Chapter 5
Projectile motion
Chapter 4:
straight line motion
that was
ONLY vertical or
ONLY horizontal motion
Chapter 5:
considers motion that follows a
diagonal path or a curved path
When you throw a baseball,
the trajectory is a curved path.
We are going to separate
the motion of a projectile into
independent x and y motions
The vertical motion is not affected
by the horizontal motion.
And the horizontal motion is not
affected by the vertical motion.
Observe:
a large ball bearing is dropped
at the same time
as a second ball bearing is fired
horizontally.
What happened?
Remember
adding 2 perpendicular vectors
horizontal
and
vertical
vectors.
When we add perpendicular vectors
we use Pythagorean theorem to find
the resultant.
Consider a vector B that is pointed
at an angle q wrt horizontal direction.
We are going to break vector B into
2 perpendicular vectors:
Bx and By
if you ADD vectors Bx + By
you get vector B.
Graphically, we can say:
Draw a rectangle
with vector B as the diagonal.
the component vectors
Bx and By
are the sides of the rectangle
Application:
A Boat in a river
How can we describe the motion
of a boat in a river?
The motion is affected
by the motor of the boat
and by the current of the river
Imagine a river 120 meters wide
with a current of 8 m/sec.
Imagine a river 120 meters wide
with a current of 8 m/sec.
If a boat is placed in the river
[motor is off] ,
how fast will the boat drift downstream?
If the boat is drifting, the total
speed of the boat just equals
the speed of the current.
for a boat drifting with the current:
Vtotal = Vboat + Vcurrent
Vtotal = -0 + Vcurrent = 8 m/sec
Now suppose this boat can travel at
a constant 15 m/sec when the
motor is on .
What is the total speed of the boat
downstream when the motor is on?
The boat is traveling in the same
direction as the current.
V total downstream =
Vtotal = Vboat + Vcurrent
Vtotal = 15↓ + 8↓ = 23 m/sec ↓
What is the total speed of the boat
traveling upstream
[against the current] ?
The boat and current now move
in opposite directions
Vtotal = Vboat +Vcurrent
Vtotal = 15 ↓+ ( 8 ↑)
Vtotal = 7↓ m/sec
Summary:
traveling downstream:
Vboat + Vcurrent
Traveling upstream:
Vboat + [-Vcurrent]
Crossing the river.
If there was no current,
how many seconds needed for this
boat to travel 120 meters
from A to B?
Velocity = distance
time
so
time = distance
velocity
time = distance
velocity
time = 120 m
15 m/sec
time = 8 seconds
But there if IS a current.
what happens when you try to go
straight across the river from A to B?
The boat will travel from A to C.
Every second
the boat travels ACROSS 15 meters
and
AT THE SAME TIME
every second
the boat will be pushed
DOWNSTREAm 8 meters
by the current .
Vboat = 15 
and
Vcurrent = 8↓
These velocities are perpendicular
The RESULTANT velocity of the boat is
Vresultant2 = Vboat2 + Vcurrent2
Vresultant2 = 152 + 82
Vresultant2 = 225+ 64 =289
Vresultant = 17 m/sec
The boat still crosses the river
in 8 seconds ,
but it lands downstream at point C
not at point B.
How far downstream is point C?
Since the boat travels for 8 seconds
the current pushes the boat
for 8 seconds
Vcurrent = 8 m/sec
Velocity = distance/time
so
Distance = Velocity• time
Distance = 8 m/sec • 8 sec
distance downstream = 64 meters
What is the total distance the boat
travels?
D2 = Dx2 + Dy2
D2 = 1202 + 642
D2 = 14400+4096
D2 = 18496
D = 136 meters
The triangles are similar:
REMEMBER
Every second
the boat travels 15 meter across
in the x direction
IT ALSO TRAVELS
8 meter in the y direction
What if you want to travel from
point A to point B? Can you do that?
You can cross from A to B if you point
the boat in the correct direction.
Remember:
Two perpendicular vectors can be
added to produce a single
resultant vector that is pointed in
a specific direction.
SIMILARLY
ANY vector at angle q can be broken
into the sum of two perpendicular
vectors:
one vector only in x direction
and one vector only in y direction.
The magnitude of the component
vectors is given by
Vx = Vocosq
Vy = Vosinq
If you want to travel from A to B,
you must direct the boat so that the
“y component” of the boat’s velocity
cancels the velocity of the current.
