Post on 28-Dec-2015
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