Do Now
Starting from rest, a car undergoes a constant
acceleration of 10. m/s/s.
How far will the car travel in 3.0 s?
How fast will the car move in 3.0 s?
Do Now
Starting from rest, a car undergoes a constant
acceleration of 10 m/s/s.
How far will the car travel in 3 s?
How fast will the car move in 3 s?
Given:
Solution:
v0  0
a  10
1
m
s
t  3 .0 s
Find:
x  ?
v?
2
 x  v 0 t  at
1 2
2
 x  0  10 ( 3 )  45 m
2
2
v  v 0  at
v  0  10 ( 3 )  30 m / s
Do Now
A basketball is dropped from rest from height of
1 m toward motion detector located on the
floor.
Draw x vs. t, v vs. t, and a vs. t graphs of the
motion of the ball.
Unit 4:
Kinematics in Two Dimensions
Unit Plan
•Free Fall
•Projectile Motion
• Solving Problems Involving Projectile Motion
• Projectile Motion Is Parabolic
• Aristotle (382BC-322BC)Greek natural
philosopher.
• A student of Plato and
teacher of Alexander the
Great
• Believed that more
massive objects fall faster.
• Did not account air
resistance.
A detail of The School of Athens, a fresco by
Raphael.
• Galileo Galilei (1564-1642) –
Italian physicist.
• Reexamined motion of falling
objects
• Has been called the “Father of
Modern Physics”(used models
and experimentation)
• Postulated that all objects
would fall with the same
constant acceleration in the
absence if air resistance.
d t
2
Free Fall
• Freely falling objects are affected
only by gravity.
• At a given location on the Earth
and in absence of air resistance,
all objects fall with the same
constant acceleration.
• Acceleration due to gravity,
or acceleration of free fall
a  g  9 .8
m
s
2
Air Resistance
• A feather an a coin
accelerate equally when
there is no air around
them (in a vacuum).
• For compact objects the
effect of air resistance is
small enough to be
neglected.
Accelerated Motion Due to Gravity
• We can choose y to be positive in the upward
direction or in the downward direction.
• Consider motion up to be positive.
m
• a   g   9 . 8 s 2 acceleration due to gravity
m
• For problem solving, we will approximate g   10 . 0 2
s
• On Earth, acceleration due to gravity always has
downward direction(towards center of Earth).
Vertical Motion with Gravity
• Start with the key equations for 1-dimensional
motion. Assume that the motion is only up and
down. Since motion is vertical  y ->  x , add the
subscript y to the velocity, and substitute –g for a.
v  v 0  at
 x  v0t 
x 
1
at
2
v  v0
2
2
 vt
v y  v 0 y  gt
 y  v0 yt 
y 
v
y
1
2
2
 v0 y
2
gt
 v yt
Object Thrown Up
A rock is thrown upward with initial velocity 30
m/s.
v
Time, t
Velocity,
y
v y  v 0  gt
v y  30  10 t
After 0 seconds
After 1 second
After 2 seconds
After 3 seconds
After 4 seconds
After 5 seconds
After 6 seconds
Position
y  y0  v0t 
y  30 t  5 t
1
2
2
gt
2
Object Thrown Up
A rock is thrown upward with initial velocity 30
m/s.
v
Time, t
Velocity,
y
v y  v 0  gt
v y  30  10 t
After 0 seconds
30 m/s
After 1 second
20 m/s
After 2 seconds
10 m/s
After 3 seconds
0 m/s
Reached top
After 4 seconds
-10 m/s
After 5 seconds
-20 m/s
After 6 seconds
-30 m/s
Displacement
y  y0  v0t 
y  30 t  5 t
1
2
2
gt
2
Object Thrown Up
A rock is thrown upward with initial velocity 30
m/s.
v
Time, t
Velocity,
y
v y  v 0  gt
v y  30  10 t
Displacement
y  y0  v0t 
y  30 t  5 t
1
2
2
After 0 seconds
30 m/s
0m
After 1 second
20 m/s
25 m
After 2 seconds
10 m/s
40 m
After 3 seconds
0 m/s
Reached top
45 m
After 4 seconds
-10 m/s
40 m
After 5 seconds
-20 m/s
25 m
After 6 seconds
-30 m/s
0m
gt
2
Velocity vs. Time Graph
Object Thrown Up. Graphs
Position vs. Time Graph
Object Thrown Up
• What is the instantaneous speed at
the highest point?
