Mechanics Lecture 2, Slide 1 Vectors and 2d-Kinematics Continued Comments on Homework Summary of Vectors and 2-d Kinematics Homework Solutions.

Mechanics Lecture 2, Slide 1

Vectors and 2d-KinematicsContinued

Comments on Homework Summary of Vectors and 2-d Kinematics Homework Solutions

Homework Results Vectors and 2-d kinematics


Average=87%No attempt = 14

Awesome Job!

See Me or TAs

Homework 1 results


Average=62.5%

Among those who made an attemptAverage=74.7%

Great Suggestions for Success


Questions/Suggestions


22

2

r

MG

mr

MmG

m

Fag

r

MmGF

m

FamaF

g

g

2

26

2421311

2 /8.91037.6

1097.51067.6 sm

m

kgskgm

R

MGg earth

Each component can be treated separately. Remember that they are component of a vector

Kinematic equations for displacement,velocity and acceleration are the source for the derived equations .

For this situation the horizontal component of the velocity remains constant. The vertical component of velocity changes due to the gravitational acceleration.


Vectors and 2d-kinematics – Main Points



Vectors and 2d-kinematics Fundamental Equations


Source of Projectile Trajectory Equations


Horizontal Vertical Boring

Ballistic Projectile Motion Quantities


Initial velocityspeed,angle

Maximum Height of trajectory, h=ymax

Range of trajectory, D

Height of trajectory at arbitrary x,t

“Hang Time”Time of Flight, tf

Derived Projectile Trajectory Equations


2

000 cos2

1

cossin)(

v

xg

v

xvxy

200 2

1)( gttvyty y

Height of trajectory as f(x), y(x)

Height of trajectory as f(t) , y(t)

g

vD

2sin20

g

v

g

vt y

f

sin2200

g

vyh

2

sin220

0

Range of trajectory

Time of Flight (“Hang Time”)

Maximum height

Homework Solutions-Baseball


Homework Solutions-Baseball


Homework Hints- Baseball Stadium Wall


Homework Hints – Stadium Wall


Calculate time to reach wall using vx:

cos// 00 vxvxt wallwallwall x

Calculate y position at time to reach wall:

200

20000

200

cos/2

1tan

cos/2

1cos/sin

2

1

vxgxyy

vxgvxvyy

tgtvyy

wallwallwall

wallwallwall

wallwallwall y

Homework Solutions-Baseball Stadium


200

20000

200

00

cos/2

1tan

cos/2

1cos/sin

2

1

cos//

vxgxyy

vxgvxvyy

tgtvyy

vxvxt

wallwallwall

wallwallwall

wallwallwall

wallwallwall

y

x

ftftftfty

sftftsftftfty

vxgxyy

ysftgftx

wall

wall

wallwallwall

37.15124.24761.3953

)8192.0/176/565)/1.16()7002)(.565(3

cos/2

1tan

0;/2.32;35;565

22

200

020

Homework Hints-Catch




cos00 vvx

sin00 vvx

g

vy

g

vyy y

2

sin

2

)( 20

0

20

0max

g

v

g

vvtvx

g

vt

tvtgtgtv

tgtvyyyy

yx

x

y

yy

y

ff

f

ffff

ffff

sincos22

22

1;

2

10

2

1;

200

0

0

022

0

2000

0



20000

0000

0

max1

0

max

/2

1/

cos;sin

)(cos;

)(cos

xjuliejuliejulie vxgvxvyy

vvvv

v

yyv

v

yyv

xy

xy

max2

0max0

20

20

20

max00max2

0

2

cossin

cos;2sin

yyvyygv

vvv

yyvvyygv

Homework Solutions-Catch


20000

0000

0

max1

0

max

/2

1/

cos;sin

)(cos;

)(cos

xjuliejuliejulie vxgvxvyy

vvvv

v

yyv

v

yyv

xy

xy

max2

0max0

20

20

20

max00max2

0

2

cossin

cos;2sin

yyvyygv

vvv

yyvvyygv



smsmvvx

/92.13)35)(cos/17(cos00

smsmvvy

/75.9)5736)(./17(sin00

m

sm

smm

g

vy

g

vyy y 34.6

)/81.9(2

/75.95.1

2

sin

2

)(2

220

0

20

0max

smyyvyygv /92.192 max2

0max0

mg

vvtvx yx

x ff 67.272 00

0



mmmmy

smmsmsmmsmmy

vxgvxvyy

julie

julie

xjuliejuliejulie xy

40.599.1290.165.1

)/17(/67.27)/81.9(2

1)/17(/67.27)/38.10(5.1

/2

1/

22

20000

smsmvv

smsm

y/38.10/92.195215.0sin

5215.0)8534.0(1sin

41.318534.0cos;8534.0/92.19//17cos

00

2

01

Homework Hints-Catch 2




cos0vvx vVx is constant !

g

vgvtgvttv y

yy ffy

0

00

2)(

g

vt

tvtgtgtv

y

yy

f

ffff

0

022

0

22

1;

2

10

Kinetic energy should be same as when ball was thrown. Y-component of velocity would be downward.



