Transcript Slide 1

SOLUTION STEPS
MOTION IN TWO DIMENSIONS
For problems involving vector addition or subtraction:
1. Use a protractor and ruler to accurately represent each vector involved in the
problem. Use an appropriate scale in representing the vector.
A) Graphical Method:
Two Vectors: Use the Parallelogram Method and measure both the magnitude
and direction of the resultant.
Three or more Vectors: Use the Polygon Method and measure both the
magnitude and direction of the resultant.
B) Trigonometric Component Method:
1)
break each vector into x and y components.
2)
use the sign convention and assign a positive sign or a negative sign
to the magnitude.
3)
determine the sum of the x components and repeat for the sum of the
y components.
4)
use the Pythagorean theorem and simple trigonometry to solve for the
magnitude and direction of the resultant.
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For problems involving projectile motion:
1. Complete a data table with the information given. Use the proper sign for the
quantity represented by the symbol in the data table depending on whether the object
was initially moving upward or downward. Use the same sign convention as in free
fall problems.
2. Use the trigonometric component method to determine the x and y components of
the initial velocity.
3. Determine which formula or combination of formulas can be used to solve the
problem.
QUAD EQUATION
• READ GIANCOLI P.A-6
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FUNDAMENTALS
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• 5. For projectile motion the
velocity at the maximum height is
zero and the acceleration is g.
• 6. For projectile motion the
velocity on the x-axis is constant.
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Comments: The ball carrier is attached to a Dynamics Cart. The ball is shot straight up when the cart passes a
photogate. The ball then follows a trajectory and lands back in the cart. The cart must be moving at a constant
velocity for this demonstration to work. This demonstration is more showy if the Dynamics Cart goes through a
tunnel after the ball is shot upwards.
Name: Ball on a String
Category: Mechanics
Subcategory: 2-D Motion
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e of a Projectile
echanics
: 2-D Motion
Use the PASCO projectile launcher to fire a small plastic ball. If the starting and ending heights are the same, the range of the projectile is
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Name: The Monkey and the Hunter
Category: Mechanics
Subcategory: 2-D Motion
Classes:
Comments: The blow gun is a long metal tube with a tubing connected on one end and a trigger on the other
end. Aim the blow gun by looking down the barrel. The trigger is wired to the electromagnet so that when the
marble shoots out of the blow gun the connection between the battery and the electromagnet is broken. The
electromagent is a reasonable magnet for a while after the current is cut as well, so it is necessary to weigh the
monkey.
Projectiles
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A+50+0 Dropped and shot balls hit bench simultaneously.
A+50+5 Water projector: Adjustable angle water jet in front of grid.
A+50+10 Monkey and the hunter.
A+50+15 Reaction jet: "L" tube rotates as water flows through it.
A+50+20 Rocket is filled with water and compressed air and launched vertically.
A+50+25 Carbon dioxide propelled rocket flies across room on wire.
A+50+26 Carbon dioxide propelled rotational device.
A+50+35 Ballistics car: Ball ejected from rolling car drops back in.
Applet:
Newton's cannon: View projectile motion with the earth's curvature taken into account. http://galileo.phys.virginia.edu/classes/
Vectors
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A+70+0 X,Y,Z-coordinate system with vector arrows.
A+70+5 Vector arrows of various sizes and colors fit in wooden bases.
A+70+10 Relative velocity: Three electric cars on tracks make chalk line.
A+70+20 Rope with slug(unit of mass) in center is lifted from ends.
A+70+25 Film loop: "Vector addition: Velocity of a boat", 3:35 min.
Applet:
Manipulate the magnitudes and directions of two vectors, and see the effect this has on their vector cross product. http://www.phy
http://home.messiah.edu/~barrett/mpg/cake.mpg
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The teacher should locate or create a projectile motion laboratory activity of the
following manner – Use an inclined plane to launch a ball bearing or similar object
from desk level into a soup can placed on the floor. A popular version of this
activity is called “Bull’s Eye.” Remind students to have read this activity the night
prior to coming to this class session so that groups and tasks will be
predetermined before entering the lab. Have students perform this activity as the
teacher circulates and provides assistance where necessary. Allow 10 minutes for
clean up of materials. The students have discussed the independence of
horizontal and vertical components of motion prior to doing this activity. This
activity requires them to predict the landing point of a projectile. It will be apparent
which students are prepared and which are not. This lab is very popular. Some
teachers have made an addition that they call Bet Your Grade. In this add-on, the
teacher sets up the inclined plane and students are permitted to make
measurements and calculations for the landing of a ball in a can. They are given
only one opportunity to land the ball in the can.
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• KICK BALL DEMO
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CHAPTER TARGETS
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Two-Dimensional Kinematics
• be able to use velocity vectors to analyze constant
velocity relative motion.
• know how to treat motion with constant velocity/
acceleration in two dimensions.
• be able to apply the equations for two-dimensional
motion to a projectile.
• be able to calculate positions, velocities, and times for
various types of projectile motion.
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Vectors
• General discussion.
• Vector - A quantity which has magnitude and
direction. Velocity, acceleration, Force, E Field,
Mag Field,
• Scalar - A quantity which has magnitude only.
(temp, pressure, time)
• This chapter: We mainly deal with
Displacement  D and Velocity = v
– Discussion is valid for any vector!
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• Adding vectors in same direction:
SEEMS SIMPLE?
Graphical Method
• For 2 vectors NOT along same line,
adding is more complicated:
Example: D1 = 10 km East, D2 = 5 km
North. What is the resultant (final)
displacement?
• 2 methods of vector addition:
– Graphical (2 methods of this also!)
– Analytical (TRIGONOMETRY)
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Graphical Method
• 2 vectors NOT along the
same line:
D1 = 10 km E, D2 = 5 km
N. Resultant = ?





