### Homework Hints

```Vectors and 2d-Kinematics
Continued



Summary of Vectors and 2-d Kinematics
Homework Solutions
Mechanics Lecture 2, Slide 1
Homework Results Vectors and 2-d kinematics
Awesome Job!
Average=87%
No attempt = 14
See Me or TAs
Mechanics Lecture 1, Slide 2
Homework 1 results
Average=62.5%
Among those who
Average=74.7%
Mechanics Lecture 1, Slide 3
Great Suggestions for Success
Mechanics Lecture 1, Slide 4
Questions/Suggestions
F  ma  a 
Fg  G
ga
Mm
r2
Fg
m
G
F
m
g G
24
M earth
11 3
1  2 5.97  10 kg



6
.
67

10
m
kg
s
 9.8m / s 2
2
2
6
R
6.37  10 m 
Mm
M
G 2
2
mr
r
For this situation the horizontal component of the velocity remains constant. The vertical component of velocity changes due
to the gravitational acceleration.
Kinematic equations for displacement,velocity and acceleration are the source for the derived equations.
Each component can be treated separately. Remember that they are component of a vector
Mechanics Lecture 1, Slide 5
Vectors and 2d-kinematics – Main Points
Mechanics Lecture 2, Slide 6
Vectors and 2d-kinematics – Main Points
Mechanics Lecture 2, Slide 7
Vectors and 2d-kinematics
Fundamental Equations
Mechanics Lecture 2, Slide 8
Source of Projectile Trajectory Equations
Horizontal
Vertical
Boring
Mechanics Lecture 1, Slide 9
Ballistic Projectile Motion Quantities
Initial velocity
speed,angle
Maximum Height of trajectory, h=ymax
“Hang Time”
Time of Flight, tf
Range of trajectory, D
Height of trajectory at arbitrary x,t
Mechanics Lecture 2, Slide 14
Derived Projectile Trajectory Equations
Maximum height
v02 sin 2 
h  y0 
2g
Time of Flight (“Hang Time”)
tf 
2v0 y
g

2v0 sin 
g
Range of trajectory
v02 sin 2
D
g
Height of trajectory as f(t) , y(t)
y (t )  y0  v0 y t 
1 2
gt
2
Height of trajectory as f(x), y(x)
 x  1  x 
  g 

y ( x )  v0 sin  
 v0 cos   2  v0 cos  
2
Mechanics Lecture 1, Slide 15
Homework Solutions-Baseball
Mechanics Lecture 1, Slide 16
Homework Solutions-Baseball
Mechanics Lecture 1, Slide 17
Mechanics Lecture 1, Slide 18
Calculate time to reach wall using vx:
twall  xwall / v0 x  xwall / v0 cos  
Calculate y position at time to reach wall:
ywall  y0  v0 y twall 
1
2
g twall 
2
ywall  y0  v0 sin   xwall / v0 cos   
ywall  y0  xwall tan  
1
2
g  xwall / v0 cos  
2
1
2
g  xwall / v0 cos  
2
Mechanics Lecture 1, Slide 19
twall  xwall / v0 x  xwall / v0 cos  
ywall  y0  v0 y twall 
1
2
g twall 
2
ywall  y0  v0 sin   xwall / v0 cos   
ywall  y0  xwall tan  
1
2
g  xwall / v0 cos  
2
1
2
g  xwall / v0 cos  
2
x  565 ft;  350 ; g  32.2 ft / s 2 ; y0  0
1
2
g  xwall / v0 cos  
2
2
 3 ft  (565 ft )(.7002)  (16.1 ft / s 2 )565 ft / 176 ft / s  0.8192) 
ywall  y0  xwall tan  
ywall
ywall  3 ft  395.61 ft  247.24 ft  151.37 ft
Mechanics Lecture 1, Slide 20
Homework Hints-Catch
Mechanics Lecture 1, Slide 21
Homework Hints-Catch
v0 x  v0 cos 
v0 x  v0 sin 
ymax  y0 
( v0 y ) 2
2g
 y0
2

v0 sin  

2g
y f  y0 ; y f  y0  v0 y t f 
0  v0 y t f 
 tf 
1
2
g t f 
2
1
2 1
2
g t f  ; g t f   v0 y t f
2
2
2v0 y
g
x f  v0 x t f 
2v0 x v0 y
g

2v 02 cos  sin 
g
