜𝑦 = 𝑚/𝑠
• 𝑎𝑦 = −𝑔
3.3 Projectile Motion
Practice Problem # 2 - A Falling Care Package
The airplane is moving horizontally with a constant velocity of
+115 m/s at an altitude of 1050m.
a) Determine the time required for the Care package to hit the ground.
b) What are the magnitude and direction of the final velocity of the Care
package?
Practice Problem # 1
Step 1: Write down what’s Given and your unknowns. (remember to
separate your X and Y components using a table)
X- Component
Y - Component
x =
y = -1050.0 m (negative due to
downward direction of fall)
vox = 115 m/s
voy = 0 (horizontal projectile)
ax = 0
ay = -9.81 m/s2
Step 2: Write down equations that might
solve it.
•
We can use the y equation to solve for the
time of the jump.
• Since 𝑣𝑜𝑦 = 0
y  voy t  ( g )( t )
1
2
y  ( g )( t )
1
2
t 
2y
g
2
2
Step 3: Carry out the solution.
Time for the jump
• Therefore, using the
modified y equation to solve
for the time of the jump.
2y
t 
g
2(-1050 )
t 
- 9.81
t  14.6 s
3.3 Projectile Motion
Practice Problem # 2 - A Falling Care Package
b) What are the magnitude and direction of the final velocity
of the Care package?
Practice Problem # 1
Step 1: Write down what’s Given and your unknowns. (remember to
separate your X and Y components using a table)
X- Component
Y - Component
Dx =
Dy = -1050.0 m (negative due to
downward direction of fall)
vox = 115 m/s
voy = 0 (horizontal projectile)
ay = -9.81 m/s2
t = 14.6 s
t = 14.6 s
Step 2: Write down equations that might
solve it.
•
We then use the time in the y equation to
solve for the final vertical velocity.
• Since 𝑣𝑜𝑦 = 0𝑚/𝑠
v y  voy  a y t
3.3 Projectile Motion
Step 3: Carry out the solution.
Final vertical velocity of the package
v y  voy  a y t


 0   9.80 m s 14.6 s 
 143 m s
2
Step 4: Carry out the solution.
Magnitude and direction of final velocity
v  vx 2  vy 2
v
2
2
115  ( 143 )
v  184m / s
 143 
  tan 

 115 
1
  51.2
I will not accept late work
CLASS ASSIGNMENT NOT H/W
3.3 Projectile Motion
Conceptual Example 5
I Shot a Bullet into the Air...
Suppose you are driving a convertible with the top down.
The car is moving to the right at constant velocity. You point
a rifle straight up into the air and fire it. In the absence of air
resistance, where would the bullet land – behind you, ahead
of you, or in the barrel of the rifle?
Explain 
Section 3.
Projectiles at an Angle
Cont.
Projectiles Launched at an angle
As shown below, projectiles launched at an
angle have an initial vertical and horizontal
component of velocity.
Projectiles Launched at an angle
• Suppose the initial velocity vector makes an angle θ with the
horizontal
• The sine and cosine functions can be used to find the horizontal
and vertical components of the initial velocity
• 𝑣𝑜𝑥 = 𝑣𝑜 (𝑐𝑜𝑠𝜃)
• 𝑣𝑜𝑦 = 𝑣𝑜 (𝑠𝑖𝑛𝜃)
3.3 Projectile Motion
Example 6 The Kickoff
• A placekicker kicks a football at and angle of 40.0
degrees and the initial speed of the ball is 22 m/s.
Ignoring air resistance, calculate the following:
a) determine the maximum height that the ball attains.
Step 1: Write down what you know and don’t
know (and resolve the vector provided)
=22 m/s
=40o
X – Component
Y - Component
x = ?
y = ?
t = ?
t = ?
ax = 0m/s2
ay = -9.81 m/s2
vox  vo cos( )
voy  vo sin( )
Step 2: Decide on a plan or equation
to use.
• For the horizontal velocity:
vox  vo cos   22 m s  cos 40  16.9 m s

• For the vertical velocity:
voy  vo sin   22 m s sin 40  14.1 m s

Step 3: Carry out the plan.
y
ay
vy
voy
?
-9.81 m/s2
0
14.1 m/s
v  v  2a y y
2
y
y
2
oy
t
v v
2
y
2
oy
2a y
0  14.1 m s 
y


10
.
1
m
2
2  9.81 m s
2


3.3 Projectile Motion
Example 6 The Kickoff
• A placekicker kicks a football at and angle of 40.0
degrees and the initial speed of the ball is 22 m/s.
Ignoring air resistance, calculate the following:
a) The maximum height that the ball attains = 10 m
b) The time between kickoff and landing
Step 1: Write down what you know and don’t
know (and resolve the vector provided)
40o
vox
X – Component
voy
Y - Component
x = ?
y = 10.1 m
t = ?
t = ?
ax = 0m/s2
ay = -9.81 m/s2
vox  16.9m / s
voy  14.1m / s
Step 2: Decide on a plan or equation
to use.
• We can calculate tine for the vertical displacement:
v y  voy  a y t
Step 3: Carry out the plan.
• For the vertical displacement (to maximum height):
v y  voy  a y t
t
v y  voy
a
0  14.1
t (
) x2
 9.81
• = 2.87 s
 14.1  14.1
t
 9.81
3.3 Projectile Motion
Example 6 The Kickoff
• A placekicker kicks a football at and angle of 40.0
degrees and the initial speed of the ball is 22 m/s.
Ignoring air resistance, calculate the following:
a) The maximum height that the ball attains = 10 m
b) The time between kickoff and landing = 2.86 s
c) The range ‘R’ of the projectile?
Step 1: Write down what you know and don’t
know (and resolve the vector provided)
40o
vox
X – Component
voy
Y - Component
x = ?
y = 10 m
t = 2.86 s
t = 1.43 s
ax = 0m/s2
ay = -9.81 m/s2
vox  16.9m / s
voy  14.1m / s
Step 2: Decide on a plan or equation
to use.
• For the horizontal displacement:
x  voxt  a x t  voxt
1
2
2
Step 3: Carry out the plan.
• For the horizontal displacement:
x  voxt  a x t  voxt
1
2
2
 16.9m / s 2.87 s 
• = 48.5 m