• 0
• How does velocity change during
the upward part of its motion?
• Decreasing from v
to 0.
0
• How much does its speed decrease
each second?
• The speed decreases 10 m/s each
second.
Object Thrown Up
• What is the instantaneous speed of
the object at points of equal
elevation?
• The same.
• Are velocities same or different at
points of equal elevation?
• Same magnitude, opposite
directions.
• Is acceleration different when the
object moving upward or
downward?
• The same 10 m/s/s downwards.
Dropped Object
A rock is dropped from the top of the cliff. How
far did it travel in 1s, 2s, and 3s?
v0 y  0
 y  v0 yt 
1
2
gt
2
y  
1
2
gt
2
y  
1
2
(10 ) t   5 t
2
2
Dropped Object
A rock is dropped from the top of the cliff. How
far did it travel in 1s, 2s, and 3s?
v0 y  0
 y  v0 yt 
1
2
Drop Time
gt
2
y  
1
gt
y  
2
1
2
2
 y   5t
2
1 second
 y   5 (1)   5 m
2 seconds
 y   5 ( 2 )   20 m
3 seconds
 y   5 ( 3 )   45 m
2
2
2
(10 ) t   5 t
2
2
Dropping with v  0 .
Find time if you know Δy.
0y
1
 y  v0 yt 
y  
1
2
10 t
2
 y   5t
t
y
5
gt
2
2
2
Time Up = Time Down
• Since for the object thrown upward the
motion up and down is symmetrical, you can
use the same formula to find the time to go
up a certain distance.
• If you throw a ball upwards with just enough
velocity to go up a distance of 35 m, how long
will it take to reach the top?
t
y
5

35
5

7  2 .6 s
Exercise 1
A ball is thrown upward with an initial velocity
of 20 m/s. How long will it take for the ball to
reach its maximum height?
y
Given:
Solution:
v 0 y  20
vy  0
m
v y  v 0 y  gt
s
g  10 . 0
m
s
Find:
t?
2
0  20  10 t
t  2s
v 0 y  20
m
s
g  10 . 0
m
s
2
Dropping With Initial Velocity
Exercise 2
• A ball is thrown downward from the top of a
roof with a speed of 25 m/s. Find the
instantaneous velocity of the ball in 2 seconds.
y
Given :
v 0 y   25 m / s
Solution :
t  2s
v y  v y 0  gt
g  10 . 0 m / s
Find :
vy  ?
2
v 0 y   25
m
s
v y   25  10 ( 2 )   45 m / s
g  10 . 0
m
s
2
ConcepTest 2.8b
When throwing a ball straight up,
Acceleration II
1) both v = 0 and a = 0
which of the following is true
2) v  0, but a = 0
about its velocity v and its
3) v = 0, but a  0
acceleration a at the highest point
4) both v 0 and a  0
in its path?
5) not really sure
ConcepTest 2.8b
When throwing a ball straight up,
Acceleration II
1) both v = 0 and a = 0
which of the following is true
2) v  0, but a = 0
about its velocity v and its
3) v = 0, but a  0
acceleration a at the highest point
4) both v 0 and a  0
in its path?
5) not really sure
At the top, clearly v = 0 because the ball has
momentarily stopped. But the velocity of the
ball is changing, so its acceleration is definitely
not zero! Otherwise it would remain at rest!!
Follow-up: …and the value of a is…?
y
ConcepTest 2.9a
You throw a ball straight
up into the air. After it
leaves your hand, at what
point in its flight does it
have the maximum value
of acceleration?
Free Fall I
1) its acceleration is constant
everywhere
2) at the top of its trajectory
3) halfway to the top of its trajectory
4) just after it leaves your hand
5) just before it returns to your hand
on the way down
ConcepTest 2.9a
You throw a ball straight
up into the air. After it
leaves your hand, at what
point in its flight does it
have the maximum value
of acceleration?