Mechanics Lecture 1, Slide 22
Homework Hints-Catch
v0 sin  2  2 g  ymax  y0 ; v0 cos  v  y  ymax 
v0 sin  2  v0 cos 2  v02
 v0  2 g  ymax  y0   v 2  y  ymax 
cos  
v ( y  ymax )
 v ( y  ymax ) 

;  cos1 
v0
v0


v0 y  v0 sin  ; v0 x  v0 cos 

 
y julie  y0  v0 y x julie / v0 x 
1
2
g x julie / v0 x 
2
Mechanics Lecture 1, Slide 23
Homework Solutions-Catch
v0 sin  2  2 g  ymax  y0 ; v0 cos  v  y  ymax 
v0 sin  2  v0 cos 2  v02
 v0  2 g  ymax  y0   v 2  y  ymax 
cos  
v ( y  ymax )
 v ( y  ymax ) 

;  cos1 
v0
v0


v0 y  v0 sin  ; v0 x  v0 cos 

 
y julie  y0  v0 y x julie / v0 x 
1
2
g x julie / v0 x 
2
Mechanics Lecture 1, Slide 24
Homework Solutions-Catch
v0 x  v0 cos   (17m / s )(cos 35)  13.92m / s
v0 y  v0 sin   (17m / s )(.5736)  9.75m / s
ymax  y0 
( v0 y ) 2
2g
 y0
2

v0 sin  

2g
 1.5m 
x f  v0 x t f 
9.75m / s 2
2(9.81m / s 2 )
2v0 x v0 y
g
 6.34m
 27.67m
v0  2 g  ymax  y0   v 2  y  ymax   19.92m / s
Mechanics Lecture 1, Slide 25
Homework Solutions-Catch
cos   17m / s / 19.92m / s  0.8534;  cos1 0.8534   31.410
sin   1  (0.8534) 2  0.5215
 v0 y  v0 sin   0.521519.92m / s   10.38m / s

 
y julie  y0  v0 y x julie / v0 x 
1
2
g x julie / v0 x 
2
1
2
y julie  1.5m  (10.38m / s )27.67m / (17m / s )   (9.81m / s 2 )27.67m / (17m / s ) 
2
y julie  1.5m  16.90m  12.99m  5.40m
Mechanics Lecture 1, Slide 26
Homework Hints-Catch 2
Mechanics Lecture 1, Slide 27
Homework Hints-Catch 2
v x  v0 cos 
 2v0 
v y (t t f )  v0 y  g t f   v0 y  g  y 
 g 
0  v0 y t f 
 tf 
Vx is constant !
v
Kinetic energy should
be same as when ball
was thrown. Ycomponent of velocity
would be downward.
1
2 1
2
g t f  ; g t f   v0 y t f
2
2
2v0 y
g
Mechanics Lecture 1, Slide 28
Homework Hints-Catch 2
x f  v0 x t f 
v0 x 
2v0 x v0 y
g
x julie
Same conditions as before
t julie
v0  2 g  ymax  y0   v 2  y  ymax 

 
y julie  y0  v0 y x julie / v0 x 
1
2
g x julie / v0 x 
2
Mechanics Lecture 1, Slide 29
Homework Hints – Soccer Kick & Cannonball
Mechanics Lecture 1, Slide 30
Homework Hints – Soccer Kick & Cannonball
v0  v02x  v02y
 v0 y
  tan 
 v0
 x
1
ymax  y0 
x f  v0 x t f 
D




( v0 y ) 2
2g
2v0 x v0 y
g
v 02 sin 2 
g
Mechanics Lecture 1, Slide 31
Homework Hints – Soccer Kick & Cannonball
v y (t  t given )  v0 y  g t given ; v x (t  t given )  v0 x
v (t  t given )  v0 x  v 2y (t  t given )
2
y t  t given   y0  v0 y t given  
1
2
g t given 
2
Mechanics Lecture 1, Slide 32
Homework Solutions-Catch
Mechanics Lecture 1, Slide 33
Homework Hints-Catch
v0 x  v0 cos 
v0 x  v0 sin 
ymax  y0 
( v0 y ) 2
2g
 y0
2

v0 sin  

2g
y f  y0 ; y f  y0  v0 y t f 
0  v0 y t f 
 tf 
1
2
g t f 
2
1
2 1
2
g t f  ; g t f   v0 y t f
2
2
2v0 y
g
x f  v0 x t f 
2v0 x v0 y
g

2v 02 cos  sin 
g
Mechanics Lecture 1, Slide 34
Homework Hints-Catch 2
Mechanics Lecture 1, Slide 35
Homework Solutions-Catch 2
v x  v0 cos   (17m / s )(cos 35)  13.92m / s
Vx is constant !
v
 2v0 y 
  v0 y
v y (t t f )  v0 y  g t f   v0 y  g 

g


v y (t t f )  9.75m / s  (9.81m / s 2 )(1.9878s )  9.75m / s
Kinetic energy should
be same as when ball
was thrown. Ycomponent of velocity
would be downward.
