8free Fall Motion

8/14/2019 8free Fall Motion

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2. Motion Along A Straight Line

2005 Pearson Education South Asia Pte Ltd

2.5 Freely Falling Bodies

Most familiar example of motion with (nearly)constant acceleration is that of a body fallingunder the influence of the earths gravitationalattraction.

To discuss this concept further, we use anidealized model of which we neglect the effects ofthe air, the earths rotation, and the decrease ofacceleration with increasing altitude. We call itfree fall.

The constant acceleration of a freely falling bodyis called the acceleration due to gravity and isdenoted by the letterg.


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2.5 Freely Falling Bodies

We use the approximate value ofgat or near theearths surface:

g= 9.8 m/s2 = 980 cm/s2 = 32 ft/s2

Becausegis the magnitude of a vector quantity, it

is always a positive number.

1.7 (Ballonist Drops Lemonade)

1.10 (Pole-Vaulter Lands)


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Example 2.6 A freely-falling coin

A one-euro coin isdropped from the

Leaning Tower of Pisa.

It starts from rest and

falls freely. Compute itsposition and velocity

after 1.0, 2.0, and 3.0 s.


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Example 2.6 (SOLN)

Identify:"Falls freely" means "has a constant acceleration due

to gravity," so we can use the constant-acceleration

equations to determine our target variables.

Set Up:We take the origin O at the starting point, the

coordinate axis as vertical, and the upward direction

as positive. Because the coordinate axis is vertical,

let's call the coordinateyinstead ofx. We replace allthexsin the constant-acceleration equations byys.

The initial coordinatey0and the initial velocity 0yare

both zero.


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Example 2.6 (SOLN)

Set Up:The acceleration is downward (in negativey-direction),

so ay = g= 9.8 m/s2. (Remember that, by defn,gitself

is always positive.) Hence our target variables are the

values ofyand yat the 3 given times. To find these,we use Eqns. 2.12 and 2.8 withxreplaced byy.

Execute:

At an arbitrary time t, the position and velocity are

( ) ( )

( ) ( )ttgtay

ttgtaty

yy

yy

20

22220

m/s81.90

m/s9.42

10

2

1

=+=+=

=+=+=


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Example 2.6 (SOLN)

Execute:When t= 1.0 s,y= (4.9 m/s2)(1.0 s)2 = 4.9m and

y= (9.8 m/s2)(1.0 s) = 9.8 m/s; after 1 s, the coin is

4.9 m below the origin (yis negative) and has a

downward velocity ( yis negative) with magnitude9.8 m/s. The position and velocity at 2.0 s and 3.0 s

are found in the same way.

Evaluate:

All our answers for yare negative because we chose

the positivey-axis to pt upward. But we could just as

well have chosen the positivey-axis to pt downward.


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Example 2.6 (SOLN)

Evaluate:In that case the acceleration would have been ay = +g

and all our answers for y would have been positive.

Either choice of axis is fine; just make sure that you

state your choice explicitly in your solution and confirmthat the acceleration has the correct sign.


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Example 2.7 Up and down motion in free fall

You throw a ball vertically upward from the roof of atall building. The ball leaves your hand at a pt even

with the roof railing with an upward speed of 15.0 m/s;

the ball is then in free fall. On its way back down, it just

misses the railing. At the location of the building,g=9.80 m/s2.

Find (a) the position and velocity of the ball 1.00 s and

4.00 s after leaving your hand; (b) the velocity when

the ball is 5.00 m above the railing; (c) the maximumheight reached and the time at which it is reached; and

(d) the acceleration of the ball when it is at its

maximum height.


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Example 2.7 (SOLN)

Identify:The words "free fall" in the statement of the problem

mean that the acceleration is constant and due to

gravity. Our target variables are position [in parts (a)

and (c)], velocity [in parts (a) and (b)], and acceleration[in part (d)].

Set Up:

In the figure, the downward path is displaced a little to

the right of its actual position for clarity. Take the originat the roof railing, at the pt where the ball leaves your

hand, and take the positive direction to be upward.


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Example 2.7 (SOLN)

Set Up:

First, let's collect our data. The initial positiony0is

zero. The initial velocity 0yis +15.0 m/s, and the

acceleration is ay= g= 9.80 m/s2. We'll again use

Eqns. 2.12 and 2.8 to find the position and velocity,respectively, as functions of time. In part (b) we're

asked to find the velocity at a certainposition rather

than at a certain time, so it will also be convenient to

use Eqn. (2.13) for that part.


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Execute:

The positionyand velocity yat any time tafter the

ball leaves your hand are given by Eqns. 2.12 and 2.8

withx'sreplaced byy's, so

When t= 1.00 s, these equations give

Example 2.7 (SOLN)

( )

( ) ( ) ( )( )

( )t

tgta

tt

tgtytatyy

yyyy

yyy

2

00

22

200

200

m/s80.9m/s0.15

m/s9.4m/s0.150

2

1

2

1

+=

+=+=

++=

++=++=

m/s2.5m0.10 +=+= yy


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Execute:

The ball is 10.1 m above the origin (yis positive), and

it is moving upward ( yis positive) with a speed of 5.2

m/s. This is less than the initial speed of 15.0 m/s, as

expected. When t= 4.00 s, the equations foryand yas functions of time tgive

The ball has passed its highest pt and is 18.4 m below

the origin (yis negative). It has a downwardvelocity( yis negative) with magnitude 24.2 m/s. This speed

is greater than the initial speed, as we should expect

for pts below the ball's launching point.