Free Fall I
1) its acceleration is constant
everywhere
2) at the top of its trajectory
3) halfway to the top of its trajectory
4) just after it leaves your hand
5) just before it returns to your hand
on the way down
The ball is in free fall once it is released. Therefore, it is entirely under
the influence of gravity, and the only acceleration it experiences is g,
which is constant at all points.
ConcepTest 2.9b
Alice and Bill are at the top of a
building. Alice throws her ball
downward. Bill simply drops
his ball. Which ball has the
greater acceleration just after
release?
Free Fall II
1) Alice’s ball
2) it depends on how hard
the ball was thrown
3) neither -- they both have
the same acceleration
4) Bill’s ball
Alice
v0
vA
Bill
vB
ConcepTest 2.9b
Alice and Bill are at the top of a
building. Alice throws her ball
downward. Bill simply drops
his ball. Which ball has the
greater acceleration just after
release?
Both balls are in free fall once they are
released, therefore they both feel the
Free Fall II
1) Alice’s ball
2) it depends on how hard
the ball was thrown
3) neither -- they both have
the same acceleration
4) Bill’s ball
Alice
v0
Bill
acceleration due to gravity (g). This
acceleration is independent of the initial
vA
velocity of the ball.
Follow-up: Which one has the greater velocity when they hit
the ground?
vB
ConcepTest 2.10a
Up in the Air I
You throw a ball upward with
1) more than 10 m/s
an initial speed of 10 m/s.
2) 10 m/s
Assuming that there is no air
resistance, what is its speed
when it returns to you?
3) less than 10 m/s
4) zero
5) need more information
ConcepTest 2.10a
Up in the Air I
You throw a ball upward with
1) more than 10 m/s
an initial speed of 10 m/s.
2) 10 m/s
Assuming that there is no air
resistance, what is its speed
when it returns to you?
3) less than 10 m/s
4) zero
5) need more information
The ball is slowing down on the way up due to
gravity. Eventually it stops. Then it accelerates
downward due to gravity (again). Since a = g on
the way up and on the way down, the ball reaches
the same speed when it gets back to you as it had
when it left.
ConcepTest 2.10b
Up in the Air II
Alice and Bill are at the top of a cliff of
height H. Both throw a ball with initial
speed v0, Alice straight down and Bill
straight up. The speeds of the balls when
they hit the ground are vA and vB. If there
is no air resistance, which is true?
1) vA < vB
2) vA = vB
3) vA > vB
4) impossible to tell
Alice v0
v0
Bill
H
vA
vB
ConcepTest 2.10b
Up in the Air II
Alice and Bill are at the top of a cliff of
height H. Both throw a ball with initial
speed v0, Alice straight down and Bill
straight up. The speeds of the balls when
they hit the ground are vA and vB. If there
is no air resistance, which is true?
Bill’s ball goes up and comes back
down to Bill’s level. At that point, it is
moving downward with v0, the same
as Alice’s ball. Thus, it will hit the
ground with the same speed as
Alice’s ball.
1) vA < vB
2) vA = vB
3) vA > vB
4) impossible to tell
Alice v0
v0
Bill
H
vA
vB
Follow-up: What happens if there is air resistance?
Projectile Motion
2-D Kinematics
Projectile Motion
• A projectile is any object that is given an initial velocity
or dropped and then follows a path determined entirely
by the effects of gravity.
• Projectiles - batted baseball, a thrown football, a
package dropped from an airplane, a bullet shot from a
rifle.
• The path followed
by a projectile
is called its
trajectory.
• The trajectory of a
projectile is
a parabola.
Horizontally Launched Projectile
Horizontal and Vertical Motion
• We can analyze projectile motion as a
combination of horizontal motion with constant
velocity and vertical motion with constant
acceleration.
3-5 Projectile Motion
It can be understood by
analyzing the horizontal and
vertical motions separately.
