Chapter 5 Projectile motion. Chapter 4: straight line motion that was ONLY vertical or ONLY...

135
Chapter 5 Projectile motion

Transcript of Chapter 5 Projectile motion. Chapter 4: straight line motion that was ONLY vertical or ONLY...

Page 1: Chapter 5 Projectile motion. Chapter 4: straight line motion that was ONLY vertical or ONLY horizontal motion.

Chapter 5 Projectile motion

Page 2: Chapter 5 Projectile motion. Chapter 4: straight line motion that was ONLY vertical or ONLY horizontal motion.

Chapter 4: straight line motion

that was ONLY vertical or

ONLY horizontal motion

Page 3: 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

Page 4: Chapter 5 Projectile motion. Chapter 4: straight line motion that was ONLY vertical or ONLY horizontal motion.

When you throw a baseball,the trajectory is a curved path.

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We are going to separate the motion of a projectile into independent x and y motions

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The vertical motion is not affected by the horizontal motion.

And the horizontal motion is not affected by the vertical motion.

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Observe: a large ball bearing is dropped

at the same time as a second ball bearing is fired horizontally.

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What happened?

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Remember

adding 2 perpendicular vectors

horizontal and

vertical vectors.

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When we add perpendicular vectors we use Pythagorean theorem to find the resultant.

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Consider a vector B that is pointed at an angle q wrt horizontal direction.

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We are going to break vector B into 2 perpendicular vectors:

Bx and By

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if you ADD vectors Bx + By

you get vector B.

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Graphically, we can say:

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Draw a rectangle with vector B as the diagonal.

the component vectors Bx and By

are the sides of the rectangle

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Page 21: Chapter 5 Projectile motion. Chapter 4: straight line motion that was ONLY vertical or ONLY horizontal motion.

Application:

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A Boat in a river

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How can we describe the motion of a boat in a river?

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The motion is affected by the motor of the boat

and by the current of the river

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Imagine a river 120 meters wide with a current of 8 m/sec.

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Imagine a river 120 meters wide with a current of 8 m/sec.

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If a boat is placed in the river [motor is off] ,

how fast will the boat drift downstream?

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If the boat is drifting, the total speed of the boat just equals

the speed of the current.

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for a boat drifting with the current:

Vtotal = Vboat + Vcurrent

Vtotal = -0 + Vcurrent = 8 m/sec

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Now suppose this boat can travel at a constant 15 m/sec when the

motor is on .

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What is the total speed of the boat downstream when the motor is on?

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The boat is traveling in the same direction as the current.

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V total downstream = Vtotal = Vboat + Vcurrent

Vtotal = 15↓ + 8↓ = 23 m/sec ↓

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What is the total speed of the boat traveling upstream

[against the current] ?

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The boat and current now move in opposite directions

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Vtotal = Vboat +Vcurrent Vtotal = 15 ↓+ ( 8 ↑)

Vtotal = 7↓ m/sec

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Summary:

traveling downstream: Vboat + Vcurrent

Traveling upstream:Vboat + [-Vcurrent]

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Crossing the river.

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If there was no current, how many seconds needed for this boat to travel 120 meters

from A to B?

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Page 41: Chapter 5 Projectile motion. Chapter 4: straight line motion that was ONLY vertical or ONLY horizontal motion.

Velocity = distance time

so time = distance velocity

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time = distance velocity

time = 120 m 15 m/sec

time = 8 seconds

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But there if IS a current. what happens when you try to go straight

across the river from A to B?

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The boat will travel from A to C.

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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 .

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Vboat = 15 and

Vcurrent = 8↓

These velocities are perpendicular

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The RESULTANT velocity of the boat is

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Vresultant2 = Vboat

2 + Vcurrent2

Vresultant2 = 152 + 82

Vresultant2 = 225+ 64 =289

Vresultant = 17 m/sec

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The boat still crosses the river in 8 seconds ,

but it lands downstream at point C not at point B.

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How far downstream is point C?

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Since the boat travels for 8 seconds

the current pushes the boat for 8 seconds

Vcurrent = 8 m/sec

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Velocity = distance/timeso

Distance = Velocity• time

Distance = 8 m/sec • 8 sec

distance downstream = 64 meters

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What is the total distance the boat travels?

D2 = Dx2 + Dy

2

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D2 = 1202 + 642 D2 = 14400+4096

D2 = 18496

D = 136 meters

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The triangles are similar:

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REMEMBEREvery second

the boat travels 15 meter across in the x direction IT ALSO TRAVELS

8 meter in the y direction

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What if you want to travel from point A to point B? Can you do that?

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You can cross from A to B if you point the boat in the correct direction.

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Remember: Two perpendicular vectors can be

added to produce a single resultant vector that is pointed in

a specific direction.

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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.

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The magnitude of the component vectors is given by

Vx = Vocosq

Vy = Vosinq

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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.

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Point the boat so that the component of the boat’s velocity “cancels” the river

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Choose VboatY so that it is equal and opposite

to the Vcurrent

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VboatX = Vboatcosq

VboatY = Vboatsinq

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How do we find angle , q the direction to point the boat?

