### Homework Hints

```Vectors and 2d-Kinematics
Continued



Relevant Equations
How to use them
Homework Hints
Mechanics Lecture 2, Slide 1
Hyperphysics-Trajectories
http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html
Mechanics Lecture 1, Slide 2
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 3
Hyperphysics-Trajectories
Mechanics Lecture 1, Slide 4
Maximum Height of Trajectory
v y ( t )  v 0 y  gt
Height of trajectory,h=ymax
y max  v y  0  t y max 
v0 y
g
h  y ( t y max )
h  y 0  v 0 y t y max 
h  y0 
2
2
gt y max
2
2
v0 y
1

1 v0 y
2 g
g
2
h  y0 
v0 y
2g
v 0 sin 
2
2
h  y0 
2g
Mechanics Lecture 2, Slide 5
Time of Flight
Mechanics Lecture 1, Slide 6
Time of Flight, “Hang Time”
y (t )  y0  v0 y t 
1
gt
2
2
t  t f  y (t  t f )  y0
y0  y0  v0 y t f 
v0 y t f 
1
2
2
gt f
1


2
gt f  t f  v 0 y  gt f   0
2
2


1
t f  0; t f 
2 v0 y
g

2 v 0 sin 
g
Mechanics Lecture 2, Slide 7
Hyperphysics-Trajectories
Mechanics Lecture 1, Slide 8
Range of trajectory
x (t )  x0  v0 xt
t  t f  y (t  t f )  y0
D  x (t  t f )  0  v0 xt f
D 
2 v0 x v0 y
2 v 0 cos  sin 
2

g
g
v 0 sin 2 
2
D 
g
Mechanics Lecture 2, Slide 9
Angle for Maximum Range
f ( )  sin( 2 )
df ( )
MAXIMUM range OCCURS AT 450
d
df ( )
d
 2 cos( 2 )
 0  cos( 2 )  0
 2  90
   45
0
0
Mechanics Lecture 2, Slide 10
Will it clear the fence
Mechanics Lecture 1, Slide 11
Height of Trajectory at time t or position x
y (t )  y0  v0 yt 
Height of trajectory, y(t)
1
gt
2
2
Height of trajectory, y(x)
x  x0  v0 xt
t
x  x0
v0 x
 x  x0
y ( x )  y 0  v 0 y 
 v0 x
 1  x  x0
  g
 2  v
0x






2
x 0  0; y 0  0
 x  1  x
  g
y ( x )  v 0 y 


v
 0 x  2  v0 x




2

 1 

x
x
  g

y ( x )  v 0 sin  
 2  v cos  
v
cos

 0

 0

2
Mechanics Lecture 2, Slide 12
Projectile Trajectory Equations
Maximum height
v 0 sin 
2
h  y0 
2
2g
Time of Flight (“Hang Time”)
tf 
2 v0 y

2 v 0 sin 
g
g
Range of trajectory
v 0 sin 2 
2
D 
g
Height of trajectory as f(t) , y(t)
y (t )  y0  v0 yt 
1
gt
2
2
Height of trajectory as f(x), y(x)

 1 

x
x
  g

y ( x )  v 0 sin  



 v 0 cos   2  v 0 cos  
2
Mechanics Lecture 1, Slide 13
Where will it land?
Mechanics Lecture 1, Slide 14
Launch Velocity-Given R and 
Mechanics Lecture 1, Slide 15
Launch Angle
Mechanics Lecture 1, Slide 16
Launch Velocity –Given R and h
Mechanics Lecture 1, Slide 17
Field Goal Example
A field goal kicker can kick the ball 30 m/s at an angle of 30 degrees w.r.t. the
ground. If the crossbar of the goal post is 3m off the ground, from how far
away can he kick a field goal?
y
x
3m
D
y-direction
x-direction
voy = vo sin(30o) = 15 m/s
vox = vo cos(30o) = 26 m/s
y = yo + voyt + ½ at 2
D = xo + vox t + ½ at 2
3 m = 0 m + (15 m/s) t – ½ (9.8 m/s2) t 2
= 0 m + (26 m/s)(2.8 s) + 0 m/s2 (2.8 s )2
t = 2.8 s or t = 0.22 s.
= 72.8 m
Illini Kicks 70 yard Field Goal
Mechanics Lecture 2, Slide 18
Homework Hints-Baseball
Mechanics Lecture 1, Slide 19
Mechanics Lecture 1, Slide 20
Calculate time to reach wall using vx:
t wall  x wall / v 0 x  x wall /  v 0 cos 

Calculate y position at time to reach wall:
y wall  y 0  v 0 y t wall 
1
2
g t wall
2
y wall  y 0  v 0 sin   x wall /  v 0 cos 
y wall  y 0  x wall tan  
1
2
 
1
2
g  x wall /  v 0 cos 
g  x wall /  v 0 cos 
 2
 2
Mechanics Lecture 1, Slide 21
Homework Hints-Catch
Mechanics Lecture 1, Slide 22
Homework Hints-Catch
v 0 x  v 0 cos 
v 0 x  v 0 sin 
(v0 y )
y max  y 0 
2g
v 0 sin  2
2
 y0 
2g
1
y f  y0 ; y f  y0  v0 y t f 
0  v0 y t f 
 tf 
1
2
g t f
2
 ; 1 g t 
2
2
2
f
g t f

2
 v0 y t f
2 v0 y
g
x f  v0x t f 
2 v0x v0 y
g
2 v 0 cos  sin 
2

g
Mechanics Lecture 1, Slide 23
Homework Hints-Catch
v 0 sin  2
 2 g  y max  y 0 ; v 0 cos   v  y  y max
v 0 sin  2  v 0 cos  2
 v0 
cos  
 v0
2
2 g  y max  y 0   v
v ( y  y max )

2
y
;   cos
 y max
1
v0

 v ( y  y max ) 




v
0


v 0 y  v 0 sin  ; v 0 x  v 0 cos 

   1 g  x
y julie  y 0  v 0 y x julie / v 0 x
2
/ v 0 x 
2
julie
Mechanics Lecture 1, Slide 24
Homework Hints-Catch 2
Mechanics Lecture 1, Slide 25
Homework Hints-Catch 2
v x  v 0 cos 
Vx is constant !
v
 2 v0 y 

v y ( t  t f )  v 0 y  g t f   v 0 y  g 
 g 


0  v0 y t f 
 tf 
1
2
g t f
 ; 1 g t 
2
2
2
f
Kinetic energy should
be same as when ball
was thrown. Ycomponent of velocity
would be downward.
 v0 y t f
2 v0 y
g
Mechanics Lecture 1, Slide 26
Homework Hints-Catch 2
x f  v0x t f 
v0 
x
v0 
2 v0x v0 y
g
x julie
Same conditions as before
t julie
2 g  y max  y 0   v

2
y
 y max

   1 g  x
y julie  y 0  v 0 y x julie / v 0 x
2
/ v 0 x 
2
julie
Mechanics Lecture 1, Slide 27
Homework Hints – Soccer Kick & Cannonball
Mechanics Lecture 1, Slide 28
Homework Hints – Soccer Kick & Cannonball
v0 
v0x  v0 y
  tan
2
2
 v0 y

v
 0x




1
(v0 y )
y max  y 0 
x f  v0x t f 
D 
2g
2 v0x v0 y
g
v 0 sin  2 
2
2

g
Mechanics Lecture 1, Slide 29
Homework Hints – Soccer Kick & Cannonball
v y ( t  t given )  v 0 y  g t given ; v x ( t  t given )  v 0 x
v ( t  t given ) 
2
v 0 x  v y ( t  t given )
2
y t  t given   y 0  v 0 y t given  
1
2
g t given

2
Mechanics Lecture 1, Slide 30
Trigonometric Identity for range equation
sin  
e
i
e
 i
2i
cos  
e
i
e
 i
2
 e i  e  i
sin  cos   
2i

sin  cos  
e
i (   )
e
  e i  e  i
 
2

i (   )
e
 e i e i  e i e  i  e  i e i  e  i e  i
 
4i

i (   )
e
i (   )
4i
i (   )
i (   )
i (   )
i (   )
1e
e
e
e
sin  cos   

2
2i
2i
sin  cos  
1
2
sin( 



  )  sin(    ) 
   
 sin  cos  
1
2
sin(    )  sin(    )  
http://mathworld.wolfram.com/Cosine.html
1
sin( 2  )
2
http://mathworld.wolfram.com/Sine.html
Mechanics Lecture 2, Slide 31
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 32
Question 2
Mechanics Lecture 2, Slide 33
Question 2
Mechanics Lecture 2, Slide 34
Hyperphysics-Trajectories
http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html
Mechanics Lecture 1, Slide 35
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