D R  D1  D 2

D R  D1  D 2
If D1 is perpendicular to D2
D R  ( D 1 )2  ( D1 )2
DR = 11.2 km
Graphical Method
Vector V is 2.50 cm long,
scale is 1 cm = 12.2 m/s, find magnitude .
V
 12.2 m
s
2.50 cm 
 1 cm

= 30.5 m/s




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Graphical Method (TIP TO TAIL) S
V = V1 + V2
1. Draw V1 & V2 to scale.
2. Place tail of V2 at tip of V1
3. Draw arrow from tail of V1 to tip of V2
This arrow is the resultant V (measure length and
the angle it makes with the x-axis)
Graphical Method (TIP TO TAIL) S
Tip
Tail
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Graphical Method (TIP TO TAIL)
• Consider the vectors A and B. Find
A + B.
B
A
A
A
B
C=A+B
B

We can arrange the vectors as we want, as long as we
maintain their length and direction!!
Graphical Method
• Adding (3 or more) vectors
V = V1 + V2 + V3
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Graphical Method (Parallelogram)
Graphical Method
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• Second graphical method of adding vectors
(parallelogram).
V = V1 + V2
1. Draw V1 & V2 to scale from common origin.
2. Construct parallelogram using V1 & V2 as 2 of the 4
sides.
Resultant V = diagonal of parallelogram from
common origin (measure length and the angle it makes
with the x-axis)
Correct Graphical Method
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Tip To Tail
Parallelogram
Graphically determine the resultant of the following three
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vector displacements: (1) 34 m, 25º north of east; (2) 48 m,
33º east of north; and (3) 22 m, 56º west of south.
The vectors for the problem are drawn approximately to scale.
The resultant has a length of 58 m and a direction 48o north
of east. If you actually measured, the actual resultant should be
22
57.4 m at 47.5o north of east.
48
34
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Subtraction of Vectors
B
A
R=A+B
R
A
B
R’ = A - B
A
R’
-B
Subtraction of Vectors
• First, define the negative of a vector:
- V  vector with the same magnitude (size)
as V but with opposite direction.
Math:
V + (- V)  0
• For 2 vectors, A - B  A + (-B)
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Subtraction of Vectors
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Graphical Method (TIP TO TAIL)
• Deck of Card Activity
0 degrees from x axis
90 degrees from x axis
180 degrees from x axis
270 degrees from x axis
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• To do the graphical method we have to
measure with ruler and protractor. Takes a
lot of time
• We don’t always have an graphic designer
on staff to draw our vectors for us.
• So instead we resolve our vectors into its
components (x,y)
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• V is resolved into components: Vx & Vy
V  Vx + Vy (Vx || x axis, Vy || y axis)
Trig Functions to Decompose Vector
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SIN PROJECTES ALONG ONLY Y
Vy = sin q * V
COS PROJECTES ALONG ONLY X
Vx = cos q * V
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• The magnitude (length) of r is found using
the Pythagorean theorem:
r
y
x
V  v  Vx  Vy
2
2
Decomposition Example S
A coast guard cutter has taken a heading of 30 north of
east to find a missing diver. Break up into it’s X and Y
d = displacement 500 m, 30º N of E
Vx  6.80 m/s
Vy  7.40 m/s,
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determine the magnitude and direction of the resultant vector
V  V V 
2
x
2
y
6.80   7.40   10.0 units
2
2
y
Vx
q
The direction is given by an angle of
Vy
tan q 
Vx
q  tan
1
7.40
6.80
Vy
 47 o
47o below the positive x-axis.
V
x
An airplane is travelling 735 km/hr in a direction 41.5º of north of west
(a) Find the components of the velocity vector in the northerly and
westerly directions. (b) How far north and how far west has the plane
travelled after 3.00 h?


vnorth   735 km h  cos 41.5o  550 km h


vwest   735 km h  sin 41.5o  487 km h
dnorth  vnortht   550 km h 3.00 h   1650 km
dwest  vwest t   487 km h 3.00 h   1460 km
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Components are Independent S
A ball is thrown 23.0 with a velocity of 22.5 m/s.
What is its vertical and horizontal velocity components?
vx  v cos q
= 22.5 m/s cos 23.0
vy  v sin q
=
20.1 m/s
= 22.5 m/s sin 23.0
=
8.79 m/s
if vy is 12.4 m/s and vx is 15.4 m/s, find the angle q.
 12.4 m
s
q  tan 1 
 15.4 m
s





= 38.8
A ship is steaming north at 5.0 km/h directly across a river that is 5.0 km acros
The current is swirling to the east (west to east) at 3.0 km/h.
What is its resultant velocity? How far down the bank does it drift at landing?
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1. Find resultant v:
c  a b
2
2
c  a b
2
2
2

 1
t  500.0 m 
km
 5.0
h

2. Find time to cross:
v
x
t
t
2
km  
km 

vr   3.0

5.0
 

h  
h 

x
v
2
Vr = 5.8 km/h

  1 km 


1
000
m

 

T = 0.100 h
3. Find distance in horizontal direction:
x
v
t
x  vt
x  3.0
km
0.10 h 

h
x = 0.30 km
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Vector Multiplication – Cross
Product
B

