PROBLEM-SOLVING

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PROBLEM-SOLVING

• GENERAL GUIDELINES

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GENERAL PROBLEM SOLVING STRATEGY   Solving
problems require three major steps Prepare So
lve Assess
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PREPARE   The "Prepare" step of a solution is
where you identify important elements of the
problem and collect information you will need to
solve it. It's tempting to jump right to the
"solve" step, but a skilled problem solver will
spend the most time on this step, the
preparation.
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Preparation includes   1. Identifying the
Physics Principle(s)



Read the problem carefully
and identify what is the underlying physics
principle of the problem, then, write down the
principle using an acronym such as the ones in
the following list. If the problem has several
steps, write down the principle(s) as
appropriate. Newton's First Law
(N1L) Newton's Second Law (N2L) Kinematics in
One-Dimension (UAM) Kinematics in Two-Dimension
(K2D) Conservation of Energy (COE) Conservatio
n of Momentum (COM)
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2. Data Given and Unknown   Make a table of the
quantities whose values you can determine from
the problem statement or that can be found
quickly with simple geometry or unit conversions.
Any relevant constants should be written here.
All units should be consistent with the SI values
(i.e. kg, m, s). All unit conversion should take
place in this section. Also, identify the
quantity or quantities that will allow you to
answer the question.
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3. Sketch, graph, FBD In many cases, this is the
most important part of a problem. The picture
lets you model the problem and identify the
important elements. As you add information to
your picture, the outline of the solution will
take shape. If appropriate, select a coordinate
axis. If the quantities involved are vectors, be
sure to draw an arrow with the tip of the arrow
clearly indicating the direction of the
vector. When drawing a free-body-diagram (FBD),
be sure to draw and clearly label only the forces
acting on the system.  
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SOLVE   The "Solve" step of a solution is where
you actually do the math or reasoning necessary
to arrive to the solution needed. This is the
part of the problem-solving strategy that you
likely think of when you think of "solving
problems". But don't make the mistake starting
here! If you just choose an equation and plug in
numbers, you will likely go wrong and will waste
time trying to figure out why. The "Prepare" step
will help you be certain you understand the
problem before you start putting numbers in
equations.
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Solving the problem includes   4. Equation
(always solve for unknown) Write the relevant
equation or equations that will allow you to
solve for the unknown. Be sure to always solve
the equation for the unknown instead of just a
'plug and chug' approach. 5. Substitution Once
you have solved the equation algebraically,
substitute the appropriate values. 6. Answer
with Units Write down the answer with the
appropriate units. Remember that 'naked' numbers
make no sense in Physics!
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ASSESS The "Assess" step of your solution is
very important. When you have an answer, you
should check to see if it makes sense.
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7. Check the Answer  Ask yourself - Does my
solution answer the question that was asked?
Make sure that you have addressed all parts of
the question and clearly written down your
solutions. - Does my answer have the correct
units and number of significant figures? - Does
the value I computed make physical sense? - Can
I estimate what the answer should be to check my
solution? - Does my final solution make sense in
the context of the material I'm learning?
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MOTION   An object is in motion if its position
changes. The mathematical description of motion
is called kinematics. The simplest kind of
motion an object can experience is uniform motion
in a straight line. The object experiences
translational motion if it is moving without
rotating.
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Describing Motion The study of one-dimensional
kinematics is concerned with the multiple means
by which the motion of objects can be
represented. Such means include the use of words,
graphs, equations, and diagrams.
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One-dimensional motion means that objects are
only free to move back and forth along a single
line. As a coordinate system for one-dimensional
motion, choose this line to be an x-axis together
with a specified origin and positive and negative
directions.
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DISTANCE AND DISPLACEMENT Distance is the length
between any two points in the path of an object.
Displacement is the length and direction of the
change in position measured from the starting
point.
