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Projectile Motion Projectile HW #1: Handout

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Projectile Motion

Projectile HW #1: Handout

Projectile Motion:

Introduction: Projectile motion refers to freefall motion in ______ dimensions. The motion of the object will have a ______________ component and a _______________ component.

There is a constant acceleration due to gravity, , points downwards only. Here the g refers to:

twohorizontal

vertical

g

There are two coordinates to describe projectile motion. The _____ coordinate and component refer to _____________ motion and the _____ coordinate and component refer to _____________ motion.

gg downwards 280.9

sm downwards

xhorizontal

verticaly

We need to define the same concepts used in Ch. 2 for our study of two dimensional motion.

Position: An object must be given a location in space. A two dimensional coordinate system is used:

+y = up

+x = right, or any other horizontal direction

The object’s location can be described in relation to the origin. The origin can be chosen to be any place convenient.

The position can be represented by a vector whose coordinates are (x,y).

yxx ,

The coordinates (x,y) also represent the x and y components of the vector position.

Displacement: The displacement of an object is the ____________ of ____________ of an object. The displacement is a ____________! The displacement is represented as:

changeposition vector

oxxx

+y

+x

Initial position

Final position

yxx ,

Average velocity: The average velocity of an object represents the __________ at which position ____________.rate changes

The average velocity is the displacement of the object divided by the ______________ time.elapsed

t

xx

t

xv o

To simplify the equations, we will always take the initial time to be __________ seconds.zero

0. otie

With this change, the ______ time always equals the ________ time.final elapsed

Instantaneous velocity: The instantaneous velocity of an object represents the __________ at one ____________ of time. The instantaneous velocity is represented by the letter v, and it is also a vector.

velocity instant

Vx

Vy

θ To find the resultant, apply Pythagoras and tan-1 θ

Average acceleration: The average acceleration of an object represents the __________ at which velocity ____________.rate changes

The average acceleration is the change of the ___________ of the object divided by the ______________ time.elapsed

t

vv

t

va o

velocity

Instantaneous acceleration: The instantaneous acceleration of an object represents the _____________ at one ____________ of time. The instantaneous acceleration is represented by the letter a, and it is also a vector.

acceleration instant

Uniformly Accelerated Motion: This kind of motion has a constant ________________. Since the acceleration is constant, the _____________ and the _________________ acceleration are equal.

accelerationaverage instantaneous

Freefall: The acceleration of an object will be due to gravity only. Gravity pulls with a constant acceleration towards the ground, or _____________ downward. The acceleration can be written as a vector as follows:

vertically

gaaa yx ,0,

As before, g just represents the numeric value of the acceleration of gravity.

22 2.3280.9s

fts

mg

The – sign shows the direction points downwards!

2 – dimensional motion difficulty: The motion of a projectile in two dimensions is quite complex!

The way to solve these problems is to resolve all the motion into components.

All the motion completely separates into components. The motion in the x direction is independent of the motion in the y direction.

By separating the motion into x and y components, the motions become independent of one another. The motion in the x – direction is not affected by the motion in the y – direction.

Equations of Motion: The motion of a projectile in each dimension can be represented by the uniform motion equations from Ch. 2. Apply the equations to each coordinate axis separately. The general equations for uniform motion from Ch. 2 are:

221 attvx o

atvv o

xavv o 222

Apply these equations to each coordinate axis. For the x – direction, the acceleration is _________. The equations from Ch. 2 can be adapted to 2 –dimensional motion by adding subscripts to the variables that are vectors. The subscripts give the component name:

zero

0xa The acceleration in the x – direction, ax, is zero.

tvx xo

x is the horizontal displacement.vox is the initial velocity in the x direction.vx is the final velocity.

oxx vv

The x component of the velocity is constant and distance is rate times time.

For the y – direction, the acceleration is _________. Substitute y for x and get the equations of motion for the vertical part of the motion:

ay = – g

221 gttvy yo gtvv yoy

ygvv yoy 222

The motion in the x – direction and the y – direction are independent of one another. The motion of a projectile is that of a constant velocity in the x – direction and a uniformly accelerated motion in the y – direction.

To solve projectile problems, just follow the basic guidelines given in Ch. 2. Read the problem, draw a picture, write down what is given and unknown, choose equations, and solve.