A
vectors MUST be tail to tail
CAxB
Magnitude of C = ABsin
Direction of C found using
right hand rule
Order of cross does matter
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Physics applications of vector
products
1) Scalar multiple of a vector
F  m  a  2kg  (3m / s 2i  4m / s 2 j )
Newton’s 2nd Law
2) Work – Dot Product
q
F
s
3) Torque on a wheel – Cross Product
F
r
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Work  F cosq  s
Torque  r x F
Multiplication by a Scalar S
• A vector V can be multiplied by a scalar C
V' = C V
V'  vector with magnitude CV & same
direction as V
I.E a force over a period of time is a greater force
In relation to the time passed.
Multiplication by a Scalar
Vector for velocity has the components from the origin of (4,3,0)
After three seconds how far have we moved?
The vector (4,3,0) has a magnitude:
v  42  32  02
v  25  5m / s
v  3sx5m / s  15m
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“x” and “y” components of
motion are independent.
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Cart
• A man on a train tosses a ball straight up
– View this from two reference frames:
Reference frame
on the moving train.
Reference frame
on the ground.
Conceptual Example 3-6 S
v0x 
• Demonstration!!
Projectile Motion
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• no air resistance
• ax = 0, vx = constant
• projectile’s weight is only force
acting
• ay = g =  9.8 m/s2
• vy = 0 at max height
• x, y motions are independent of
each other
• resolve initial velocity into vx, vy
and use our basic kinematic
equations to solve problem
Projectile Motion
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• PHYSICS: y part of motion:
vy = gt , y = (½)g t2
SAME as free fall motion!!
 An object projected horizontally will
reach the ground at the same time
as an object dropped vertically from
the same point!
(x & y motions are independent)
SINCE G IS ALWAYS DOWN, MAKE IT NEGATIVE FOR ALL PROJ MOTION
AND POSITIVE FOR FREE FALL
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Vertical:
v y  v yo  gt
v  vo  gt
1 2
y  y0  v yot  gt
2
1
y  vo t  gt 2
2
v  vo  2gy
2
2
Horizontal:
x  v0t
v  v  2g y 
2
2
y
yo
Your book says negative, just remember in most cases g will be
opposite the initial velocity (so -9.8 m/s) (Except free fall)
vx  vxo
(no acceleration)
x  x0  vxot
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DEMO
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Ball Rolls Across Table & Falls Off
In this case I take down as positive since vyo=0
vyo=0
vy=gt
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• Summary: Ball rolling across the table &
falling.
• Vector velocity v has 2 components:
vx  vx 0
v y  gt
• Vector displacement D has 2 components:
x  vx0t
y  vo t 
1 2
gt
2
For problems like this if we set y at the top = 0 then gravity is – 9/8 m/s2
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A rookie soldier in basic training jumps over an obstacle, and
fires a shot from his rifle at the same time. The barrel of the
rifle is exactly horizontal at the moment of the shot.(no air res.)
Who touches the ground first, the soldier or the bullet?
The soldier and the bullet touch down at the same time.
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Driving off a cliff!!
negative y
1 2
y  gt
2
2y
t
g
2( 50m)
t
2
 9.8m / s
3.19 s  t
(no acceleration)
90
x
 v0 3.19  v0
t
y is positive upward, y0 = 0 at top. Also vy0 = 0
• How fast must the motorcycle leave the cliff to land at
x = 90 m, y = -50 m? vx0 = ?
28.2m / s  vx 0
A flag pole ornament falls off the top of a 25.0 m flagpole.
How long would it take to hit the ground?
1 2
y  gt
2
2y
t 
g
2
Now assume that a bullet causes it to
Fall with a horizontal velocity of 5 m/s
How far does it travel from the pole?
X=vxt
X= 5m/s x 2.26s = 11.3 m
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2y
t
g
2  25.0 m 
t
m
9.80 2
s
t = 2.26 s
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A stone is thrown horizontally from the top of a cliff that is
40.0 m high. It has a horizontal velocity of 15.0 m/s.
We want to find how long it takes the stone to fall to the deck and
how far it will travel from the base of the cliff.
1 2
y  gt
2
2y
t
g
Plug in the given values:
2  40.0 m 
t
m
9.80 2
s
t =2.86 s
Find x :
x = vx t
m
x  15.0  2.86 s 
s
x = 42.9 m
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B-17 is flying level at 375 km/h. The bombs travel a
horizontal distance of 5250 m. What was the
altitude of the plane at the time they called the old
"bombs away"?
km  1000 m  1 h 
vx  375


 = 104 m/s
h  1 km  3600 s 
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x = v xt
t
x
vx

 1
t  5 250 m 
 104 m
s






t = 50.5 s
Find vertical distance (altitude) to take 50 sec:
1 2
y  gt
2
1
m
2
  9.80 2   50.5 s 
2
s 
y = 12 509 m
A diver running 1.8 m/s dives out horizontally from the edge of a
vertical cliff and 3.0 s later reaches the water below. 1) How high was
the cliff, and 2)how far from its base did the diver hit the water?
Choose downward to be the positive y direction.
The origin will be at the point where the diver dives from the cliff.
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In the horizontal direction,
vx 0  1.8 m sand
ax  0
In the vertical direction
t  3.0 s
vy 0  0 ay  9.80 m s2
y0  0
y  y0  v y 0 t  12 a y t 2

y  00
1
2
 9.80 m s   3.0 s 
2
2
 44 m
The distance from the base of the cliff to where the diver hits the water is found from the
horizontal motion at constant velocity:
x  vxt  1.8m s 3 s   5.4 m
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A ball thrown horizontally at 22.2 m/s from the roof of
a building lands 36.0 m from the base of the building. How tall is
the building?
vy 0  0
vx  22.2m / s
x  36 .0m
x  vxt
,
ay  9.80 m s
2
y0  0
 t  x vx  36.0 m 22.2 m s  1.62 s
The vertical displacement, which is the height of the building, is
found by applying.
1 22
2
2
1 gt
1
yy  yy0 
v
t

