Home  Work  Solutions            PHYS111.02  SFSU  Eradat                                    [WA4  CHAPTER  4]   1    Chapter  4  7.     Picture  the  Problem:  The  paths  of  the  two  canoeists  are  shown  

at  right.  

  Strategy:  Canoeist  1’s  45°  path  determines  an  isosceles  right  triangle  whose  legs  measure  1.0  km.  So  canoeist  2’s  path  determines  a  right  triangle  whose  legs  measure  1.0  km  north  and  0.5  km  west.  Use  the  right  triangle  to  find  the  angle  θ.  Find  the  distance  traveled  by  each  canoeist  and  set  the  times  of  travel  equal  to  each  other  to  determine  the  appropriate  speed  of  canoeist  2:

 

 

  Solution:  1.  (a)  Find  the  angle  θ    from  the  right  triangle  of  canoeist  2:  

θ = tan−1 0.5 km

1.0 km⎛⎝⎜

⎞⎠⎟= 27°

 

  2.  (b)  Set  the  travel  times    equal  to  each  other:  

t1 = t2 ⇒

Δr1

v1

=Δr2

v2

 

 

 3.  Use  the  resulting  ratio  to  find  the    appropriate  speed  of  canoeist  2:  

v2 =d2

d1

v1 =0.5km( )2

+ 1.0 km( )2

1.0 km( )2+ 1.0 km( )2

1.35 ms

⎛⎝⎜

⎞⎠⎟= 1.1 m/s  

  Insight:  There  are  other  ways  to  solve  this  problem.    For  instance,  because  the  motions  are  independent,  we  could  set  the  time  it  takes  canoeist  1  to  travel  1.0  km  horizontally  equal  to  the  time  it  takes  canoeist  2  to  travel  0.5  km  horizontally.  

 16.     Picture  the  Problem:  The  trajectory  of  the  clam  is  indicated  in  the  

figure  at  right.  

  Strategy:  Use  equation  4-­‐7  to  analyze  the  motion  of  the  clam  as  it  is  launched  horizontally  and  falls  to  the  rocks  below.    Let    upward  be  the  positive  y  direction.  

  Solution:  1.  (a)  The  horizontal  velocity  remains  constant  at  2.70  m/s.  

  2.  (b)  The  vertical  speed  increases  due  to  the  acceleration  of  gravity:  

vy = −gt = − 9.81 m/s2( ) 2.10 s( )= –20.6 m s

 

  3.  (c)  If  the  speed  of  the  crow  were  to  increase,  the  speed  of  the  clam  in  the  x-­‐direction  would  increase,  but  the  speed  in  the  y-­‐direction  would  stay  the  same.    

 

  Insight:  The  speed  of  the  crow  determines   vx  and  gravity  determines vy .    As  in  most  of  the  other  

problems  in  this  chapter,  the  answers  are  only  true  as  long  as  we  neglect  the  effects  of  air  friction.    25.     Picture  the  Problem:  The  ball  falls  down  along  a  parabolic  arc,  maintaining  its  horizontal  velocity  but  

gaining  vertical  speed  as  it  falls.  

  Strategy:  The  ball  is  accelerated  only  by  gravity.    The  initial  height  of  the  ball  determines  the  vertical  component  of  its  final  velocity,  which  together  with  the  final  speed  can  be  used  to  find  the  horizontal  component  of  the  velocity.    The  initial  speed  equals  the  horizontal  component  because  it  remains  the  same  throughout  the  flight.  Equation  4-­‐7  can  be  used  in  each  case.  

  2  

  Solution:  1.  (a)  The  acceleration  is  due  only  to  gravity:  

a = 9.81 m/s2 downward  

  2.  (b)  Find  the  vertical  component  of  the  final  velocity:  

vy2 = −2gΔy

vy = −2 9.81 m/s2( ) – 0.75 m( ) = 3.84 m/s

 

  3.  Find  the  horizontal  component  of  the  ball’s  velocity,  which  is  also  its  initial  speed:  

vx = v2 − vy2 = 4.0 m/s( )2

− 3.84 m/s( )2= 1.1 m/s  

 4.  (c)  Repeat  for  a  final  speed  of  5.0  m/s:  

vx = v2 − vy

2 = 5.0 m/s( )2− 3.84 m/s( )2

= 3.2 m/s  

  Insight:  The  vertical  motion  of  an  object  dropped  from  rest  or  launched  horizontally  is  determined  solely  by  the  initial  height  and  the  acceleration  of  gravity.    That  is  why  the  vertical  component  of  the  final  velocity  is  the  same  in  part  (c)  as  it  is  in  part  (b).  

