3c free fall kinematics 3d

8/19/2019 3c Free Fall Kinematics 3D

1/28

Physics 111: Lecture 2, Pg 1

Physics: Lecture 2Physics: Lecture 2

Today’s AgendaToday’s Agenda

Recap of 1-D motion with constant acceleration

1-D free fall example

3-D Kinematics

hoot the mon!ey

"ase#all $n%epen%ence of x an% y components


2/28


Review:Review:

&or constant acceleration we foun%:

at v v 0 +=

2 0 0 at 2

1t v x x ++=

const a =

x

a

v t

t

t

v)(v 2

1v

) x 2a(x v v

0 av

0 2 0

2

+=

−=−

&rom which we %eri'e%:


3/28


Recall what you saw:Recall what you saw:

12

2 2

3 2

(2

2 0 0

at 2

1t v x x ++=


4/28

Physics 111: Lecture 2, Pg (

1-D Free-Fall1-D Free-Fall

)his is a nice example of constant acceleration *gra'ity+: $n this case, acceleration is cause% #y the force of gra'ity:

sually pic! y-axis upwar%.

/cceleration of gra'ity is %own.:

y

ay 0 − g

gt v v y

0 y −=

2 y 0 0

t g 2

1t v y y −+=

y

a

v

t

t

t

g ay −=


5/28

Physics 111: Lecture 2, Pg

Gravity acts:Gravity acts:

g %oes not %epen% on the nature of the material

alileo *14(-14(2+ figure% this out without fancy cloc!s5 rulers

%emo - feather 5 penny in 'acuum

6ominally, g 0 7891 ms2

/t the e;uator g 0 78


6/28


Pro!le":Pro!le":

)he pilot of a ho'ering helicopter%rops a lea% #ric! from a heightof 1>>> m8 ?ow long %oes it ta!eto reach the groun% an% how fastis it mo'ing when it gets there@*neglect air resistance+

1000 m


7/28Physics 111: Lecture 2, Pg <

Pro!le":Pro!le":

&irst choose coor%inate system8

Arigin an% y -%irection8

6ext write %own position e;uation:

RealiBe that v 0y = 0 8

2 0y 0 gt

2 1t v y y +=

2 0 gt

2

1y y −=

1000 m

y = 0

y


8/28Physics 111: Lecture 2, Pg 9

Pro!le":Pro!le":

ol'e for time t when y = 0 gi'en that y 0 =1000 m.

Recall:

ol'e for v y :

y 0 = 1000 m

y

s314sm819

m1000 2

g

y 2 t

2 0 8

8=

×==

2 0

gt 2

1y y -=

y = 0

+* 0 2

y 0 2 y y y a2 v v -- =

sm140

gy 2 v 0 y

−=

±=



Lecture 2#Lecture 2# Act 1 Act 1

1D ree all1D ree all Alice and $ill are standing at the to% o a cli o heightAlice and $ill are standing at the to% o a cli o height H H & $oth throw a !all with initial s%eed& $oth throw a !all with initial s%eed v v 0 0 # Alice straight# Alice straight downdown and $ill straightand $ill straight u%u%& The& The

s%eed o the !alls when they hit the ground ares%eed o the !alls when they hit the ground are v v A A andand v v BB res%ect ively&res%ectively& 'hich o the ollowing is true:'hich o the ollowing is true:

(a)(a) v v A A ** v v BB (!)(!) v v A A ++ v v BB (c)(c) v v A A ,, v v BB

v v 0 0

v v 0 0

$ill$illAliceAlice

H H

v v A A v v BB


10/28Physics 111: Lecture 2, Pg 1>

Lecture 2#Lecture 2# Act 1 Act 11D Free all1D Free all

ince the motion up an% #ac! %own is symmetric, intuition shoul% tell you that v = v 0

Ce can pro'e that your intuition is correct:

v v 0 0

$ill$ill

H H

v v = v = v 0 0

( ) 0 H H g 2 v v 2 0 2

=−−=− +*.uation:.uation:

This loo/s 0ust li/e $ill threwThis loo/s 0ust li/e $ill threwthe !all down with s%eedthe !all down with s%eed v v 0 0 # so# so

the s%eed at the !otto" shouldthe s%eed at the !otto" should!e the sa"e as Alice’s !all&!e the sa"e as Alice’s !all&

y = 0 y = 0



Lecture 2# Act 1Lecture 2# Act 11D Free all1D Free all

'e can also 0ust use the e.uation directly:'e can also 0ust use the e.uation directly:

( )H 0 g 2 v v 2 0 2

−−=− +*Alice:Alice:

v v 0 0

v v 0 0

AliceAlice $ill$ill

y = 0 y = 0

( )H 0 g 2 v v 2 0 2 −−=− +*$ill:$ill:sa"e sa"e



-D 3ine"atics-D 3ine"atics

)he position, 'elocity, an% acceleration of a particle in 3%imensions can #e expresse% as:

r r = x i i + y j j + z k k

v v = v x i i + v y j j + v z k k *i i , j j , k k unit vectos )

aa = a x i i + ay j j + az k k

Ce ha'e alrea%y seen the 1-D !inematics e;uations:

a !v

!t

! x

!t = =

2

2 v

!x

!t = x x(t = )




&or 3-D, we simply apply the 1-D e;uations to each of thecomponent e;uations8

Chich can #e com#ine% into the 'ector e;uations:

r r = r r (t) v v = ! r r " !t aa = ! 2 r r " !t 2

a ! x

!t x =

2

2

v !x

!t x =

x x(t = )

a ! y

!t y =

2

2

v !y

!t y =

y y t = ( )

a ! z

!t z =

2

2

v !z

!t z =

z z t = ( )


14/28Physics 111: Lecture 2, Pg 1(


o for constant acceleration we can integrate to get:

aa = const

v v = v v 0 + aa t

r r = r r 0 + v v 0 t + 1 " 2 aa t 2

*where aa, v v , v v >, r r , r r >, are all 'ectors+



2-D 3ine"atics2-D 3ine"atics

=ost 3-D pro#lems can #e re%uce% to 2-D pro#lems whenacceleration is constant:

hoose y axis to #e along %irection of acceleration

hoose x axis to #e along the other. %irection of motion

4a"%le4a"%le: )hrowing a #ase#all *neglecting air resistance+

/cceleration is constant *gra'ity+

hoose y axis up: ay = #g

hoose x axis along the groun% in the %irection of thethrow

lost mar#les



5546 and 5y6 co"%onents o "otion are46 and 5y6 co"%onents o "otion areinde%endent&inde%endent&

/ man on a train tosses a #all straight up in the air8

Eiew this from two reference frames:

Reference frameon the mo'ing train8

Reference frame

on the groun%8

art


17/28Physics 111: Lecture 2, Pg 1<

Pro!le":Pro!le":

=ar! =cwire clo##ers a fast#all towar% center-fiel%8 )he#all is hit 1 m *y o + a#o'e the plate, an% its initial 'elocity is348 ms *v + at an angle of 3>o * + a#o'e horiBontal8 )hecenter-fiel% wall is 113 m *D+ from the plate an% is 3 m *h+high8

Chat time %oes the #all reach the fence@Chat time %oes the #all reach the fence@

Does =ar! get a home run@Does =ar! get a home run@

θ

v v

h

D

y 0



Pro!le"&&&Pro!le"&&&

hoose y axis up8

hoose x axis along the groun% in the %irection of the hit8

hoose the origin (0,0) to #e at the plate8

ay that the #all is hit at t = 0 , x = x 0 = 0

F;uations of motion are:

v x = v 0x v y = v 0y # gt

x = v x t y = y 0 + v 0y t # 1 " 2 gt 2




se geometry to figure out v 0x an% v 0y :

y

x

g

θ

v v

v 0x

v 0y

&in% v 0x = $v $ cos θ8an% v 0y = $v $ sin θ8

y 0


20/28Physics 111: Lecture 2, Pg 2>


)he time to reach the wall is: t = % " v x *easy+

Ce ha'e an e;uation that tell us y(t) = y 0 + v 0y t + a t 2 " 2

o, weGre %one8888now we Hust plug in the num#ers:

&in%:

v x 0 348 cos*3>+ ms 0 3184 ms

v y 0 348 sin*3>+ ms 0 1982 ms

t 0 *113 m+ *3184 ms+ 0 389 s

y(t) 0 *18> m+ I *1982 ms+*389 s+ -

*>8+*789 ms2+*389 s+2

0 *18> I 483 - 4289+ m 0 &7 ""

ince the wall is " high, =ar! gets the homer



Lecture 2#Lecture 2# Act 3 Act 38otion in 2D8otion in 2D

)wo foot#alls are thrown from the same point on a f lat fiel%8 "oth are thrown at an angle of 30 o a#o'e the horiBontal8 &a'' 2 has twice the initialspee% of a'' 18 $f a'' 1 is caught a %istance %1 from the thrower, how far away from the thrower %2 will the recei'er of a'' 2 #e when he catches it@

*a+ %2 = 2%1 *#+ %2 = 4%1 *c+ %2 = 8%1



Lecture 2#Lecture 2# Act 3 Act 39olution9olution

)he %istance a #all will go is simply x = (oizonta' s*ee!) x (time in ai) = v 0x t

2 y 0 0

t g

2

1t v y y −+=

)o figure out time in air., consi%er the

e;uation for the height of the #all:

0 t g 2

1t v 2

y 0 =− Chen the #all is caught, y = y 0

0 t g 2

1v t

y 0 =

− g

v 2 t y 0 =

t = 0 *time of throw+

*time of catch+

twosolutions



Lecture 2#Lecture 2# Act 3 Act 39olution9olution

g

v 2 t y 0 = o the time spent in the air is proportional to v 0y :

ince the angles are the same, #oth v 0y an% v 0x for a'' 2

are twice those of a'' 18

a'' 1

a'' 2

v 0y ,1

v 0x ,1

v 0y ,2

v 0x ,2

v 0,1

v 0,2

"all 2 is in the air tice as long as #all 1, #ut it also has tice the horiBontal spee%, so it will go times as far

x = v 0x t


24/28

Physics 111: Lecture 2, Pg 2(

9hooting the 8on/ey9hooting the 8on/ey(tran.uili;er gun)(tran.uili;er gun)

Chere %oes the Boo!eeperaim if he wants to hit the mon!ey@

* ?e !nows the mon!ey willlet go as soon as he shoots +


25/28


9hooting the 8on/ey&&&9hooting the 8on/ey&&&

$f there were no gra'ity, simply aim

at the mon!ey

r = r 0

r =v 0 t


26/28


9hooting the 8on/ey&&&9hooting the 8on/ey&&&

r r = v v 0 t #1 " 2 g g t

2

Cith gra'ity, still aim at the mon!ey r r = r 0 # 1 " 2 g g t 2

Dart hits the

mon!ey


27/28

Physics 111: Lecture 2, Pg 2<

Reca%:Reca%:9hooting the "on/ey&&&9hooting the "on/ey&&&

x x == x x 0 0

y y = #1 " 2 g g t 2

)his may #e easier to thin! a#out8

$tGs exactly the same i%ea

x x == v v 00 t t

y y = #1 " 2 g g t 2


28/28

Reca% o Lecture 2Reca% o Lecture 2

Recap of 1-D motion with constant acceleration8 *)ext: 2-3+

1-D &ree-&all *)ext: 2-3+

example

3-D Kinematics *)ext: 3-3 5 3-(+

hoot the mon!ey *Fx8 3-11+

"ase#all pro#lem

$n%epen%ence of x an% y components

3c free fall kinematics 3d

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