yo
 v y 0t  2 a y t
 y  0  0  2  9.80 m s  1.62 s   12.9 m
0
2
The pilot of an airplane traveling 180 km/hr wants to drop
supplies to flood victims isolated on a patch of land 160 m
below. The supplies should be dropped how many seconds
before the plane is directly overhead?
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vy 0  0
ay  9.80 m s2
The time of flight
1
y  y0  v0t  gt 2
2
y0  0
1 2
y  gt
2
y  160 m
2( 160m)
t
2
 9.8m / s
5.71s  t
Note that the speed of the airplane does not enter into this calculation.
Projectile Motion
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• PLAY JB 5 CLIP HERE
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Projectile Motion
• General Case:
a = g (down always)
Vyo=Vosinqo
Vxo=Vocosq0
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If Bubba is traveling at 20m/s and the ramp is 30 deg will he make a
ramp distance of
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http://www.youtube.com/watch?v=_GAGjkkFBv4&feature=related&safety_mode=true&pe
rsist_safety_mode=1&safe=active
The football leaves punters foot at S
angle of 37o and 20.0 m/s.
θ0 = 37º, v0 = 20 m/s
 vx0= v0cos(θ0) = 16 m/s, vy0= v0sin(θ0) = 12 m/s
a. Max height? b. Time when hits ground?
c. Total distance traveled in the x direction?
d. Velocity at top?
e. Acceleration at top?
A ball is given an initial velocity of 22. 7 m/s at an angle of
.
66.0 to the horizontal. Find how high the
ball will go?
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vy0 = v0 sin q0
Find y:
vy0 = 20.7 m/s
vy0 = 22.7 m/s (sin 66.0) vy0 = 20.7 m/s
and
vy = 0.
vy2  vy20  2gy
2 gy  vy20
2 possible
Formulas
2
y
v y20
2g
m

  20.7 
s
dy  
m

2  9.80 2 
s 


m2 
  428.5 2 
s 

dy 
m


19.60

s 2 

y = 21.9 m
The biker has a hang time of 4.5 s.
He lands 48 m from the ramp,
(assume no ramp height)
what was the ramp angle?
NEED TO FIND Vy0 and Vx0
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48 m vx0 = 10.67 m/s
dx
vx 
vx 
t
4.5 s
1
NEED TO FIND t at max height
t = 2.25 s
t  4.5s 
2
NEED TO FIND Vy
vy  vy 0  gt
vy 0  gt
0  vy 0  gt
m

vi    9.80 2  2.25 s
s 

Vy0 = 22.05 m/s
NEED TO FIND t at max height
 vy 0 
q  tan  
 vx 0 
1
 22.05 m
q  tan 
 10.67 m
1
s

s
q = 64 
• A confused soldier fires his artillery piece at angle that maximizes
Vy. His enemy utilizes an angle of 40o. If both are in the air for 3
seconds (same height) what was the velocity of the enemy's ball.
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At Vy Max the Vxo =0, t to the top = 1.5s
Vy  v y 0  gt
 Vyo  gt Vyo  (9.8m / s2)(1.5s)
Vy 0  14.7m / s
Since both are in the air for the same amount of time, both have same Vyo
Vy 0  14.7m / s
14.7m / s  vy (sin 40)
y
q40
Vy  22.8m / s
A naval gun fires a 16 inch projectile. The muzzle
velocity (speed of the bullet) is 345 m/s with an
angle of 32.0 . What is the range of the shot?
Hint since both are at y=o then you don’t need the quadratic to solve for time
Find the vertical velocity:
vy0
m
 345 sin 32.0
s
Find the time:
vy  vy 0  gt
(182 .8m / s )
 tup
2
9.8m / s
18.6s  tup
(18.6s)2  ttotal
t = 37.27 s
vy 0  182.8m / s
Find range:
x   v0 cosq0  t
m

x   345 cos 32.0
s


 37.27 s

x = 10904 m
S
You throw a potato at an angle of 22.2 . If it is in the
air for 1.55 s, how far did it go?
S
Find the vertical velocity:
vy  vy 0  gt
vy 0  gt
m  1.55 s 
v y 0  9.80 2 

s  2 
v y 0  7.60m/s
Find the horizontal velocity:
tan q 
vy 0
vx 0
vx 0 
vy 0
tan q
m
s
vx 0 
tan 22.2
7.60
vx 0  18.62m/s
Find the horizontal displacement:
x  vx0t
x  18.62
m
1.55 s 
s
x  28.9 m
A ball is thrown at some angle. The ball is in the air for 4.50 seconds
before it hits. If it travels 45.0 meters before it hits the ground, what
was the initial velocity of the ball (magnitude and direction please)?
x
vx 
t
45 m

4.50 s
m
 10
s
v  vo  at v  vo  at 2v  at v 
at
2
2
v  v y 2  vx 2
tan q 
vy
vx
9.8
m  m