Example #1: A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 18.0 m/s. The cliff is 50.0 m above a flat, horizontal beach, as shown below. a. How long after being released does the stone strike the beach below the cliff?b. How far from the base of the cliff does the stone land? c. With what speed and angle of impact does the stone land?

a. Use the y – direction to find the time to fall.

y = – 50.0 m voy = 0

221 gttvy yo

vox

22100.50 gtm

280.9

0.5020.502

sm

m

g

mt s19.3

solve for t:

b. Use the x – direction to find the horizontal distance.

18.0 3.19mox sx v t s m5.57

c. Solve for each component of the final velocity just before impact.

For the x – direction:sm

oxx vv 0.18

For the y – direction:y o yv v gt

0 (9.80 / / )3.19secyv m s s sm3.31

Note: The minus sign on vy is to show it points downwards!

vx = +18.0 m/sv y

= –

31.

3 m

/s

v

Use Pythagorean theorem:

222yx vvv

22 3.310.18 sm

smv

smv 1.36

Use inverse tangent:

x

y

v

vtan

sm

sm

0.18

3.31tan 1 o1.60

below horizontal

Example #2: A ball is launched horizontally from a height h above the ground. At the same moment, an identical ball is dropped from rest from the same height. Which ball will hit the ground first?

Solution: Both balls hit the ground at the same time. For the ball launched horizontally, the motion in the x – direction and the motion in the y – direction separate. The motion in the y – direction only depends on the y – direction, and is independent of the x – direction. That means the ball dropped straight down and the ball launched horizontally have the same vertical motion. They should arrive at the bottom at the same time with the same vertical speed.

Video demonstration:

Example #3: A hunter aims his banana cannon directly at a monkey hanging on a branch. If the monkey lets go of the branch at the moment the cannon is fired, will the monkey catch the banana?

Both banana and monkey are accelerated equally gravity. No matter how slow the banana is fired, or how far the monkey falls, the two will always make contact.

This can be shown a with a couple of different scenarios. First what would happen if gravity were “turned off”?

If gravity were “off”, the banana would travel directly to the monkey.

If gravity is restored, and the cannon points directly to the monkey, the banana will still arrive to the monkey. Both fall the same distance due to gravity as they move. Here is a high projectile speed example:

Here is a low projectile speed example:

Here is a high projectile speed aimed too high:

Example #4: A student stands at the edge of a building and throws a stone horizontally over the edge with a speed of 12.0 m/s. The stone lands 2.41 seconds after it is thrown. a. How tall is the building?b. How far from the base of the building does the ball land? c. With what speed and angle of impact does the stone land?

vox = 12.0 m/s

voy = 0 m/s

yo = h = ?

y = 0

y = 0 – h

Look for an equation with y and t and solve.

221 gttvy yo

2210 gth

2

21 41.280.9 2 sh

sm m5.28

b. Solve for the x – direction displacement:

stvx sm

xo 41.20.12 m9.28

c. Solve for each component of the final velocity just before impact.

For the x – direction:sm

oxx vv 0.12

For the y – direction: gtvv yoy

svs

my 41.280.90 2 s

m6.23

The last step is to combine these into magnitude and direction. Draw the components and use Pythagorean theorem and inverse tangent.

vx = +12.0 m/sv y

= –

23.

6 m

/s

v

Use Pythagorean theorem:

222yx vvv

22 6.230.12 sm

smv

smv 5.26

Use inverse tangent:

x

y

v

vtan

sm

sm

0.12

6.23tan 1 o1.63

below horizontal

Example #5: Bubba stands at the edge of a building and throws an opossum horizontally over the edge with a speed of 6.50 m/s. The opossum lands 24.1 meters horizontally from the base of the building. How tall is the building? How long for the opossum to land? With what speed and angle of impact does the opossum land?

vox = 6.50 m/s

voy = 0 m/s

yo = h = ?

y = 0

y = 0 – h

x = 24.1 m

You do not have to solve the questions in the order given. Start with the x – direction and solve for time.

tvx xoxov

xt

sm

tsm

71.350.6

1.24

Now that the time is given, follow the last example to find height and final velocity.

Look for an equation with y and t and solve.2

21 gttvy yo

2210 gth

2

21 71.380.9 2 sh

sm m4.67

Solve for each component of the final velocity just before impact.

For the x – direction:sm

oxx vv 50.6

For the y – direction: gtvv yoy

svs

my 71.380.90 2 s

m3.36

The last step is to combine these into magnitude and direction. Draw the components and use Pythagorean theorem and inverse tangent.

vx = +6.50 m/sv y

= –

36.