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DISTANCE AND DISPLACEMENT
The distance an object travels tells you nothing
about the direction of travel, while displacement
tells you precisely how far, and in what
direction, from its initial position an object is
located.   Distance is the total length of
travel and displacement is the net length of
travel accounting for direction.
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The displacement is written
Displacement is positive.
Displacement is negative.
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3.1 You leave your home and drive 4.83 km North
on Preston Rd. to go to the grocery store. After
shopping, you go back home by traveling South on
Preston Rd. a. What distance do you travel during
this trip?
UAM
x1 4.83 km x2 4.83 km
distance x1 x2 4.83 4.83
9.66 km
b. What is your displacement?
x1 4.83 km, N x2 4.83 km, S
displacement ?x x2 - x1
4.83 - 4.83
0 km
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SPEED If an object takes a time interval t to
travel a distance x, then the average speed of
the object is given by  
Units m/s
3.2 A ship steams at an average speed of 30 km/h.
a. What is the speed in m/s?
UAM
v 30 km/h
8.33 m/s
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b. How far in km does it travel in a day?
t 24 h
30(24) 720 km
c. How long in hours does it take to travel 500
km?
x 500 km
16.67 h
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AVERAGE SPEED AND AVERAGE VELOCITY   Average
velocity is the displacement divided by the
amount of time it took to undergo that
displacement. The difference between average
speed and average velocity is that average speed
relates to the distance traveled while average
velocity relates to the displacement.
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Speed how far an object travels in a given time
interval
Velocity includes directional information
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3.3 A car travels north at 100 km/h for 2 h, at
75 km/h for the next 2 h, and finally turns
south at 80 km/h for 1 h. a. What is the cars
average speed for the entire journey?
UAM
v1 100 km/h t1 2 h v2 75 km/h t2 2 h v3
80 km/h t3 1 h
Total time tT 2 2 1 5 hours
Total distance traveled x v t x1 v1t1 100
(2) 200 km x2 75 (2) 150 km x3 80 (1)
80 km xT 200 150 80 430 km
86 km/h
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b. What is the cars average velocity for the
entire journey?
x1 200 km, N x2 150 km, N x3 80 km,
S Displacement 200 150 80 270 km
Total time 5 h
54 km/h, N
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3.4 Give a qualitative description of the motion
depicted in the following x-versus-t graphs
x
a.
Object starts at the origin and moves in the
positive direction with constant velocity.
t
x
b.
Object starts to the right of the origin and
moves in the negative direction with constant
velocity ending at the origin.
t
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x
c.
Object starts to the right of the origin and
moves in the positive direction with constant
velocity.
t
x
d.
Object starts to the left of the origin and moves
in the positive direction with constant velocity
ending at the origin.
t
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The slope of the graphs yields the average
velocity. When the velocity is constant, the
average velocity over any time interval is equal
to the instantaneous velocity at any time.
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3.5 Give a qualitative description of the motion
depicted in the following v-versus-t graphs
a.
v
Object moving to the right at a slow constant
speed.
t
b.
v
Object moving to the left at a fast constant
speed.
t
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ACCELERATION  Acceleration is the rate of change
of velocity. The change in velocity ?v is the
final velocity vf minus the initial velocity
vo.
Acceleration happens when   An object's velocity
increases An object's velocity decreases An
object changes direction
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The acceleration of an object is given by

Units m/s2
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EQUATIONS FOR MOTION UNDER CONSTANT ACCELERATION
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3.6 An object starts from rest with a constant
acceleration of 8 m/s2 along a straight line. a.