  22.05   10 
s  s


22.05 m
10 m
s
s
q
m
4.5 s 
2 
m
s
 22.05
2
s
2

 24.21
m
s
65.6o
S
S
But what if the height is not the
same after landing?
A husky kicker wants to provide as much hang time as possible For
his coverage team to get down the field. If he kicks the ball with a
velocity 40m/s at an angle of 35 degrees above horizontal, how
much time does his team have to get down field and pound a cougar?
At what distance from kick will we find a pounded cougar returner?
Hint since both are at y=o then you don’t need the quadratic to solve for time
S
35
vy0 = v0 sin q
vy0 = 40 sin 35
vy0 = 22.9 m/s
Vy  v y 0  gtup
vx0 = v0 cos q
vx0 = 40 cos35
vx0 = 32.7 m/s
0  (22.9m / s)  (9.8m / s)tup
( 22 .9m / s )
 tup
tup  2.34s
2
9 .8 m / s
ttotal  (2.34s)(2)  4.68s
x  v cos35(t )
x  32.7(4.68s)
x  153 .4m
S
A football leaves a punters foot 1.0 m above the
ground at 20.0 m/s. θ0 = 37º. Find t then x.
Vy0 = 12.0 m/s, θ0 = 37º
1 2
y  yo  vot  gt
2
m 1
m
t  (9.8 2 )t 2
s
2
s
m 2
m
0  4.9 2 t  12 t  1.0m
s
s
Quadratic Equation
0m  1.0m  12.0
t  0.081s,2.53s
x
vx0 
t
x
16 m / s 
2.53
40.5m  x
A stone is thrown off the top of a building from a height of 45.0 m.
The stone has a launch angle of 62.5 and a speed of 31.5 m/s.
How long is the stone in flight
S
What is its speed just before it hits the ground?
What angle does it hit?
Vy 0  31.5 sin 62.5
Vx0  31.5 cos62.5
vx 0  vxf
Vy 0  27.9m / s
y
vxf  14.5m / s
v y  v y 0  gt
0  27.9m / s  (9.8m / s 2 )(tup )
tup  2.84s
tarch  (2.84s)(2)  5.68s
45.0 m
Quadratic Equation
Vf at bottom from 45 meters
v f  v0  2 gy
 (27.9m / s) 2  2(9.8m / s 2 )(45m)
2
x
vf
vf
2

2
1666 m 2 / s 2 v f
 40.8m / s
Vy  v y 0  gt
t1/ 2down  1.32s
 40.7m / s  27.9m / s  (9.8m / s 2 )(t1/ 2down )
S
ttotal  1.32  5.68s  7.0s
What is its speed just before it hits the ground?
v  v y  vx
2
2
v   40.82 14.52
v  43.2m / s
vx
What angle does it hit?
tanq 
vy
vx
q
 40.7m / s
tan q 
14.5m / s
q  70.3o
q  tan1 (2.8)
Remember that this 2.8 is negative
(so will be below horizon)
vy
q
q
FORMULAS GIVEN ON AP
S
EXAM TIME
S
• Ichiro clobbers a fastball toward Safeco center-field. The ball
is hit at 1 m (yo ) above the plate, and it’s initial velocity is
36.5 m/s (v) at an angle of 30o above horizontal. The
center-field wall is 113 m from the plate and is 3 m (y) high.
• How high does it fly?
• What time does the ball reach the fence?
• At What height does it clear the fence?
• Does he get a home run?
v
y0
q
X0
v
q
h
v0y
v0x
D
Note: t to wall is not necessarily twice the tup
S
• How high does it fly ?
vyo  (36.5m / s) sin 30
vy  vy 0  gt
0  18.25m / s  (9.8m / s)(t )
tup  1.86s
1 2
y  yo  v yot  gt
2
vyo  18.25m / s
y
P
v0
h
q
L
1
2
2
y  1  (18.25m / s )(1.86 s )  (9.8m / s )(1.86 s )
2
y  1m  33.95m  16.95m
y  18.0m
S
x
• How long does it take for ball to reach the wall ?
vxo  (36.5m / s) cos30
x
vx0 
t
113m
31.6m / s 
t wall
vxo  31.6m / s
S
twall  3.57s
Not the total time of flight!!!!
•What is the balls height at the wall ?
1 2
y  yo  v yot  gt
2
1
y  1m  (18.25m / s )( 3.57 s )  (9.8m / s )( 3.57 ) 2
2
y
y  1m  65.15m  62.45m
y  3.7m
v0
q
P
h
L
x
S
A depth charge is shot from the top of an aircraft
carrier125 m above sea level with an initial speed
of 65.0 m/s at an angle of 37.0º with the horizontal
(a) Determine the time taken by the projectile to hit point P at
ground level.
(b) Determine the range X of the projectile as measured from the
base of the cliff. At the instant just before projectile hits point P.
(c) the horizontal and the vertical components of its velocity,
(d) the magnitude of the velocity, and
(e) the angle made by the velocity vector with the horizontal.
(f) Find the maximum height above the cliff top reached
S
(a) Time to hit P
v0  65.0 m s
ay   g
1 2
y  yo  v yot  gt
2
t

3982

 9.8
 39.1
y0  125 vy 0  v0 sin q0
vyo  65sin 37  35.6m / s
0  125m  (35.6m / s)t  (4.9m / s2 )t 2
y
t  2.45s,10.4s
45.0 m
x
S
(b) Determine the range
vxo  (65.0m / s) cos37.0
x
vx0 
t
54 .3m / s
x  565m
x  54.3m / s(10.4s)
(c) the horizontal and the vertical components of its velocity, not
where it is abeam launch site. Hint: Just find Vy at 10.4s. Vx is same

v y  v y 0  at  v0 sin q 0  gt   65.0 m s  sin 37.0o  9.80 m s 2
 10.4 s 
 63.1m s
vx  v0 cos q. 0   65.0 m s  cos 37.0o  51.9 m s
(d) The magnitude of the velocity is found from the x and y componen
calculated in part c) above.
v  vx2  v y2 
 51.9 m s    63.1m s   81.7 m s
2
2
(e) the angle made by the velocity vector with the horizontal.
51.9
v
63.1
q  tan 1
-63.1
y
vx
 tan 1
51.9
 50.6o
50.6o below the horizon
Can use y=Vot+2gy as well
v  v  2 a y  y  y0 
2
y
2
y0
ymax 
v sin q 0
2
0
0  v sin q 0  2 gymax
2
2
o
65.0
m
s
sin
37.0