3 m

/s

v

Use Pythagorean theorem:

222yx vvv

22 3.3650.6 sm

smv

smv 9.36

Use inverse tangent:

x

y

v

vtan

sm

sm

50.6

3.36tan 1 o9.79

below horizontal

Example #6: Bubba stands at the edge of a building and throws an armadillo horizontally over the edge with a speed of 10.0 m/s. The armadillo lands on the ground with a final velocity directed at 55.0o below the horizontal. What is the height of the building?

vox = 10.0 m/s

voy = 0 m/s

yo = h = ?

y = 0

y = 0 – h

vy = ?

Think of what is given to solve…..

sm

oxx vv 0.10

x

y

v

vtan

tanxy vv

osm

yv 0.55tan0.10 sm3.14

Next use the vy to solve for the height:

ygvv yoy 222 hgvy 202

280.92

3.14

2

22

sm

sm

y

g

vh

m4.10

Projectile Motion Day 2 & 3

Projectile HW #2

Projectiles launched at an angle to the horizontal: Yesterday’s notes involved projectiles launched horizontally. In this case the projectiles had a horizontal component to the velocity but not a vertical component. Today the initial velocity will have a magnitude, vo, and an initial angle, o. The components to the initial velocity are as follows:

Angle above the horizontal…..

v

vx

The x – component is adjacent, so use cosine.

v

vxcos cosvvx so

vy

The y – component is opposite, so use sine.

v

vysin so sinvvy

Angle below the horizontal…..

v

vx The x – component is adjacent, so use cosine.

v

vxcos cosvvx so

vy

The y – component is opposite, so use sine. This component also points downwards, so it is negative.

v

vysin so

sinvvy

Example #7: A ball is launched at 25.0 m/s at an initial angle of 36.9o above the horizontal. (a) What are the x and y components of the initial velocity?

vo

vox

voy

ooox vv cos

osm

oxv 9.36cos0.25

sm

oxv 0.20

osm

oooy vv 9.36sin0.25sin sm0.15

(b) What is the greatest height reached by the ball?

voy = +15.0 m/s

vy,top = 0 m/s

y = ?

ygvv yoy 222 ygv yo 20 2

g

vy yo

2

2

280.92

0.15 2

sm

sm

m5.11

+

-

Recall: Vo = 25.0 m/s

Note: This greatest height can be written as:

g

vy yo

2

2

g

vo2

sin 20

g

vo2

sin 022

At what angle should the object be thrown to reach the greatest height?o90

Reason: The largest value that sin() can have is +1, and that occurs only when the angle is 90o.

(c) How long did it take the ball to reach the highest point?

voy = +15.0 m/s vy,top = 0 m/s t = ?

gtvv yoy gtv yo 0

g

vt youp

280.9

0.15

sm

sm

s53.1

(d) What was the total time the ball was in the air?

voy = +15.0 m/s yo = 0 m t = ?yf = 0 m y = 0 m

221 gttvy yo 2

210 gttv yo

221 gttv yo

g

vt yotot

2

280.9

0.152

sm

sm

t s06.3

Think of the ball being thrown up vertically with Voy only.

Note: The total time in the air is just twice the time to the top. This is the same result as in chapter 2. There is the same symmetry here as in the purely vertical motion in Ch. 2: When an object starts and stops at the same vertical height, the time to travel to the top is equal to the time to fall back down.

totaldownup ttt 21

(e) What is the final velocity of the ball?

Start with the x – component: oooxx vvv cos

For the y – direction: gtvv yoy

oooyoy

oyy vvg

vgvv sin

2

The last step is to combine these into magnitude and direction. Draw the components and use Pythagorean theorem and inverse tangent.

vx = + vocosov y

= –

vos

in o

v

Use Pythagorean theorem:

222yx vvv

22 sincos oooo vvv

ovv Use inverse tangent:

x

y

v

vtan

oo

oo

v

v

cos

sintan 1 o

below horizontal

(f) What is the range of the ball?

The range of a projectile is the horizontal distance the particle travels.

stvx sm

xo 06.30.20 m2.61

(g) Write a general formula for the range using only vo, o, and g.

tvx xog

vt yotot

2

ooox vv cos oooy vv sin

Start substituting values in:

tvx xog

vv yoxo

2

oooo vvg

x sincos2

ooo

g

vx cossin2

2

From your Trig class, you learned (or will learn) of an identity:

ooo cossin22sin

ooo

g

vx cossin2

2

oo

g

vx 2sin

2

Note that 45o gives the greatest range.

