Chapter 5 Projectile motion. Chapter 4: straight line motion that was ONLY vertical or ONLY...
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Transcript of Chapter 5 Projectile motion. Chapter 4: straight line motion that was ONLY vertical or ONLY...
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 secondthe 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 = Vboat
2 + 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/timeso
Distance = Velocity• time
Distance = 8 m/sec • 8 sec
distance downstream = 64 meters
What is the total distance the boat travels?
D2 = Dx2 + Dy
2
D2 = 1202 + 642 D2 = 14400+4096
D2 = 18496
D = 136 meters
The triangles are similar:
REMEMBEREvery 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 directionY = 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 occursat elevation 45o
And for complementary angles
40 and 50 degrees30 and 60 degrees15 and 75 degrees10 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 = 0Acceleration = 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 = 0g = constant acceleration
distance h = ½ gt2
Equations in x-direction
acceleration = 0[constant velocity]
V= VoRange 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 metersR = 3.6 meters
h = ½ g t2
R = Vo t
h = ½ g t2
2.0 = ½ [10] t2
2.0 = ½ [10] t2
2.0 = 5 t2
2/5 = 0.40 = t2
t = 0.63 sec
For equation: R = Vo tuse
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.