2
2
2g

2
0


2 9.80 m s
2

 78.1 m
S
1-D Velocity Trick #3
S
“I Hate the Quadratic Approach”
If both points are at a y=0 on path
• Velocity at beginning is + Vyo
Velocity at
at
yo and Velocity
landing is –Vyo
• So we can figure out the time by subtracting
the time it takes gravity to move us from Vyo to
+ –Vyo
vy  vy 0  gt
 v y 0  v y 0  (9.8m / s)t
 vy0  vy0
t
2
 9.8m / s
-
31. A projectile is shot from the edge of a cliff 125 m above ground
level with an initial speed of 65.0 m/s at an angle of 37.0º with the
horizontal
(a) Determine the time taken by the projectile to hit point P at
ground level.
(b) Determine the range X of the projectile as measured from the
base of the cliff. At the instant just before projectile hits point P.
(c) the horizontal and the vertical components of its velocity,
(d) the magnitude of the velocity, and
(e) the angle made by the velocity vector with the horizontal.
(f) Find the maximum height above the cliff top reached
S
y
45.0 m
(a) Time to hit P
x
v0  65.0 m s
ay   g
1 2
y  yo  v yot  gt
2
t

3982

 9.8
 39.1
y0  125 vy 0  v0 sin q0
vyo  65sin 37  35.6m / s
0  125m  (35.6m / s)t  (4.9m / s2 )t 2
t  2.45s,10.4s
S
(b) Determine the range
vxo  (65.0m / s) cos37.0
x
vx0 
t
54 .3m / s
x  565m
x  54.3m / s(10.4s)
(c) the horizontal and the vertical components of its velocity, not
where it is abeam launch site. Hint: Just find Vy at 10.4s. Vx is same

v y  v y 0  at  v0 sin q 0  gt   65.0 m s  sin 37.0o  9.80 m s 2
 10.4 s 
 63.1m s
vx  v0 cos q. 0   65.0 m s  cos 37.0o  51.9 m s
(d) The magnitude of the velocity is found from the x and y componen
calculated in part c) above.
v  vx2  v y2 
 51.9 m s    63.1m s   81.7 m s
2
2
(e) the angle made by the velocity vector with the horizontal.
51.9
v
63.1
q  tan 1
-63.1
y
vx
 tan 1
51.9
 50.6o
50.6o below the horizon
v y2  v y20  2a y  y  y0 
ymax 
v sin q 0
2
0
2
2g


 65.0 m s 

0  v02 sin 2 q 0  2 gymax
2
2
sin 37.0
2 9.80 m s
2

o
 78.1 m
S
Leading the Reciever
• Demonstration!!
S
Shooting the Monkey
(tranquilizer gun)

S
Where does the zookeeper
aim if he wants to hit the monkey?
( He knows the monkey will
let go as soon as he shoots ! )
NOTES
2 derivations
• vertical displacement as a function of
horizontal range
– shape of trajectory is inverted parabola
• horizontal range as a function of launch
angle
S
S
WHICH ONE IS CORRECT?
Shooting the Monkey...

If there were no gravity, simply aim
at the monkey
r =v0t
r = r0
S
Shooting the Monkey...

With gravity, still aim at the monkey!
r = v0 t - 1/2 g t2
S
r = r0 - 1/2 g t2
Dart hits the
monkey!
Both Ball And Monkey Fall at the Same Rate
Recap:
Shooting the monkey...
x = v0 t
y = -1/2 g t2

This may be easier to think about.
It’s exactly the same idea!!
x = x0
y = -1/2 g t2
S
32. A shot-putter throws the shot with an initial speed of
15.5 m/s at a 34.0º angle to the horizontal. Calculate the
horizontal distance travelled by the shot if it leaves the
athlete’s hand at a height of 2.20 m above the ground.
S
v yo  v0 sin q

ay  9.80m s2
y  0m
y0  2.20m

v y 0  15.5 sin 34.0o m s  8.67 m s
1 2
2

(
8
.
67
m
/
s
)
t

4
.
9
t
y yyyo vv, yott  1 a gtt 2  1 a t 2  0v t 2.y2m
0 
0
y0
y
y0
22 y
2
t
v y 0 , v y20  4  12 a y    y 
vxo  v0 sin q
2 12 a y
2

t 8.67

8.67


 8.67
8.67 2(29.89.80
(2.2) 2.20 


 1.
2(4.99.80
)
t  0.225 s,1.99 s
vxo  (15.5m / s) cos34.0
13 .3m / s
x
x  13.3m / s(1.99s)
vx0 
t
x  25.6m
34.0
2.2m
MAX RANGE
S
• For a given range, R < Rmax there are 2
different firing angles that will land the
projectile in the same place. If you fire a shell
at 30 degrees, it will land at exatly the same
spot as if you fire it at 60 degrees.
• The 2 angles are related to each other by q2 =
90 – q1, so at Rmax, q2= q1 = 45o
Vo sin 2qo
R
g
2
S
Example 3-8
• Range (R) of projectile  Maximum horizontal
distance before returning to ground. Derive a
formula for R.
S
MAX RANGE
• Range R  the x where y = 0!
• Use
vx = vx0 , x = vx0 t , vy = vy0 - gt
y = vy0 t – (½)g t2, (vy) 2 = (vy0)2 - 2gy
• First, find the time t when y = 0
0 = vy0 t - (½)g t2
t = (2vy0)/g
 t = 0 (of course!) and t = (2vy0)/g
• Put this t in the x formula: x = vx0 (2vy0)/g  R
R = 2(vx0vy0)/g, vx0= v0cos(θ0), vy0= v0sin(θ0)
R = (v0)2 [2 sin(θ0)cos(θ0)]/g
R = (v0)2 sin(2θ0)/g (by a trig identity)
S
19.(II) A fire hose held near the ground shoots water at a speed of
6.8m/s. At what angle(s) should the nozzle point in order that the
water land 2.0 m away? Why are there two different angles?
Sketch the two trajectories.
S
R
v02 sin 2q 0
g
sin 2q 0 
Rg
v02