Example 2.7 (SOLN)

m/s2.24m4.18 == yy


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Execute:

(a) Note that to get these results, we don't need to find

the highest pt reached or the time at which it was

reached. The eqns of motion give the position and

velocity at any time, whether the ball is on the way upor on the way down.

(b) The velocity yat any positionyis given by Eqn.

2.13 withx'sreplaced byy's:

Example 2.7 (SOLN)

( ) ( ) ( )

( ) ( )y

ygyya yyyy

22

200

20

2

m/s80.92m/s0.15

022

+=

+=+=


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Execute:

(a) When the ball is 5.00 m above the origin,

y= +5.00 m, so

We get two values of y: 1 positive and 1

negative. As shown in figure, the ball

passes this pt twice, once on the wayup and again on the way down. The

velocity on the way up is +11.3 m/s,

and on the way down it is 11.3 m/s.

Example 2.7 (SOLN)

( ) ( )( )m/s3.11

/sm127m00.5m/s80.92m/s0.15 22222

=

=+=

y

y


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Execute:

(c) At the highest pt, the ball stops going up (positive

y) and starts going down (negative y). At the instant

when it reaches the highest pt, y = 0. The maximum

heighty1can then be found in 2 ways. The first is touse Eqn. 2.13 and substitute y = 0,y0 = 0, and ay=

-g:

Example 2.7 (SOLN)

( ) ( )

( )( )

m5.11m/s80.92

m/s0.152

020

2

22

01

12

0

+===

+=

gy

yg

y

y


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Execute:

(c) The second way is find the time at which = 0

using Eqn. 2.8, y = 0y + ayt, and then substitute this

value oftinto Eqn. 2.12 to find the position at this time.

From Eqn. 2.8, the time t1 when the ball reaches thehighest point is given by

Example 2.7 (SOLN)

( )

s53.1m/s80.9m/s0.15

0

201

10

===

+==

gt

tg

y

yy


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Execute:

(c) Substituting this value oftinto Eq. (2.12), we find

Notice that with the first way of finding the maximum

height, it's not necessary to find the time first.

(d) CAUTION: It's a common misconception that at thehighest pt of the motion the velocity is zero andthe

acceleration is zero. If this were so, once the ball

reached the highest point it would hang there in mid-

air forever!

Example 2.7 (SOLN)

( ) ( ) ( )

( )( ) m5.11s53.1m/s8.921

s53.1m/s1502

1

22

200

+=+

+=++= tatyy yy


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Execute:

(d) To see why, remember that acceleration is the rate

of change of velocity. If the acceleration were zero at

the highest point, the ball's velocity would no longer

change, and once the ball was instantaneously at restit would remain at rest forever. The truth of the matter

is that at the highest point, the acceleration is still ay=

g= 9.80 m/s2, the same value as when the ball is

moving up and when it's moving down. The ball stopsfor an instant at the highest pt, but its velocity is

continuously changing, from positive values through

zero to negative values.

Example 2.7 (SOLN)


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Evaluate:

A useful way to check any motion problem is to draw

the graphs of position versus time and velocity versus

time. Figure shows these graphs for the situation in

this problem. Note thatthe y-tgraph has a

constant negative slope.

This means that the

acceleration is negative(downward) on the way

up, at the highest pt,

and on the way down.

Example 2.7 (SOLN)


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Example 2.8 Two solutions or one?

Find the time when the ball in Example 2.7 is 5.00 m

below the roof railing.

Solution:

Identify and Set Up:

We again choose they-axis, soy0, 0y, and ay= ghave the same values as in Example 2.7. The position

y as a function of time tis again given by Eqn 2.12:

We want to solve this for the value oftwhen

y= 5.00 m. Since this equation involves t2, it is a

quadraticequation fort.

( ) 2

00

2

00 2

1

2

1tgtytatyy

yyy

++=++=


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Example 2.8 (SOLN)

( ) ( ) ( ) 021 2002 =++=++ CBtAtyyttg y

Execute:

We first rearrange our equation into the same form as

the standard quadratic equation for an unknownx,

Ax2+Bx + C = 0:

soA =g/2,B = 0y and C =yy0. Using the quadratic

formula (Appendix B), we find that this eqn has two

solutions:


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Example 2.8 (SOLN)

( ) ( ) ( ) ( )

( )( )

g

yygt

g

yyg

A

ACBBt

yy

yy

02

00

02

00

22/2

2/4

2

42

=

=

=

Execute:

Substituting the valuesy0 = 0, 0y= +15.0 m/s, g =9.80 m/s2, andy= 5.00 m, we find


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Example 2.8 (SOLN)

Execute:

Substituting the valuesy0 = 0, 0y= +15.0 m/s, g =

9.80 m/s2, andy= 5.00 m, we find

Which is the right answer? The key qn to ask is, "Are

these answers reasonable?" The 2nd answer,t= 0.30 s, is not; it refers to a time 0.30 s before the

ball left your hand! The correct answer is t= +3.36 s.