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Use arcsin or arctan

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arcsin = sin-1

Arcsine means “ the angle whose sine is” :

Sin-1 [VboatY/Vboat] = q

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Arctan = Tan-1 Arctan means

“ the angle whose tangent is”

Page 77: Chapter 5 Projectile motion. Chapter 4: straight line motion that was ONLY vertical or ONLY horizontal motion.

Arctan = Tan-1 Arctan means

“ the angle whose tangent is”

tan-1[Vboaty/Vboatx] = q

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Remember

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PROJECTILE MOTION

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Projectile motion:

A projectile that has horizontal motion has a parabolic trajectory

We can separate the trajectory into x motion and y motion.

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In the x direction:

constant velocity

Vx = constant

distance in x direction X = Vx • t

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In y direction: free fall = constant acceleration.

Velocity in y direction : V = Vo – g t

Distance in y directionY = Yo + Vot – ½ g t2

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The range of a projectile is the maximum horizontal distance.

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Range and maximum height depend on the initial elevation angle.

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If you throw a projectile straight up,

the range = 0 height is maximum.

0 degrees : the minimum range but the maximum height.

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The maximum range occursat elevation 45o

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And for complementary angles

40 and 50 degrees30 and 60 degrees15 and 75 degrees10 and 80 degrees

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The range is identical for complementary angles

BUT the larger elevation angle gives a greater maximum height.

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Remember:

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For a horizontal launch: Vo = initial horizontal velocity

0 = initial vertical velocity

in x direction: velocity is constant

in y direction: acceleration is constant

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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?

Page 94: Chapter 5 Projectile motion. Chapter 4: straight line motion that was ONLY vertical or ONLY horizontal motion.

Horizontal launch:in x [horizontal] direction

velocity is constant Vx = Vo

acceleration = 0

range = Vo • t [t = total time ]

Page 95: Chapter 5 Projectile motion. Chapter 4: straight line motion that was ONLY vertical or ONLY horizontal motion.

Horizontal launch:

In y direction:projectile is free falling.

Voy = 0Acceleration = g = 10 m/sec2↓

V = gt ↓d = ½ gt2

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Projectile motion lab:

part 1

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Part 1: determine the velocity Vo of the projectile.

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Projectile: - fired horizontally from height h.- follows parabolic path - Range R is where projectile hits the floor.

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Equations: in y direction

Voy = 0g = constant acceleration

distance h = ½ gt2

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Equations in x-direction

acceleration = 0[constant velocity]

V= VoRange R = Vo ▪ t

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Part 1: fire projectile horizontally.

Measure all distances in METERS.

Measure starting height , h.

Measure range R.

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Part I Calculations:

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distance h = ½ gt2

measure h [ in METERS]

use g = 10 m/sec2

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solve equation to find t [ in seconds ]

distance h = ½ gt2

h= 5 t 2

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Measure value for R, the range in x direction

in METERS

Page 109: Chapter 5 Projectile motion. Chapter 4: straight line motion that was ONLY vertical or ONLY horizontal motion.

Use equation: R = Vo t∙

use measured value of R and calculated value for T

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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.

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Given: H = 2.0 metersR = 3.6 meters

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h = ½ g t2

R = Vo t

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h = ½ g t2

2.0 = ½ [10] t2

Page 115: Chapter 5 Projectile motion. Chapter 4: straight line motion that was ONLY vertical or ONLY horizontal motion.

2.0 = ½ [10] t2

2.0 = 5 t2

2/5 = 0.40 = t2

Page 116: Chapter 5 Projectile motion. Chapter 4: straight line motion that was ONLY vertical or ONLY horizontal motion.

t = 0.63 sec

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For equation: R = Vo tuse

R = 3.6 m and

t = 0.63 sec

Page 118: Chapter 5 Projectile motion. Chapter 4: straight line motion that was ONLY vertical or ONLY horizontal motion.

R = 3.6 m = Vo [.63 sec] Vo = 3.6 m

0.63 sec

Vo = 5.7 m/sec

Page 119: Chapter 5 Projectile motion. Chapter 4: straight line motion that was ONLY vertical or ONLY horizontal motion.

The height of a projectile at any time along the path can be calculated.

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First calculate the height if there was no gravity.

If that case, a projectile would follow a straight line path

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the projectile is always a distance 5t2 below this line.

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Y = voy t – ½ gt2 Y = voy t – 5t2

i

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summary

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Vectors have magnitude and direction

Scalars have only magnitude

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The resultant of 2 perpendicular vectors

is the diagonal of a rectangle that has the 2 vectors as the sides.

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The perpendicular components of a vector are independent

of each other.

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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]

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The path of a boat crossing a stream is diagonal

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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.

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The vertical component of a projectile is same as for an object in free fall.

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The vertical motion of a horizontally fired projectile is the same as free fall.

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For a projectile fired at an angle, the projectile will be 5t2 below where it would be if there was

no gravity.

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