 2.0 m   9.8 m

2
 6.8 m s 
2q 0  sin 1 0.4239

s2
  0.4239
q 0  13o , 77 o
2.5
2
1.5
1
0.5
0
0
-0.5
0.5
1
1.5
2
S
S
A flaming load is catapulted at an angle of 40o from the
horizontal. With what speed must the projectile be
launched so that it lands 1500m from the launching point?
NEED TO FIND t through Vy0
Find t at zenith
vyo  v0 sin 40
vxo  v0 cos40
vyo  0.64v0
vxo  .77v0
Vy  v y 0  gt
m
m
0  .64 v0  ( 9.8 2 )t
s
s
Total time in flight is the double of the time in flight for the zenith
t  .065vo
t  .13vo
1500m
x
.77v0 
vx0 
(.77v0 )(.13vo )  1500m
.13vo
t
(0.10vo )  1500
2
q
Vy0
40o
Vxo
S
1500
vo  15000
m
vo  122 .4
s
S
Ch. 3 p.65 : 4, 5, 6, 9,10, 18, 19, 20, 21, 23, 27, 30, 31, 32, 39,40, 41, 44, 45
 v0 sin q0
.
(a)
The maximum height is found from Eq. 2-11c,
 vy20  2ay 30.
 y  y0 
with
(II) A projectile is fired with an initial speed of
S
at an angle of 34.5º above the horizontal on a long flat firing range. Determin
0
at
the maximum height.
x
65.2 m s
 0
v y2  v y20
2a y

v02 sin 2 q 0
2 g

v02 sin 2 q 0
2g

 65.2 m s 

2
sin 2 34.5o
2 9.80 m s 2

 69.6 m
(b)The total time in the air is found from Eq. 2-11b, with a total vertical displacement
of 0 for the ball to reach the ground.
 y0  v y 0t  12 a y t 2
2v0 sin q 0
g


0  v0 sin q 0t  12 gt 2
2  65.2 m s  sin 34.5o
 9.80 m s 
2

 7.54 s and t  0
The time of 0 represents the launching of the ball.
(c)The total horizontal distance covered is found from the horizontal motion at constant velocity.

 vx t   v0 cos q 0  t   65.2 m s  cos 34.5o
  7.54 s  
405 m
(d)The velocity of the projectile 1.50 s after firing is found as the vector sum of the horizontal
and vertical velocities at that time. The horizontal velocity is a constant


cos q 0   65.2 m s  cos 34.5o  53.7 m s
. The vertical velocity is found from
Eq. 2-11a.
o
2
S
S
VECTORS
S
Example 1: Find the height of a
building if it casts a shadow 90 m long
and the indicated angle is 30o.
The height h is opposite 300 and
the known adjacent side is 90 m.
opp
h
tan 30 

adj 90 m
0
h
300
h = (90 m) tan 30o
90 m
h = 57.7 m
Finding Components of Vectors
S
A component is the effect of a vector along
other directions. The x and y components of
the vector (R,q are illustrated below.
x = R cos q
R
q
x
y
y = R sin q
Finding components:
Polar to Rectangular Conversions
Example 2: A person walks 400 m in a
direction of 30o N of E. How far is the
displacement east and how far north?
N
N
R
q
x
400 m
y
30o
E
y=?
x=?
The x-component (E) is ADJ:
x = R cos q
The y-component (N) is OPP:
y = R sin q
E
S
Example 2 (Cont.): A 400-m walk in a
direction of 30o N of E. How far is the
displacement east and how far north?
N
S
Note: x is the side
400 m
30o
y=?
x=?
E
x = (400 m) cos 30o
= +346 m, E
adjacent to angle 300
ADJ = HYP x Cos 300
x = R cos q
The x-component is:
Rx = +346 m
Example 2 (Cont.): A 400-m walk in a
direction of 30o N of E. How far is the
displacement east and how far north?
N
S
Note: y is the side
400 m
30o
y=?
x=?
E
opposite to angle 300
OPP = HYP x Sin 300
y = R sin q
y = (400 m) sin 30o
The y-component is:
= + 200 m, N
Ry = +200 m
Example 2 (Cont.): A 400-m walk in a
direction of 30o N of E. How far is the
displacement east and how far north?
N
400 m
30o
Rx =
Ry =
+200 m
E
The x- and ycomponents are
each + in the
first quadrant
+346 m
Solution: The person is displaced 346 m east
and 200 m north of the original position.
S
S
RELATIVE VELOCITY
NOT ON THE AP EXAM
Section 3-8: Relative Velocity S
•
Section 3-8: Relative
A useful exampleVelocity
of vector addition!
• Example: 2 trains approaching each other (along a
line) at 95 km/h each, with respect to the Earth.
• Observers on either train see the other coming at 95
+ 95 = 190 km/h. Observer on ground sees  95
km/h.
S
Velocities not along the same S
line
• Need to use full vector addition.
– A common error is adding or subtracting
wrong
velocities
– A method to help avoid this is:
Proper subscript labeling of velocities
• CONVENTION:
– Velocities with 2 subscripts. First = object,
O,
Second = reference frame, R.
Conceptual Example 3-10:
Boat Crossing A River
vBS = vBW +
•
vWS
• Outer subscripts on both sides are the same!
S
Can extend this to more than 2
S
v’s
• Suppose, to the previous example, we
add a fisherman walking on boat with
velocity
vFB = velocity of the Fisherman with
respect to the Boat:
vFS = vFB + vBW + vWS
• Outer subscripts on both sides are the same!
• Inner subscripts are the same!
• Finally: Relative velocities obey:
vAB = -vBA
Example 3-11
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Example 3-12
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*39.
(II) A boat can travel
2.30 m s
1.20 m s,
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in still water. (a) If the boat points its prow directly across a stream whos
what is the velocity (magnitude and direction) of the boat relative to the sh
39.
(a)
vboat rel. 
shore
Call the direction of the flow of the river the x direction, and the direction the boat is headed the y direction.
2
2
vwater
 vboat
 1.202  2.302  2.59 m s
rel.
rel.
shore
water
v water rel.
q  tan
1
1.20
2.30
shore
 27.6 ,   90  q  62.4 relative to shore
o
o
o
v boat rel.
q
shore

