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
Mechanics Lecture 1, Slide 2
Average=87%No attempt = 14
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Homework 1 results
Mechanics Lecture 1, Slide 3
Average=62.5%
Among those who made an attemptAverage=74.7%
Questions/Suggestions
Mechanics Lecture 1, Slide 5
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.
Ballistic Projectile Motion Quantities
Mechanics Lecture 2, Slide 14
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
Mechanics Lecture 1, Slide 15
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 Hints – Stadium Wall
Mechanics Lecture 1, Slide 19
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
Mechanics Lecture 1, Slide 20
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
Mechanics Lecture 1, Slide 22
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 Hints-Catch
Mechanics Lecture 1, Slide 23
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
Mechanics Lecture 1, Slide 24
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
Mechanics Lecture 1, Slide 25
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
Homework Solutions-Catch
Mechanics Lecture 1, Slide 26
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
Mechanics Lecture 1, Slide 28
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.
Homework Hints-Catch 2
Mechanics Lecture 1, Slide 29
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
Mechanics Lecture 1, Slide 31
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
Homework Hints – Soccer Kick & Cannonball
Mechanics Lecture 1, Slide 32
)()(
)(;)(
220
00
givenyxgiven
givenxgivengiveny
ttvvttv
vttvtgvttvxy
200 2
1givengivengiven tgtvytty
y
Homework Hints-Catch
Mechanics Lecture 1, Slide 34
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
Mechanics Lecture 1, Slide 36
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
Mechanics Lecture 1, Slide 37
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 Solutions– soccer kick
Mechanics Lecture 1, Slide 39
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
Mechanics Lecture 1, Slide 40
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
Cannonball Solutions
Mechanics Lecture 1, Slide 42
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
Mechanics Lecture 1, Slide 43
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
Mechanics Lecture 2, Slide 45
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
Mechanics Lecture 2, Slide 46
)2sin(2
1)sin()sin(
2
1cossin
)sin()sin(2
1cossin
222
1cossin
4cossin
422cossin
2cos
2sin
)()()()(
)()()()(
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ee
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iiii
iiiiiiiiiiii
ii
ii
http://mathworld.wolfram.com/Cosine.html http://mathworld.wolfram.com/Sine.html
Trigonometric Identities relating sum and products
Mechanics Lecture 2, Slide 47
List of trigonometric identities
cossin2sincoscossin)2sin(
sincoscossin)sin(
Hyperphysics-Trajectories
Mechanics Lecture 1, Slide 48
http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html