g

vvtvx yx

x ff

00

0

2

julie

julie

t

xv

x0

Same conditions as before

max2

0max0 2 yyvyygv

20000 /2

1/ xjuliejuliejulie vxgvxvyy

xy

Homework Hints – Soccer Kick & Cannonball




20

200 yx

vvv

x

y

v

v

0

01tan

g

vyy y

2

)( 20

0max

g

vD

2sin2

0

g

vvtvx yx

x ff

00

0

2



)()(

)(;)(

220

00

givenyxgiven

givenxgivengiveny

ttvvttv

vttvtgvttvxy

200 2

1givengivengiven tgtvytty

y





cos00 vvx

sin00 vvx

g

vy

g

vyy y

2

sin

2

)( 20

0

20

0max

g

v

g

vvtvx

g

vt

tvtgtgtv

tgtvyyyy

yx

x

y

yy

y

ff

f

ffff

ffff

sincos22

22

1;

2

10

2

1;

200

0

0

022

0

2000

0



Homework Solutions-Catch 2


smsmvvx /92.13)35)(cos/17(cos0 vVx is constant !

smssmsmttv

vg

vgvtgvttv

fy

ffy y

y

yy

/75.9)9878.1)(/81.9(/75.9)(

2)(

2

0

0

00

ssm

sm

g

vt

tvtgtgtv

y

yy

f

ffff

988.1/81.9

)/75.9(222

1;

2

10

2

0

022

0

Kinetic energy should be same as when ball was thrown. Y-component of velocity would be downward.

Homework Solutions-Catch 2


mg

vvtvx yx

x ff 67.272 00

0

smsm

m

t

xv

julie

julie

x/17

/628.1

67.270 Same conditions as before

smyyvyygv /92.192 max2

0max0

mmmmy

smmsmsmmsmmy

vxgvxvyy

julie

julie

xjuliejuliejulie xy

40.599.1290.165.1

)/17(/67.27)/81.9(2

1)/17(/67.27)/38.10(5.1

/2

1/

22

20000

Homework Hints – soccer kick


Homework Solutions– soccer kick


smsmsmvvvyx

/213.21/15/15 2220

200

01

0

01 451tantan

x

y

v

v

m

sm

smm

g

vyy y 47.11

)/81.9(2

/150

2

)(2

220

0max

m

sm

smsm

g

vvtvx yx

x ff 87.45)/81.9(

/15/15222

00

0

m

sm

smsm

g

vD 87.45

)/81.9(

/15/15)2sin(2

20

Homework Solutions – soccer kick


smstv

smsmstvstvstv

smssmsmsgvstv

yx

y y

/06.17)7.0(

)/13.8()/15()7.0()7.0()7.0(

/133.8)7.0)(/81.9(/157.0)7.0(

2222

20

mssmssmmsty

sgsvystyy

097.87.0/81.92

17.0/1507.0

7.02

17.07.0

22

200

Homework Solutions - Cannonball


Cannonball Solutions


smsmsmvvvyx

/566.43/23/37 2220

200

01

0

01 87.3137

23tantan

x

y

v

v

m

sm

smm

g

vyy y 96.26

)/81.9(2

/230

2

)(2

220

0max

m

sm

sm

g

vD 51.173

/81.9

)87.31(2sin/566.432sin2

022

0

Cannonball- Solutions


smstv

smsmstvstvstv

vstvsmssmsmsgvstv

yx

xy xy

/28.39)0.1(

)/19.13()/37()0.1()0.1()0.1(

)0.1(;/19.13)0.1)(/81.9(/230.1)0.1(

2222

02

0

mssmssmmsty

sgsvystyy

09.180.1/81.92

10.1/2300.1

0.12

10.10.1

22

200




Time spend in the air depends on the maximum height

Maximum height depends on the initial vertical velocity

vtrain car

Trigonometric Identity for range equation


)2sin(2

1)sin()sin(

2

1cossin

)sin()sin(2

1cossin

222

1cossin

4cossin

422cossin

2cos

2sin

)()()()(

)()()()(

i

ee

i

ee

i

eeee

i

eeeeeeeeee

i

ee

ee

i

ee

iiii

iiii

iiiiiiiiiiii

ii

ii

http://mathworld.wolfram.com/Cosine.html http://mathworld.wolfram.com/Sine.html

http://mathworld.wolfram.com/Cosine.html

http://mathworld.wolfram.com/Sine.html

Trigonometric Identities relating sum and products


List of trigonometric identities

cossin2sincoscossin)2sin(

sincoscossin)sin(

http://www.ask.com/wiki/List_of_trigonometric_identities?o=2801&qsrc=999&ad=doubleDown&an=apn&ap=ask.com#Angle_sum_and_difference_identities

Hyperphysics-Trajectories


http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html



Mechanics Lecture 2, Slide 1 Vectors and 2d-Kinematics Continued Comments on Homework Summary of Vectors and 2-d Kinematics Homework Solutions.

Documents

tas slide

t f slide

results mechanics lecture

success mechanics lecture

d kinematics mechanics

homework results vectors

homework summary of

d height of trajectory