Independence of Horizontal and
Vertical Components
The vertical force acts perpendicular to the horizontal motion and
will not affect it since perpendicular components of motion are
independent of each other. Thus, the projectile travels with a constant
horizontal velocity and a downward vertical acceleration.
Independence of Horizontal and
Vertical Motion Demo
• Two balls released
simultaneously. One
ball dropped freely,
another projected
horizontally
• Both balls fall the same
vertical distance in
equal times.
3-5 Projectile Motion
The speed in the x-direction
is constant; in the ydirection the object moves
with constant acceleration g.
This photograph shows two balls
that start to fall at the same time.
The one on the right has an initial
speed in the x-direction. It can be
seen that vertical positions of the
two balls are identical at identical
times, while the horizontal
position of the yellow ball
increases linearly.
Projectile Motion
Vertical motion:
Horizontal motion:
Vertical downward acceleration:
a y   g   10 . 0 m / s
2
v x  v0 x
v y  v 0 y  gt
 y  v0 yt 
1
Horizontal velocity is never changing
gt
2
2
Vertical velocity is constantly
changing
 x  v0 xt
Practice Problem
The boy on a tower (h = 5m) throws a ball a distance of
20m. At what speed is the ball thrown?
Given:
Solution:
 x  20 . 0 m
v0 y  0 m / s
g   10 . 0
m
s
h  5 . 00 m
2
Find: v 0 x  ?
v0 x  ?
Vertical:
 y  v0 yt 
1
h  y 
1
Horizontal:
gt
2
gt
2
t
2h
g
t
 x  v0 xt
2
2 ( 5 . 00 )
10 . 0
t= 1.00 s
2
v0 x 
x
t

20 m
1s
v 0 x  20 m / s
Do Now
A stone is thrown horizontally at a speed of
+5.0 m/s from the top of a cliff 80.0 m high.
a. How long does it take the stone to reach the
bottom of a cliff?
b. How far from the base of the cliff does the
stone strike the ground?
Do Now
A stone is thrown horizontally at a speed of
+5.0 m/s from the top of a cliff 80.0 m high.
a. How long does it take the stone to reach the
bottom of a cliff?
b. How far from the base of the cliff does the
stone strike the ground?
v0 x  5m / s
h  80 . 0 m
x  ?
Given:
Solution:
Vertical:
Horizontal:
v0 x  5m / s
h  80 . 0 m
 y  v0 yt 
1
2
h  y 
gt
1
 x  v0 xt
2
gt
2
2
Find:
a) t  ?
b)  x  ?
t
2h
g
t
2 ( 80 . 0 m )
10 . 0 m / s
2
 4 . 00 s
 x  ( 5 . 0 m / s )( 4 . 00 s )  20 .m
Conclusion
1. A projectile is any object upon which the only force is _______,
2. Projectiles travel with a _____________ trajectory due to the
influence of gravity,
3. There are __________horizontal forces acting upon projectiles
and thus __________ horizontal acceleration.
4. The horizontal velocity of a projectile is ____________
5. There is a vertical acceleration caused by gravity; its value is
_______________
6. The vertical velocity of a projectile changes by ______ m/s each
second.
7. The horizontal motion of a projectile is _________________ of
its vertical motion.
Conclusion
• A projectile is any object upon which the only force is gravity,
• Projectiles travel with a parabolic trajectory due to the influence
of gravity,
• There are no horizontal forces acting upon projectiles and thus
no horizontal acceleration,
• The horizontal velocity of a projectile is constant (a never
changing in value),
• There is a vertical acceleration caused by gravity; its value is 9.8
m/s/s, down,
• The vertical velocity of a projectile changes by 9.8 m/s each
second,
• The horizontal motion of a projectile is independent of its
vertical motion.
Do Now
A steel ball rolls with constant velocity across a
tabletop 0.950 m high. It rolls off and hits the
ground +0.352 m horizontally from the edge of
the table. How fast was the ball rolling?
