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.