Find the speed at the end of 5 s
UAM
vo 0 m/s a 8 m/s2 t 5 s
0 8(5) 40 m/s
b. Find the average speed for the 5 s interval
20 m/s
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c. Find the distance traveled in the 5 s
0 ½(8)(5)2 100 m
or
20(5) 100 m
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3.7 A truck's speed increases uniformly from 15
km/h to 60 km/h in 20 s. a. Determine the
average speed
UAM
vo 15/3.6 4.17 m/s vf 60/3.6 16.7 m/s t
20 s
10.4 m/s
b. Determine the acceleration
0.63 m/s2
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c. Determine the distance traveled
10.4(20) 208 m
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3.8 A skier starts from rest and slides 9.0 m
down a slope in 3.0 s. In what time after
starting will the skier acquire a speed of 24
m/s? Assume that the acceleration is constant.
UAM
vo 0 m/s, x 9 m t 3 s, vf 24 m/s
2 m/s2
12 s
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3.9 A car moving at 30 m/s slows uniformly to a
speed of 10 m/s in a time of 5.0 s. a. Determine
the acceleration of the car
UAM
vo 30 m/s vf 10 m/s t 5 s
- 4 m/s2
b. Determine the distance it moves in the third
second
30 (3 - 2) ½ (- 4) (32 - 22) 20 m
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3.10 The speed of a train is reduced uniformly
from 15 m/s to 7.0 m/s while traveling a
distance of 90 m. a. Calculate the acceleration.
UAM
vo 15 m/s vf 7 m/s x 90 m
- 0.98 m/s2
b. How much farther will the train travel before
coming to rest, provided the acceleration remains
constant?
25 m
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3.11 A drag racer starts from rest and
accelerates at 7.40 m/s2. How far has it traveled
in 1.00 s, 2.00 s, and 3.00 s? Graph the results
in a position versus time graph.
UAM
a 7.4 m/s2 t 1.00 s
3.70 m
at t 2.00 s, x 14.8 m at t 3.00 s, x
33.3 m
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This example illustrates one of the key features
of accelerated motion position varies directly
with the square of the time.
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vo 0
x 0
v
x
a
t
t
t
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vo 0
x 0
v
x
a
t
t
t
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vo ? 0
x 0
v
x
a
t
t
t
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vo 0
x 0
v
x
a
t
t
t
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vo 0
x 0
v
x
a
t
t
t
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• Observation
• the sign of the velocity and the acceleration is
  the same if
• the object is speeding up and that
• the sign of the velocity and the acceleration is
  the opposite
• if the object is slowing down.

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• GRAPHICAL ANALYSIS OF MOTION
•  Graphical interpretations for motion along a
  straight line (the x-axis) are as follows
• the slope of the tangent of an x-versus-t graph
  and define instantaneous velocity,
• the slope of the v-versus-t graph and understand
  that the value obtained is the average
  acceleration,
• the area under the v-versus-t graph and
  understand that it gives the displacement,
• the area under the a-versus-t graph and
  understand that it gives the change in velocity.

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The slope of the tangent of an x-versus-t graph
yields the instantaneous velocity,
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The slope of the v-versus-t graph gives the
average acceleration,
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The area under the v-t curve yields the
displacement.
A bh 1/2 bh m/s . s m
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3.13. A boat moves slowly inside a marina with a
constant speed of 1.50 m/s. As soon as it leaves
the marina, it accelerates at 2.40 m/s2. a. How
fast is the boat moving after accelerating for
5.0 s?
UAM
vo 1.5 m/s a 2.40 m/s2 t 5.00 s
vf vo at 1.5 (2.40)(5) 13.5 m/s
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v (m/s)
b. Plot a graph of velocity versus time.
13.5
1.5
t (s)
5.0
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c. How far has the boat traveled in this time?
v (m/s)
13.5
1.5
t (s)
5.0
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c. How far has the boat traveled in this time?
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ACCELERATION DUE TO GRAVITY All bodies in free
fall near the Earth's surface have the same
downward acceleration of   g 9.8 m/s2   A
body falling from rest in a vacuum thus has a
velocity of 9.8 m/s at the end of the first
second, 19.6 m/s at the end of the next second,
and so forth. The farther the body falls, the
faster it moves. A body in free fall has the
same downward acceleration whether it starts from
rest or has an initial velocity in some direction.