Given:
Solution:
Vertical:
Horizontal:
 x  0 . 352 m
h  0 . 950 m
 y  v0 yt 
1
gt
2
1
h  y 
 x  v0 xt
2
gt
2
2
Find:
v0 x  ?
t
v0 x 
2h
g
t
2 ( 0 . 950 )
10 . 0 m / s
2
 0 . 436 s
v0 x 
0 . 352
0 . 436
x
t
 0 . 807 m / s
Horizontally Launched Projectile
Non-Horizontally Launched Projectile
• A cannonball is shot at
an upward angle.
• The cannonball falls the
same amount of
distance in every
second as it did when it
was falling down.
3-5 Projectile Motion
If an object is launched at an initial angle of θ0 with the
horizontal, the analysis is similar except that the initial velocity
has a vertical component.
Horizontal and Vertical Velocity
• The horizontal component is always the same.
• The vertical component changes.
• At the top of the parabola vertical velocity = 0.
True or False?
• The velocity of a projectile at its highest point
is zero.
• False; only vertical component is zero, not
velocity itself. The velocity at the highest point
is equal to its horizontal component.
Range and Projection Angles
• Same initial speed, neglect air resistance
• At different angles projectiles reach different
heights and have different horizontal ranges.
• Angles that add up
to 90 degrees
have the same
range.
The longest range
has a
45 degree angle.
Projectile Motion With Air Resistance
• With air resistance the range is diminished.
• The path is not a true parabola.
Follow-Up Question:
Describe the vertical and horizontal components
of a projectile launched at an angle.
Range Formula
Derive a formula for the horizontal range R of a
projectile in terms of v 0 and 0 .
y  y0  v0 yt 
Solve for t:
v0 yt 
t (v0 y 
1
2
1
gt
2
1
gt
2
2
0  0  v0 yt 
1
gt
2
2
0
gt )  0
t0
or
v0 y 
2
t
1
gt  0
2
2v0 y
g
Range Formula
 2v0 y
R  x  v 0 x t  v 0 x 
 g




v 0 x  v 0 cos  0
v 0 y  v 0 sin  0
 2 v 0 y  ( v 0 cos  0 )( 2 v 0 sin  0 )

R  v 0 x 

g
g


2 ( v 0 sin  )( v 0 cos  )  sin 2
R 
( v 0 cos  0 )( 2 v 0 sin  0 )
g
v 0 sin 2 0
2

g
v 0 sin 2 0
2
R 
g
3-6 Solving Problems Involving Projectile
Motion
Read the problem carefully, and choose the object(s)
you are going to analyze.
1.
2. Draw a diagram.
3. Choose an origin and a coordinate system.
4. Decide on the time interval; this is the same in both
directions, and includes only the time the object is
moving with constant acceleration g.
5. Examine the x and y motions separately.
6. List know and unknown quantities.
7. Plan how you will proceed. Use the appropriate
equations; you may have to combine some of them.
Projectile Launched at an Angle Problem
You shoot a rocket at an angle of 40.0°relative to
the horizontal. The rocket has an initial speed of
30.0 m/s.
a)What are the horizontal and vertical components
of the rockets initial velocity?
b) After 1.00 second of flight, how high is the
rocket?
c) After 1.00 second of flight, how far horizontally
has the rocket traveled?
d) How long will it take the rocket to travel to its
highest point?
a) Given:
v 0  30 . 0 m / s
  40 . 0
Solution:
v 0 x  v 0 cos 
0
Find: v 0 x  ?
v0 y  ?
v 0 x  ( 30 . 0 m / s ) cos( 40 . 0 )  23 . 0 m / s
0
v 0 y  v 0 sin 
v 0 y  ( 30 . 0 m / s ) sin 40 . 0  19 . 3 m / s
0
b) Given:
Solution:
v 0 y  19 . 3 m / s
 y  v0 yt 
t  1 . 00 s
1
gt
2
2
Find: Δy-?
 y  (19 . 3 m / s )(1 . 00 s ) 
1
2
(10 m / s )(1 . 00 s )  19 . 3 m  5 . 00 m  14 . 3 m
2
2
c) Given:
v 0 x  23 . 0 m / s
t  1 . 00 s
Find: Δx-?