The ball is 5.00 m below the railing 3.36 s afterit

leaves your hand.

( ) ( ) ( )( )

s30.0s36.3

m/s80.9

0m00.5m/s80.92m/s0.15m/s0.15

2

22

=+=

=tort

t


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Example 2.8 (SOLN)

Evaluate:

Where did the erroneous "solution" t= 0.30 s comefrom? Remember that we began with Eqn. 2.12 with

ay= g, that is,y =y0+ 0yt+ 0.5(g)t2. This eqn has

built into it the assumption that acceleration is constantforallvalues oft, whether positive, negative, or zero.

Taken at face value, this eqn would tell us that the ball

had been moving upward in free fall ever since the

dawn of time; it eventually passes your hand aty= 0at the special instant we chose to call t= 0, then

continues in free fall.


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Example 2.8 (SOLN)

Evaluate:

But anything that this equation describes happeningbefore t= 0 is pure fiction, since the ball went into free

fall only after leaving your hand at t= 0; the "solution"

t= 0.30 s is part of this fiction. You should repeat

these calculations to find the times at which the ball is

5.00 m above the origin (y= +5.00 m). The two

answers are t= +0.38 s and t= +2.68 s; these are

both positive values oft, and both refer to the realmotion of the ball after leaving your hand. The earlier

time is when the ball passes throughy= +5.00 m

moving upward, and the later time is when the ball

passes through this point moving downward.


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Example 2.8 (SOLN)

Evaluate:

You should also solve for the times at whichy= +15.0 m. In this case, both solutions involve the

square root of a negative number, so there are no real

solutions. This makes sense; we found in part (c) of

Example 2.7 that the ball's maximum height is only

y= +11.5 m, so it never reachesy= +15.0 m. While a

quadratic equation such as Eqn. 2.12 always has 2

solutions, in some situations one or both of thesolutions will not be physically reasonable.

2 M i Al A S i h i


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*2.6 Velocity and Position by Integration

In many physical situations, position and velocity

are not known as functions of time, while

acceleration is.

How can we find the position and velocity from the

acceleration function ax(t)? First, we consider the graphical approach.

Graph shows a body whose

acceleration is not constantbut increases with time.

2 M ti Al A St i ht Li


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Divide time interval into many smaller intervals,

each is t.

Total velocity change is represented graphically by

total area under the ax-tcurve between vertical lines

t1 and t2.

As tapproaches a limiting value, area under the

ax-tcurve is the integral ofa

x.

Then,

( )15.22

1

2

112

=

=

t

t

x

xxx

dta

dx

x



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Then,

We can carry out same procedure with curve of

velocity vs. time, where xis a general function of

t.

Thus we get,

( )15.221

2

112 ==

tt

xxxx dtadxx

( )16.22

1

2

112 ==

t

tx

x

xdtxdxx



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Ift1= 0 and t

2is any later time t, and ifx

0and

0x

are the position and velocity, at time t= 0, then

rewriting Eqns 2.15 and 2.16 as follows:

( )17.200

+=t

xxxdta

( )18.20

0 +=t

x dtxx



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Concepts Summary

When a particle moves along a straight line, we

describe its position with respect to an origin ) bymeans of a coordinate such asx.

The particles average velocity av-x

during a time

interval t= t2t1 is equal to its displacement x =x

2x

1divided by t.

Instantaneous velocity xat any time tis equal to

the average velocity for the time interval from

tto t+ tin the limit that tgoes to zero.

Equivalently, xis the derivative of the position

function with respect to time.



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Concepts Summary

The average acceleration aav-x during a time interval

tis equal to the change in velocity

= 2x 1x during that time interval divided by

t.

The instantaneous acceleration ax is the limit ofaav-x as tgoes to zero, or the derivative of xwith

respect to t.

When acceleration is constant, four equationsrelate the positionx and velocity x at any time tto

the initial positionx0, the initial velocity 1x (both

measured at time t= 0) and the acceleration ax.



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Concepts Summary

Free fall is a case of motion with constant

acceleration. The magnitude of the acceleration

due to gravity is a positive quantity,g.

The acceleration of a body in free fall is always

downward.

2 Motion Along A Straight Line


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Key Equations

)3.2(lim

0 dt

dx

t

x

t

x =

=

)2.2(1212

tx

ttxxxav

=

=

)4.2(12

12

t

v

tta xxxxav

=

=

)5.2(lim0 dt

dv

t

va xx

tx =

=

2 Motion Along A Straight Line


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Key Equations

)8.2(0 tav xxx+=

( )12.22

1 200 tatxx xx ++=

( ) )13.2(2 02

02

xxaxxx +=

)14.2(2

00 txx

xx

+=

( )17.20

0 +=t

xxx dta

( )18.2

0

0

+=

t

x dtxx

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