v boat rel.
water
(II) Two planes approach each other head-on. Each has a speed of
*40.
785 km h,
and they spot each other when they are initially 11.0 km apart. How much time do the pilots have to take evasive action?
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40.
If each plane has a speed of 785 km/hr, then their relative speed of approach is 1570 km/hr. If the planes are 11 km ap
d  vt

t
d
v
 11.0 km   3600 sec 
  25.2 s

 1570 km hr   1 hr 

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(II) An airplane is heading due south at a speed of
*41.
600 km h.
If a wind begins blowing from the southwest at a speed of
100 km h
(average), calculate: (a) the velocity (magnitude and direction) of the plane relative to the ground, and (b) how far from its intended position will it be a
Call east the positive x direction and north the positive y direction. Then the following vector velocity relationship exists.
a)


h  100 cos 45.0o ,100 sin 45.0 o km h
q v
plane rel.
km h

2
ground
  529 km h   540 km h
2
v plane
rel. air
east of south
The plane is away from its intended position by the distance the air has
vair rel.caused it to move. The wind speed is 100 km/h, so after 10 min (1/6 h), the pl
 h   17 km
1
6
ground
44.
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(II) A passenger on a boat moving at
1.50 m s
on a still lake walks up a flight of stairs at a speed of
0.50 m s
(Fig. 3–39). The stairs are angled at 45º pointing in the direction of motion as shown. What is the velocity of the passen
44.
vpassenger
Call the direction of the boat relative to the water
the x direction, and upward the y direction. Also see the diagram.
rel. water
q
v passenger  v passenger  v boat rel.
rel. water
rel. boat

water

 0.50 cos 45o , 0.50 sin 45o m s
 1.50, 0  m s  1.854, 0.354  m s
vpassenger  1.854 2  0.354 2  1.89 m s
rel. water
v boat rel.
water
vpassenger
rel. boat
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(II) A motorboat whose speed in still water is
*45.
2.60 m s
must aim upstream at an angle of 28.5º (with respect to a line perpendicular to the shore) in order to travel directly across the stream. (a) What is the s
v water rel. straight across the river the y direction. The boat is
Call the direction of the flow of the river the x direction, and the direction
45.
shore
q  28.5o
upstream, at a speed of
v boat rel.
vboat rel.  2.60 m s
shore
water
q v
boat rel.
water
.
(a)
sin q  vwater rel. vboat rel.
shore
water
 vwater rel.   2.60 m s  sin 28.5o  1.24 m s
shore
(b)
cos q  vboat rel. vboat rel.
shore
water
 vboat rel.   2.60 m s  cos 28.5o  2.28m s
shore
110 m
v water rel.
shore
v boat rel.
shore
260 m
v boat rel.
water
q
45o
Suppose a cannon is placed at the bottom of a hill that has
a slope of 15o with the horizontal.The cannon itself is at 45o
with the vertical. The cannon is fired and the cannonball as an
Initial v of 125 m/s. Where will cannonball strike the hill?
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CONCEPTUAL QUESTION
• Two footballs are thrown from the same
point on a flat field. Both are thrown at an
angle of 30o above the horizontal. Ball 2
has twice the initial speed of ball 1. If ball
1 is caught a distance D1 from the thrower,
how far away from the thrower D2 will the
receiver of ball 2 be when he catches it?
(a) D2 = 2D1
(c) D2 = 8D1
(b) D2 = 4D1

The distance a ball will go is simply
x = (horizontal speed) x (time in air) = v0x t

To figure out “time in air”, consider the
equation for the height of the ball:

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1
When
y  y the
 vball tiscaught,
g t 2 y = y0
0
0y
v0 y t 
2
1
g t2  0
2
1


t v0 y  g t   0
2


t2
two
solutions
v0 y
g
t 0
(time of catch)
(time of throw)
Lecture 2, Act 3
Solution
x = v0x t


So the time spent in the air is proportional to v0y :
t 2
v0 y
g
Since the angles are the same, both v0y and v0x for ball 2
are twice those of ball 1.
v0,2
v0,1
ball 1
v0x ,1

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v0y ,2
ball 2
v0y ,1
v0x ,2
Ball 2 is in the air twice as long as ball 1, but it also has twice
the horizontal speed, so it will go 4 times as far!!