Solution:
 x  v0 xt
 x  ( 23 . 0 m / s )(1 . 00 s )  23 . 0 m
d) Given:
Solution:
v 0 y  19 . 3 m / s
v y  v 0 y  gt
vy  0
Find: t-?
t
t
v y  v0 y
g
0  19 . 3 m / d
 10 . 0 m / s
2
 2 . 30 s
ConcepTest 3.10a
Shoot the Monkey I
You are trying to hit a friend with a
water balloon. He is sitting in the
window of his dorm room directly
across the street. You aim straight
at him and shoot. Just when you
shoot, he falls out of the window!
Does the water balloon hit him?
1) yes, it hits
2) maybe – it depends on
the speed of the shot
3) no, it misses
4) the shot is impossible
5) not really sure
Assume that the shot does have
enough speed to reach the dorm
across the street.
ConcepTest 3.10a
Shoot the Monkey I
You are trying to hit a friend with a
water balloon. He is sitting in the
window of his dorm room directly
across the street. You aim straight
at him and shoot. Just when you
shoot, he falls out of the window!
Does the water balloon hit him?
Your friend falls under the
influence of gravity, just like the
water balloon. Thus, they are
both undergoing free fall in the
y-direction. Since the slingshot
was accurately aimed at the
right height, the water balloon
will fall exactly as your friend
does, and it will hit him!!
1) yes, it hits
2) maybe – it depends on
the speed of the shot
3) no, it misses
4) the shot is impossible
5) not really sure
Assume that the shot does have
enough speed to reach the dorm
across the street.
ConcepTest 3.10b
Shoot the Monkey II
You’re on the street, trying to hit a
friend with a water balloon. He sits
in his dorm room window above
your position. You aim straight at
him and shoot. Just when you
shoot, he falls out of the window!
Does the water balloon hit him??
1) yes, it hits
2) maybe – it depends on
the speed of the shot
3) the shot is impossible
4) no, it misses
5) not really sure
Assume that the shot does
have enough speed to reach
the dorm across the street.
ConcepTest 3.10b
Shoot the Monkey II
You’re on the street, trying to hit a
friend with a water balloon. He sits
in his dorm room window above
your position. You aim straight at
him and shoot. Just when you
shoot, he falls out of the window!
Does the water balloon hit him??
This is really the same
situation as before!! The only
change is that the initial
velocity of the water balloon
now has a y-component as
well. But both your friend and
the water balloon still fall with
the same acceleration -- g !!
1) yes, it hits
2) maybe – it depends on
the speed of the shot
3) the shot is impossible
4) no, it misses
5) not really sure
Assume that the shot does
have enough speed to reach
the dorm across the street.
ConcepTest 3.10c
Shoot the Monkey III
You’re on the street, trying to hit a
friend with a water balloon. He sits in
his dorm room window above your
position and is aiming at you with HIS
water balloon! You aim straight at
him and shoot and he does the same
in the same instant. Do the water
balloons hit each other?
1) yes, they hit
2) maybe – it depends on the
speeds of the shots
3) the shots are impossible
4) no, they miss
5) not really sure
ConcepTest 3.10c
Shoot the Monkey III
You’re on the street, trying to hit a
friend with a water balloon. He sits in
his dorm room window above your
position and is aiming at you with HIS
water balloon! You aim straight at
him and shoot and he does the same
in the same instant. Do the water
balloons hit each other?
1) yes, they hit
2) maybe – it depends on the
speeds of the shots
3) the shots are impossible
4) no, they miss
5) not really sure
This is still the same situation!!
Both water balloons are aimed
straight at each other and both
still fall with the same
acceleration -- g !!
Follow-up: When would they NOT hit each other?
ConcepTest 3.4b
Now the cart is being pulled
along a horizontal track by an
external force (a weight
hanging over the table edge)
and accelerating. It fires a ball
straight out of the cannon as it
moves. After it is fired, what
happens to the ball?