1
2 1
2
g t f  ; g t f   v0 y t f
2
2
2v0 y 2(9.75m / s )
 tf 

 1.988s
2
g
9.81m / s
0  v0 y t f 
Mechanics Lecture 1, Slide 36
Homework Solutions-Catch 2
x f  v0 x t f 
v0 x 
x julie
t julie

2v0 x v0 y
g
27.67m
 17m / s
1.628m / s
 27.67m
Same conditions as before
v0  2 g  ymax  y0   v 2  y  ymax   19.92m / s

 
y julie  y0  v0 y x julie / v0 x 
1
2
g x julie / v0 x 
2
1
2
y julie  1.5m  (10.38m / s )27.67m / (17m / s )   (9.81m / s 2 )27.67m / (17m / s ) 
2
y julie  1.5m  16.90m  12.99m  5.40m
Mechanics Lecture 1, Slide 37
Homework Hints – soccer kick
Mechanics Lecture 1, Slide 38
Homework Solutions– soccer kick
v0  v02x  v02y 
15m / s 2  15m / s 2
 v0 y
 v0
 x
  tan 1 
ymax  y0 
( v0 y ) 2
2g
x f  v0 x t f 

  tan 1 1  450


 0m 
2v0 x v0 y
g

 21.213m / s
15m / s 2
2(9.81m / s 2 )
 11.47m
215m / s 15m / s 
 45.87m
(9.81m / s 2 )
v02 sin(2 ) 15m / s 15m / s 
D

 45.87m
g
(9.81m / s 2 )
Mechanics Lecture 1, Slide 39
Homework Solutions – soccer kick
v y (t  0.7 s )  v0 y  g 0.7 s   15m / s  (9.81m / s 2 )(0.7 s )  8.133m / s
v (t  0.7 s )  v x2 (t  0.7 s )  v 2y (t  0.7 s )  (15m / s ) 2  (8.13m / s ) 2
v (t  0.7 s )  17.06m / s
y t  0.7 s   y0  v0 y 0.7 s  
1
2
g 0.7 s 
2
1
2
y t  0.7 s   0m  15m / s 0.7 s   9.81m / s 2 0.7 s   8.097m
2
Mechanics Lecture 1, Slide 40
Homework Solutions - Cannonball
Mechanics Lecture 1, Slide 41
Cannonball Solutions
v0  v02x  v02y 
37m / s 2  23m / s 2
 v0 y
  tan 
 v0
 x
1
ymax  y0 
D
v 02 sin 2 
g
( v0 y ) 2
2g
 43.566m / s

  tan 1  23   31.870

 37 

 0m 
23m / s 2
2
2(9.81m / s )
 26.96m
2

43.566m / s  sin 2(31.870 ) 

 173.51m
9.81m / s 2
Mechanics Lecture 1, Slide 42
Cannonball- Solutions
v y (t  1.0s )  v0 y  g 1.0s   23m / s  (9.81m / s 2 )(1.0s )  13.19m / s; v x (t  1.0s )  v0 x
v (t  1.0s )  v x2 (t  1.0s )  v 2y (t  1.0s )  (37m / s ) 2  (13.19m / s ) 2
v (t  1.0s )  39.28m / s
y t  1.0s   y0  v0 y 1.0s  
1
2
g 1.0s 
2
1
2
y t  1.0s   0m  23m / s 1.0s   9.81m / s 2 1.0s   18.09m
2
Mechanics Lecture 1, Slide 43
Vectors and 2d-kinematics – Main Points
Mechanics Lecture 2, Slide 44
vtrain car
Time spend in the air depends on the maximum height
Maximum height depends on the initial vertical velocity
Mechanics Lecture 2, Slide 45
Trigonometric Identity for range equation
ei  e  i
sin  
2i
ei  e i
cos 
2
 ei  e i  ei  e i  ei ei  ei e i  e i ei  e i e i

 
sin  cos   
2
i
2
4i



ei (   )  ei (   )  ei (   )  e i (   )
sin  cos  
4i
1  e i (    )  e i (    ) e i (    )  e  i (    ) 

sin  cos   

2
2i
2i

sin  cos  
   
1
sin(   )  sin(   ) 
2
 sin  cos 
1
sin(   )  sin(   )   1 sin(2 )
2
2
http://mathworld.wolfram.com/Cosine.html
http://mathworld.wolfram.com/Sine.html
Mechanics Lecture 2, Slide 46
Trigonometric Identities relating sum and products
List of trigonometric identities
sin(   )  sin  cos   cos sin 
   
 sin(2 )  sin  cos  cos sin   2 sin  cos
Mechanics Lecture 2, Slide 47
Hyperphysics-Trajectories
http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html
Mechanics Lecture 1, Slide 48
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