 42.     Picture  the  Problem:  The  trajectory  of  the  

keys  is  depicted  in  the  first  figure.    The  second  figure  shows  the  geometry  of  the  initial  position  and  velocity  of  the  keys.  

  Strategy: Find the initial speed of the keys from the motion of the Ferris wheel. Then use geometry to find the position of the keys when they are released. Find the initial horizontal and vertical components of the key’s velocity upon release. Then use equation 4-10 to find the vertical speed of the keys just before they hit the ground. Use the initial and final vertical speeds to find the time of flight, and use the time of flight together with the horizontal velocity of the keys to determine the impact location.  

  Solution:  1.  Find  the  speed  of  the  keys  by  finding    the  speed  of  the  rim  of  the  Ferris  wheel.  

v = Ct= 2πr

t= 2π (5.00 m)

32.0 s= 0.982 m/s  

  2.  Each  hour  on  the  clock  is  360°/12=30°  apart.    Therefore  the  initial  position  vector  of  the  keys,    relative  to  the  center  of  the  wheel,  is  α  =  60°  counterclockwise  from  the  vertical  position.      The  x  and  y  initial  positions  are:  

x0 = −r sinα = − 5.00 m( )sin 60° = – 4.33 m (left of center)

y0 = r cosα + r + b = 5.00 m( )cos60° + 5.00 m +1.75 my0 = 9.25 m (above the ground)

 

  3.  Now  find  the  x  and  y  components  of   v0  by    

realizing  from  the  diagram  that  θ  =  α  =  60°:  

v0x = v0 cosθ = 0.982 m/s( )cos60° = 0.491 m/s

v0 y = v0 sinθ = 0.982 m/s( )sin 60° = 0.850 m/s  

  4.  Use  equation  4-­‐10  to  find    the  final  vertical  speed:  

vy = ± v0 y2 − 2gΔy = ± 0.850 m/s( )2

− 2 9.81 m/s2( ) −9.25 m( )= −13.5 m/s the keys are traveling downwards( )

 

  5.  Find  the  time  of  flight  from  the  initial    and  final  vertical  velocities:  

t =

vy − v0 y

−g= −13.5− 0.850 m/s

−9.81 m/s2 = 1.46 s  

θ

α cgv

 r  

0v  

Figure  (b):  Vector  diagram  of  the  release  point.  

   Figure  (a):  Trajectory  of  the  keys.  

Home  Work  Solutions            PHYS111.02  SFSU  Eradat                                    [WA4  CHAPTER  4]   3       6.  Use  the  time  of  flight  to  find  the    

horizontal  position  upon  landing:   x = x0 + v0xt = − 4.33 m + 0.491 m/s( ) 1.46 s( ) = −3.61 m ,    or  the  keys  land  3.61  m  left  of  the  base  of  the  Ferris  wheel.  

  Insight:  The  time  of  flight  can  also  be  found  from y = v0 sinθ( ) t − 1

2 gt2 .    Although  such  an  approach  would  not  need  step  4  above,  it  would  require  the  use  of  the  quadratic  formula.  

 43.     Picture  the  Problem:  The  trajectory  of  the  girl  is  depicted  at  right.  

 Strategy:  Use  equation  4-­‐10  and  the  given  time  of  flight,  initial  speed,  and  launch  angle  to  determine  the  initial  height  of  the  girl  at  the  release  point.  