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Galileo is alleged to have performed free-fall
experiments by dropping objects off the Tower of
Pisa
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The presence of air affects the motion of falling
bodies partly through buoyancy and partly through
air resistance. Thus two different objects
falling in air from the same height will not, in
general, reach the ground at exactly the same
time.
Because air resistance increases with velocity,
eventually a falling body reaches a terminal
velocity that depends on its mass, size, shape,
and it cannot fall any faster than that.
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FREE FALL When air resistance can be neglected,
a falling body has the constant acceleration g,
and the equations for uniformly accelerated
motion apply. Just substitute a for g.
Sign Convention for direction of motion If
the object is thrown downward then g 9.8
m/s2 If the object is thrown upward then g -
9.8 m/s2
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FREE-FALL An object thrown upward in the absence
of air resistance yields the following graphs.
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3.14 A ball is dropped from rest at a height of
50 m above the ground. a. What is its speed just
before it hits the ground?
UAM
vo 0 m/s y 50 m g 9.8 m/s2
31.3 m/s
b. How long does it take to reach the ground?
3.19 s
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3.15 A stone is thrown straight upward and it
rises to a height of 20 m. With what speed was it
thrown?
UAM
y 20 m g - 9.8 m/s2
At highest point vf 0
19.8 m/s
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3.16 A stone is thrown straight upward with a
speed of 20 m/s. It is caught on its way down at
a point 5.0 m above where it was thrown. a. How
fast was it going when it was caught?
UAM
vo 20 m/s y 5 m g - 9.8 m/s2
17.4 m/s Since it is moving down we
write it as - 17.4 m/s
b. How long did the trip take?
3.8 s
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3.17 A baseball is thrown straight upward on the
Moon with an initial speed of 35 m/s. (g 1.60
m/s2) Find a. The maximum height reached by the
ball
UAM
vo 35 m/s g - 1.6 m/s2
At highest point vf 0
382. 8 m
b. The time taken to reach that height
21.9 s
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c. Its velocity 30 s after it is thrown
t 30 s
- 13 m/s
d. Its velocity when the ball's height is 100 m
y 100 m
30 m/s
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3.18 A rock is thrown vertically upward with a
velocity of 20 m/s from the edge of a bridge 42 m
above a river. How long does the rock stay in the
air?
vo 20 m/s y - 42 m g - 9.8 m/s2
UAM
First, find the velocity of the rock at the
moment that it hits the river.
35 m/s
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Negative velocity because the rock will be moving
toward the river.
vf - 35 m/s
5.6 s
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THE MONKEY AND THE ZOOKEPER A monkey spends most
of its day hanging from a branch of a tree
waiting to be fed by the zookeeper. The zookeeper
shoots bananas from a banana cannon.
Unfortunately, the monkey drops from the tree
the moment that the banana leaves
the muzzle of the cannon and the zookeeper is
faced with the dilemma of where to aim the banana
cannon in order to feed the monkey. If the
monkey lets go of the tree the moment that the
banana is fired, where should he aim the banana
cannon?
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Banana thrown ABOVE the monkey
Wrong move!
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Banana thrown AT the monkey
Happy monkey!
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PROJECTILE MOTION An object launched into space
without motive power of its own is called a
projectile.
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If we neglect air resistance, the only force
acting on a projectile is its weight, which
causes its path to deviate from a straight line.
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The projectile has a constant horizontal velocity
and a vertical velocity that changes uniformly
under the influence of the acceleration due to
gravity.
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HORIZONTAL PROJECTION
If an object is projected
horizontally, its motion can best be described by
considering its horizontal and vertical motion
separately.
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In the figure we can see that the vertical
velocity and position increase with time as those
of a free-falling body. Note that the horizontal
distance increases linearly with time, indicating
a constant horizontal velocity.