Example #8: A football is kicked upwards at 75.0 ft/s at 70.0o above the horizontal from the top of a 90.0 foot tall building.(a) What is the maximum height of the ball above the ground?

sfto

sft

oooy vv 48.700.70sin0.75sin

0, topyv ?y

ygvv yoy 2220

g

vy yo

2

2

g

vy yo

2

2

22.322

48.702

sftsft

ft1.77

Add this onto the 90.0 ft tall starting height:

167.y ft

(b) How long does it take for the ball to hit the ground?

fty 0.90 sft

oyv 48.702

21 gttvy yo

221

22.3248.700.90 ttfts

ftsft

00.9048.701.16 2 tt

1.162

0.901.16448.7048.70 2 t

sst 41.5,03.1

(c) What is the horizontal range of the football?

tvx ox

sx sft 41.565.25

ftx 139

sfto

sft

ooox vv 65.250.70cos0.75cos

Stop Tuesday. Do Problems 20, 22, 29 independently. For problem 29 see page 62 for a useful formula.

Warm Up Example #9. 10 Minutes.

Example #9: A cannon fires a round at an angle of 65.0°, and it is in the air for 12.60 s. Find (a) the initial velocity of the projectile.

221 gttvy yo

02

21 gttv yo

21 12 2 9.80 12.60 61.7 mm

o y ssv gt s

sms

m

oooyo vvv 1.680.65sin

74.61sin

(b) What is the range of the projectile?

tvx ox

tvx oo cos

sx sm 60.120.65cos12.68

mx 363

(c) What is the velocity of the projectile, as magnitude and direction, at 10.00 s?

oooxx vvv cos

sm

sm

xv 79.280.65cos12.68

gtvgtvv ooyoy sin

svs

msm

y 00.1080.974.61 2

sm

yv 26.36

vx = +28.79 m/sv y

= –

36.

26 m

/s

v

Use Pythagorean theorem:

222yx vvv

22 26.3679.28 sm

smv

smv 3.46

Use inverse tangent:

x

y

v

vtan

sm

sm

79.28

26.36tan 1 6.51

below horizontal

Example #10: In making a record jump, the truck “Bigfoot” jumped 208 feet. If the truck left the ramp at 69.3 mph, and the landing ramp was identical in angle and height, determine the angle of the launch ramp.

Last 4 words spoken by a Redneck?

“Hey Y’all, Watch This!”

hourmile

ov 3.69

s

hour

mile

ft

3600

1

1

5280

102 fto sv

From example #7, part (g), use: oo

g

vx 2sin

2

21 11 12 2 22

32.2 208sin sin

102

fts

oft

o s

ftg x

v

2.20o

Example #11: It is the last play of the game and Troy is losing by 2 points to Sunny (cough) Hills. They decide to attempt a 55.0 yard field goal. The kicker kicks the ball straight at the 10.0 foot tall goal posts. If he kicks the ball at 52.5 mph and 40.0° above the horizontal, does Troy win?

hourmile

ov 5.52

s

hour

mile

ft

3600

1

1

5280

sft

ov 0.77

First, determine the time for the ball to travel 55.0 yard = 165 feet. Then determine the height of the ball (hopefully) above the ground at this time. Is this height greater than 10.0 feet?

tvx oo cos

sft

tsft

80.20.40cos0.77

165

221 gttvy yo

sy sft 80.20.40sin0.77

2

21 80.22.32 2 s

sft

ftfty 0.105.12

Alternatively….. See the solution to problem 29 in the textbook.

Example #12: Jürgen releases a shot-put 2.00 m above the ground at an angle of 45.0° above the horizontal. If his toss is 20.9 m, how fast did he release it?

oo

g

vx 2sin

2

Why can’t we use the range formula to find Vo?

But…. We could use formula that show the relationships between Y, X, Vo, g, and θ.

Example #12: Jürgen releases a shot-put 2.00 m above the ground at an angle of 45.0° above the horizontal. If his toss is 20.9 m, how fast did he release it?

?ov

mx 9.20 my 00.2

0.45oSubstitute x into y and solve for vo.

tvx oo cosoov

xt

cos

221sin gttvy oo

2

21

coscossin

oooooo v

xg

v

xvy

oo

ov

xgxy

22

2

cos2tan

yxv

xgo

oo

tan

cos2 22

2

oo

o yx

xgv

2

22

costan2

mphv sm

o 5.307.13

Warm UP to Test Tomorrow.

From memory, notes, text or homework write down all useful formulas

Homework Check1.Sample Test2.Problems 20, 22,29,30,31,32,35, 67, 75