    Solution:  Use  equation  4-­‐10  to  find    

the  initial  height  of  the  girl  at  the  release    point.    If  we  let  the  release  height    correspond  to  y  =  0,  then  the  landing    height  is:  

y = v0 sinθ( ) t − 12 gt2

= 2.25 m/s( ) sin 35.0°( ) 0.616 s( ) − 12 9.81 m/s2( ) 0.616 s( )2

y = −1.07 m  In  other  words,  she  was  1.07  m  above  the  water  when  she  let  go  of  the  rope.  

  Insight:  The  girl’s  speed  upon  impact  with  the  water  is  5.10  m/s  (check  for  yourself)  or  11  mi/h.    A  fun  plunge!  

 46.     Picture  the  Problem:  The  dolphin  travels  along  a  parabolic  arc  as  it  glides  through  the  air  and  lands  in  

the  water.  

  Strategy:    Because  the  dolphin  is  traveling  horizontally  as  it  passes  through  the  hoop,  we  conclude  that  

vy = 0  at  that  point  and  that  the  dolphin  must  be  at  the  peak  of  its  flight.    Use  the  formula  derived  in  Example  4-­‐7  to  find  how  high  the  center  of  the  hoop  is  above  the  surface  of  the  water.  

 Solution:  Find  the  maximum  height    of  the  dolphin  above  the  water:  

ymax =

v0 sinθ( )2

2g=

12.0 m/s( )sin 40.0°⎡⎣ ⎤⎦2

2 9.81 m/s2( ) = 3.03 m  

  Insight:  In  order  to  pass  through  a  higher  hoop  the  dolphin  must  either  increase  the  launch  angle  or  jump  with  a  higher  initial  speed.  

 56.     Picture  the  Problem:  The  hay  bale  travels  along  a  parabolic  arc,  maintaining  its  horizontal  velocity  but  

changing  its  vertical  speed  due  to  the  constant  downward  acceleration  of  gravity.     Strategy: The initial velocity of the bale is given as

v0 = 1.12 m/s( ) x + 8.85 m/s( ) y . Use the fact that the

horizontal component of the bale’s velocity never changes throughout the flight in order to find the vertical component of the velocity when the total speed is 5.00 m/s. Then find the time elapsed between the initial throw and the instant the bale has that vertical speed. For part (b) set the vertical speed equal to the (constant) horizontal speed in magnitude but negative in direction (pointing downward). Find the time elapsed between initial throw and the instant the bale has that new vertical speed.

  Solution:  1.  (a)  Determine  the  y  component    of  the  velocity  when  the  total  speed  is  5.00  m/s:  

v2 = v0x2 + vy

2

vy = ± v2 − v0x2 = ± 5.00 m/s( )2

− 1.12 m/s( )2= ±4.873 m/s

 

  4  

  2.  Use  the  positive  value  of   vy because  the  bale    

is  rising  when  the  speed  first  equals  5.00  m/s.    Find    the  time  elapsed  to  this  point  from  equation  4-­‐6:  

t =

vy − v0 y

−g= 4.87 m/s − 8.85 m/s

− 9.81 m/s2 = 0.405 s  

  3.  (b)  If  the  bale’s  velocity  points  45.0°  below  the    horizontal  then  we  know  the  vertical  velocity:  

vy = −v0x = −1.12 m/s  

 4.  Find  the  time  elapsed  from  equation  4-­‐6:  

t =

vy − v0 y

−g= −1.12 m/s − 8.85 m/s

− 9.81 m/s2 = 1.02 s  

  5.  (c)  If   v0  is  pointed  straight  upward  then  the  initial  vertical  velocity  component  will  be  larger,  so  it  will  

rise  higher  and  its  time  in  the  air  will  increase.     Insight:  The  bale  will  have  a  speed  of  5.00  m/s  again  after  1.40  s  has  elapsed.    If  it  were  thrown  straight  

upward  with  the  same  initial  speed  (8.92  m/s)  it  would  rise  to  a  height  of  4.06  m,  as  opposed  to  3.99  m  as  in  the  original  case.  

 62.     Picture  the  Problem:  The  initial  velocity  and  acceleration  vectors  of  the  

hummingbird  are  depicted  at  right.  