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3.19 A cannonball is projected horizontally with
an initial velocity of 120 m/s from the top of a
cliff 250 m above a lake. a. In what time will it
strike the water at the foot of the cliff?
UAM
v0x 120 m/s y 250 m v0y 0
7.14 s
b. What is the x-distance (range) from the foot
of the cliff to the point of impact in the lake?
UM
x vx t 120(7.14) 857 m
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c. What are the horizontal and vertical
components of its final velocity?
UM UAM
vx 120 m/s vy voy gt 0 9.8 (7.14)
70 m/s
d. What is the final velocity at the point of
impact and its direction?
UAM
139 m/s
30.2? below horizontal v (139 m/s, 30.2?)
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3.20 A person standing on a cliff throws a stone
with a horizontal velocity of 15.0 m/s and the
stone hits the ground 47 m from the base of the
cliff. How high is the cliff?
vx 15 m/s x 47 m vy 0
UM
3.13 s
y ½ gt2 ½ (9.8)(3.13)2 48 m
UAM
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PROJECTILE MOTION AT AN ANGLE The more general
case of projectile motion occurs when the
projectile is fired at an angle.
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Problem Solution Strategy 1. Upward direction
is positive. Acceleration due to gravity (g) is
downward thus g - 9.8 m/s2 2. Resolve the
initial velocity vo into its x and y components
vox vo cos ? voy vo sin ? 3. The
horizontal and vertical components of its
position at any instant is given by x voxt
y voy t ½gt2 4. The horizontal and
vertical components of its velocity at any
instant are given by vx vox vy voy
gt 5. The final position and velocity can then
be obtained from their components.
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3.23 An artillery shell is fired with an initial
velocity of 100 m/s at an angle of 30? above the
horizontal. Find a. Its position and velocity
after 8 s
UAM UM
vox 100 cos 30? 86.6 m/s voy 100 sin
30? 50 m/s
vo 100 m/s, 30? t 8 s g - 9.8 m/s2
x vox t 86.6(8) 692.8 m y voy t ½
gt2 50(8) ½ (-9.8)(8)2 86.4 m
vx vox 86.6 m/s vy voy gt 50
(-9.8)(8) - 28.4 m/s
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b. The time required to reach its maximum height
UAM
At the top of the path vy 0 vy voy gt
5.1 s
c. The horizontal distance (range)
UM
Total time T 2t 2(5.1) 10.2 s
x vox t 86.6(10.2) 883.7 m
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3.24 A baseball is thrown with an initial
velocity of 120 m/s at an angle of 40?above the
horizontal. How far from the throwing point will
the baseball attain its original level?
vox 120 cos 40? 91.9 m/s voy 120 sin
40? 77.1 m/s
vo 120 m/s, 40? g - 9.8 m/s2
At top vy 0
UM
UAM
7.9 s
x vox (2t) 91.9(2)(7.9) 1452 m
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3.25 A plastic ball that is released with a
velocity of 15 m/s stays in the air for 2.0 s.
a. At what angle with respect to the
horizontal was it released?
vo 15 m/s t 2 s
time to maximum height 1 s at the top vy 0 vy
voy gt
UAM
40.8º
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b. What was the maximum height achieved by the
ball?
UAM
y voy t ½ gt2 (15)(sin 40.8º)(1) ½
(-9.8)(1)2 4.9 m
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3.26 Find the range of a gun which fires a shell
with muzzle velocity vo at an angle ? .
K2D
At top vy 0 vy voy gt vo sin ? - gt
x vxt
Total time 2t
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sin ? cos ? ½ sin 2?
b. What is the angle at which the maximum range
is possible?
Maximum range is 45? since 2? 90?
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c. Find the angle of elevation ? of a gun that
fires a shell with muzzle velocity of 120 m/s and
hits a target on the same level but 1300 m
distant.
vo 120 m/s x 1300 m
0.885
sin-1(2?) 62? ? 31?
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