Point the boat so that the
component of the boat’s velocity
“cancels” the river
Choose VboatY
so that it is equal and opposite
to the Vcurrent
VboatX = Vboatcosq
VboatY = Vboatsinq
How do we find angle q, the
direction to point the boat?
Use arcsin or arctan
arcsin = sin-1
Arcsine means
“ the angle whose sine is” :
Sin-1 [VboatY/Vboat] = q
Arctan = Tan-1
Arctan means
“ the angle whose tangent is”
Arctan = Tan-1
Arctan means
“ the angle whose tangent is”
tan-1[Vboaty/Vboatx] = q
Remember
PROJECTILE
MOTION
Projectile motion:
A projectile that has horizontal
motion has a parabolic trajectory
We can separate the trajectory
into x motion and y motion.
In the x direction:
constant velocity
Vx = constant
distance in x direction X = Vx • t
In y direction:
free fall = constant acceleration.
Velocity in y direction :
V = Vo – g t
Distance in y direction
Y = Yo + Vot – ½ g t2
The range of a projectile is the
maximum horizontal distance.
Range and maximum height depend
on the initial elevation angle.
If you throw a projectile
straight up,
the range = 0 height is maximum.
0 degrees : the minimum range but
the maximum height.
The maximum range occurs
at elevation 45o
And for complementary angles
40 and 50 degrees
30 and 60 degrees
15 and 75 degrees
10 and 80 degrees
The range is identical for
complementary angles
BUT the larger elevation angle gives
a greater maximum height.
Remember:
For a horizontal launch:
Vo = initial horizontal velocity
0 = initial vertical velocity
in x direction: velocity is constant
in y direction: acceleration is
constant
If one object is fired horizontally
at the same time
as a second object is dropped
from the same height,
which one hits the ground first?
Horizontal launch:
in x [horizontal] direction
velocity is constant Vx = Vo
acceleration = 0
range = Vo • t
[t = total time ]
Horizontal launch:
In y direction:
projectile is free falling.
Voy = 0
Acceleration = g = 10 m/sec2↓
V = gt ↓
d = ½ gt2
Projectile motion lab:
part 1
Part 1: determine the velocity
Vo of the projectile.
Projectile:
- fired horizontally from height h.
- follows parabolic path
- Range R is where projectile hits the
floor.
Equations: in y direction
Voy = 0
g = constant acceleration
distance h = ½
2
gt
Equations in x-direction
acceleration = 0
[constant velocity]
V= Vo
Range R = Vo ▪ t
Part 1: fire projectile horizontally.
Measure all distances in METERS.
Measure starting height , h.
Measure range R.
Part I Calculations:
distance h = ½ gt2
measure h [ in METERS]
use g = 10 m/sec2
solve equation
to find t [ in seconds ]
distance h = ½ gt2
h= 5 t 2
Measure value for R,
the range in x direction
in METERS
Use equation: R = Vo ∙ t
use measured value of R and
calculated value for T
Example: A projectile is fired
horizontally from a table that is 2.0
meters tall. The projectile strikes
the ground 3.6 meters from the
edge of the table.
Given:
H = 2.0 meters
R = 3.6 meters
h = ½ g t2
R = Vo t
h = ½ g t2
2.0 = ½ [10] t2
2
t
2.0 = ½ [10]
2.0 = 5 t2
2/5 = 0.40 = t2
t = 0.63 sec
For equation: R = Vo t
use
R = 3.6 m
and
t = 0.63 sec
R = 3.6 m = Vo [.63 sec]
Vo = 3.6 m
0.63 sec
Vo = 5.7 m/sec
The height of a projectile at any time
along the path can be calculated.
First calculate the height if there was
no gravity.
If that case, a projectile would follow a
straight line path
the projectile is always a
distance 5t2 below this line.
Y = voy t – ½ gt2
Y = voy t – 5t2
i
summary
Vectors have magnitude and
direction
Scalars have only magnitude
The resultant
of 2 perpendicular vectors
is the diagonal of a rectangle
that has the 2 vectors as the sides.
The perpendicular components of a
vector are independent
of each other.
The motion of a boat in a stream is
the sum of a constant velocity of a
boat [x dir] and the constant velocity
of the stream [y dir]
The path of a boat crossing a stream
is diagonal
The horizontal component of a
projectile is constant,
like a ball rolling on a surface with
zero friction.
Objects in motion remain in motion at
constant speed.
The vertical component of a
projectile is same as for an object in
free fall.
The vertical motion of a horizontally
fired projectile is the same as free
fall.
For a projectile fired at an angle,
the projectile will be 5t2 below
where it would be if there was no
gravity.