  Strategy:    Use  the  horizontal  acceleration  to  determine  the  horizontal  component  of  the  velocity  at  time  t  =  0.72  s.    Since  there  is  no  acceleration  in  the  y  direction  the  bird’s  vertical  speed  remains  at  4.6  m/s  throughout  its  motion.    Use  the  known  components  of  the  velocity  to  determine  the  direction  of  motion.  

  Solution:  1.  (a)  Find   vx  at  t  =  0.72  s:  

vx = v0x + axt = 0 + 11 m/s2( ) 0.72 s( )= 7.9 m/s

 

  2.  (b)   vy remains  constant:  

vy = 4.6 m/s  

 

 3.  (c)  Find  the  direction  of  motion:  

θ = tan−1

vy

vx

⎛

⎝⎜

⎞

⎠⎟ = tan−1 4.6 m/s

7.9 m/s⎛⎝⎜

⎞⎠⎟= 30° above the positive x axis  

  Insight:  The  bird’s  motion  is  similar  to  projectile  motion  except  that  the  uniform  acceleration  is  in  the  horizontal  direction  and  the  constant  velocity  motion  is  along  the  vertical  direction.  

 64.     Picture  the  Problem:  The  trajectory  of  the  cork  and  its  initial  velocity  

vcg  

as  seen  by  an  observer  on  the  ground  are  depicted  at  right.  

  Strategy: Add the velocity of the balloon to the velocity of the cork as seen by an observer in the balloon to find the velocity of the cork as seen by an observer on the ground. Use the components of the initial velocity to find its magnitude and direction. Then use equation 4-10 to determine the maximum height of the cork above ground and the time of flight.

 

  Solution:  1.  (a)  Add   vbg (the  velocity  of  the    

balloon  relative  to  ground)  to   vcb (the  velocity    

of  the  cork  relative  to  the  balloon):  

vcg =vcb +

vbg

= 5.00 m/s( ) x⎡⎣ ⎤⎦ + 2.00 m/s( ) y⎡⎣ ⎤⎦vcg = 5.00 m/s( ) x + 2.00 m/s( ) y

 

cbv  

bgv  

 y0  

cgv  

Home  Work  Solutions            PHYS111.02  SFSU  Eradat                                    [WA4  CHAPTER  4]   5       2.  (b)  Use  the  components  of  

vcg    

to  find  the  speed:   vcg = 5.00 m/s( )2

+ 2.00 m/s2( ) = 5.39 m/s  

  3.  Use  the  components  of   vcg    

to  find  the  direction:   θ = tan−1

vy

vx

⎛

⎝⎜

⎞

⎠⎟ = tan−1 2.00 m/s

5.00 m/s⎛⎝⎜

⎞⎠⎟= 21.8° above horizontal

 

  4.  (c)  Use  equation  4-­‐6  to  find  the    maximum  height  of  the  cork,  noting    its  initial  height  is  6.00  m:  

vy2 = 0 = v0 y

2 − 2gΔy

Δy =v0 y

2

2g=

2.00 m/s( )2

2 9.81 m/s2( ) = 0.204 m

ymax = y0 + Δy = 6.00 + 0.204 m = 6.20 m

 

  5.  (d)  Use  equation  4-­‐6  to  find  the    time  to  reach  the  peak  of  flight:  

tup =

vy − v0 y

−g= 0 − 2.00 m/s

−9.81 m/s2 = 0.204 s  

  6.  Use  equation  4-­‐6  to  find  the  time    to  fall  from  the  peak  of  flight:  

y = y0 + v0 yt −12 gt2 ⇒ 0 = y0 + 0 − 1

2 gtdown2

tdown =2y0

g=

2 6.20 m( )9.81 m/s2 = 1.124 s

 

 7.  Add  the  times  to  find  the  total  time  of  flight:  

tflight = tup + tdown = 0.204 +1.124 s = 1.33 s  

  Insight:  Another  way  to  solve  part  (d)  is  to  use   vy

2 = v0 y2 − 2gΔy  to  find

vy ,  then  use

vy ,   v0 y  and  the  

acceleration  of  gravity  to  find  the  total  time  of  flight.    It  saves  a  step  or  two  but  doesn’t  reveal  tup  or  tdown.