Firing Balls II
1) it depends upon how much the
track is tilted
2) it falls behind the cart
3) it falls in front of the cart
4) it falls right back into the cart
5) it remains at rest
ConcepTest 3.4b
Now the cart is being pulled
along a horizontal track by an
external force (a weight
hanging over the table edge)
and accelerating. It fires a ball
straight out of the cannon as it
moves. After it is fired, what
happens to the ball?
Firing Balls II
1) it depends upon how much the
track is tilted
2) it falls behind the cart
3) it falls in front of the cart
4) it falls right back into the cart
5) it remains at rest
Now the acceleration of the cart is completely unrelated to the ball. In
fact, the ball does not have any horizontal acceleration at all (just like
the first question), so it will lag behind the accelerating cart once it is
shot out of the cannon.
ConcepTest 3.4c
The same small cart is
now rolling down an
inclined track and
accelerating. It fires a
ball straight out of the
cannon as it moves.
After it is fired, what
happens to the ball?
Firing Balls III
1) it depends upon how much the track is tilted
2) it falls behind the cart
3) it falls in front of the cart
4) it falls right back into the cart
5) it remains at rest
ConcepTest 3.4c
The same small cart is
now rolling down an
inclined track and
accelerating. It fires a
ball straight out of the
cannon as it moves.
After it is fired, what
happens to the ball?
Firing Balls III
1) it depends upon how much the track is tilted
2) it falls behind the cart
3) it falls in front of the cart
4) it falls right back into the cart
5) it remains at rest
Because the track is inclined, the cart accelerates. However, the ball
has the same component of acceleration along the track as the cart
does! This is essentially the component of g acting parallel to the
inclined track. So the ball is effectively accelerating down the incline,
just as the cart is, and it falls back into the cart.
3-6 Solving Problems Involving
Projectile Motion
Projectile motion is motion with constant
acceleration in two dimensions, where the
acceleration is g and is down.
 x  v x0 t
 y  v y0 t 
1
2
gt
2
3-6 Solving Problems Involving
Projectile Motion
1.
Read the problem carefully, and choose the object(s)
you are going to analyze.
2. Draw a diagram.
3. Choose an origin and a coordinate system.
4. Decide on the time interval; this is the same in both
directions, and includes only the time the object is
moving with constant acceleration g.
5. Examine the x and y motions separately.
6. List know and unknown quantities.
7. Plan how you will proceed. Use the appropriate
equations; you may have to combine some of them.
3-7 Projectile Motion Is Parabolic
In order to demonstrate that
projectile motion is parabolic,
we need to write y as a function
of x. When we do, we find that it
has the form:
This is
indeed the
equation for
a parabola.
Do Now
• While skiing, Ellen encounters an icy
bump, which she leaves horizontally at
12.0 m/s.
How far out , horizontally from her
starting point will Ellen land if she drops
a distance of 7.00 m in the fall?
Dropping With Initial Velocity
Exercise 1
• A ball is thrown downward from the top of a
roof with a speed of 25 m/s. Find the
instantaneous velocity of the ball in 2 seconds.
Average Speed
• During the span of a second time interval a
falling object begins at -10 m/s and ends at
-20 m/s . What is the average speed of the
object during this 1-second interval? What is
its acceleration?
v initial  v final
2

 10 m / s  (  20 m / s )
2
The acceleration is -10 m/s.
  15 m / s
1. $40
2. 40 m/s
3. $20+$10x(3s)=$50
v  v initial  gt
4. 20+10x3=50 m/s
5. $50-10x(time)=0
time=5s
6. 5s
7. 0 m/s
8. 10 m/s, 20 m/s
1. 125 m
2. 105m
3. a. 30 m/s
v
v
0 m / s  30 m / s

 15 m / s
b.
2
2
c. 45 m
1
initial
final
x
gt
2
2
4.
x  v initial t 
x  10  3 
1
2
1
at
2
2
10 ( 3 )  75 m
2
Problem 1.
A long jumper leaves the ground with an initial velocity of 12
m/s at an angle of 28-degrees above the horizontal. Determine
the time of flight, the horizontal distance, and the peak height
of the long-jumper.
Given: