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Resisted Motion
Resisted Motion
Resistance is ALWAYS in the OPPOSITE direction to the motion.
Resisted Motion
Resistance is ALWAYS in the OPPOSITE direction to the motion.
                      (Newton’s 3rd Law)
Resisted Motion
Resistance is ALWAYS in the OPPOSITE direction to the motion.
                         (Newton’s 3rd Law)
Case 1 (horizontal line)
Resisted Motion
Resistance is ALWAYS in the OPPOSITE direction to the motion.
                         (Newton’s 3rd Law)
Case 1 (horizontal line)
Resisted Motion
Resistance is ALWAYS in the OPPOSITE direction to the motion.
                         (Newton’s 3rd Law)
Case 1 (horizontal line)
                m
                 x
Resisted Motion
Resistance is ALWAYS in the OPPOSITE direction to the motion.
                         (Newton’s 3rd Law)
Case 1 (horizontal line)
                m
                 x

         R
Resisted Motion
Resistance is ALWAYS in the OPPOSITE direction to the motion.
                         (Newton’s 3rd Law)
Case 1 (horizontal line)
                  mx                      m   R
                                             x

         R
Resisted Motion
Resistance is ALWAYS in the OPPOSITE direction to the motion.
                         (Newton’s 3rd Law)
Case 1 (horizontal line)
                  mx                      m   R
                                             x
                                                    R
                                               
                                             x
          R                                         m
Resisted Motion
Resistance is ALWAYS in the OPPOSITE direction to the motion.
                         (Newton’s 3rd Law)
Case 1 (horizontal line)
                  mx                      m   R
                                             x
                                                    R
                                               
                                             x
          R                                         m

 Case 2 (upwards motion)
Resisted Motion
Resistance is ALWAYS in the OPPOSITE direction to the motion.
                         (Newton’s 3rd Law)
Case 1 (horizontal line)
                  mx                      m   R
                                             x
                                                    R
                                               
                                             x
          R                                         m

 Case 2 (upwards motion)
Resisted Motion
Resistance is ALWAYS in the OPPOSITE direction to the motion.
                         (Newton’s 3rd Law)
Case 1 (horizontal line)
                  mx                      m   R
                                             x
                                                    R
                                               
                                             x
          R                                         m

 Case 2 (upwards motion)

       m
        x
Resisted Motion
Resistance is ALWAYS in the OPPOSITE direction to the motion.
                         (Newton’s 3rd Law)
Case 1 (horizontal line)
                  mx                      m   R
                                             x
                                                    R
                                               
                                             x
          R                                         m

 Case 2 (upwards motion)

       m
        x

             mg
Resisted Motion
Resistance is ALWAYS in the OPPOSITE direction to the motion.
                         (Newton’s 3rd Law)
Case 1 (horizontal line)
                  mx                      m   R
                                             x
                                                    R
                                               
                                             x
          R                                         m

 Case 2 (upwards motion)

       m
        x

             mg R
Resisted Motion
Resistance is ALWAYS in the OPPOSITE direction to the motion.
                         (Newton’s 3rd Law)
Case 1 (horizontal line)
                  mx                      m   R
                                             x
                                                    R
                                               
                                             x
          R                                         m

 Case 2 (upwards motion)
                                        m   mg  R
                                         x
       m
        x

             mg R
Resisted Motion
Resistance is ALWAYS in the OPPOSITE direction to the motion.
                         (Newton’s 3rd Law)
Case 1 (horizontal line)
                  mx                      m   R
                                             x
                                                    R
                                               
                                             x
          R                                         m

 Case 2 (upwards motion)
                                        m   mg  R
                                         x
       m                                          R
        x                                   g 
                                         x
                                                    m
             mg R
Resisted Motion
Resistance is ALWAYS in the OPPOSITE direction to the motion.
                         (Newton’s 3rd Law)
Case 1 (horizontal line)
                  mx                      m   R
                                             x
                                                    R
                                               
                                             x
          R                                         m

 Case 2 (upwards motion)
                                        m   mg  R
                                         x
       m                                          R
        x                                   g 
                                         x
                                                    m
             mg R
                              NOTE: greatest height still occurs
                                    when v = 0
Case 3 (downwards motion)
Case 3 (downwards motion)
Case 3 (downwards motion)




      m
       x
Case 3 (downwards motion)




      m
       x    mg
Case 3 (downwards motion)

            R

      m
       x    mg
Case 3 (downwards motion)
                            m  mg  R
                             x
            R

      m
       x    mg
Case 3 (downwards motion)
                            m  mg  R
                             x
            R                         R
                               g 
                             x
                                      m
      m
       x    mg
Case 3 (downwards motion)
                                      m  mg  R
                                       x
            R                                   R
                                         g 
                                       x
                                                m
      m
       x    mg
                            NOTE: terminal velocity occurs
                                  when   0
                                        x
Case 3 (downwards motion)
                                                m  mg  R
                                                 x
                R                                         R
                                                   g 
                                                 x
                                                          m
         m
          x     mg
                                    NOTE: terminal velocity occurs
                                              when   0
                                                    x
e.g. (i) A particle is projected vertically upwards with a velocity of u m/s
         in a resisting medium.
         Assuming that the retardation due to this resistance is equal to kv 2
         find expressions for the greatest height reached and the time
         taken to reach that height.
Case 3 (downwards motion)
                                                m  mg  R
                                                 x
                R                                         R
                                                   g 
                                                 x
                                                          m
         m
          x     mg
                                    NOTE: terminal velocity occurs
                                              when   0
                                                    x
e.g. (i) A particle is projected vertically upwards with a velocity of u m/s
         in a resisting medium.
         Assuming that the retardation due to this resistance is equal to kv 2
         find expressions for the greatest height reached and the time
         taken to reach that height.
Case 3 (downwards motion)
                                                m  mg  R
                                                 x
                R                                         R
                                                   g 
                                                 x
                                                          m
         m
          x     mg
                                    NOTE: terminal velocity occurs
                                              when   0
                                                    x
e.g. (i) A particle is projected vertically upwards with a velocity of u m/s
         in a resisting medium.
         Assuming that the retardation due to this resistance is equal to kv 2
         find expressions for the greatest height reached and the time
         taken to reach that height.

        m
         x
Case 3 (downwards motion)
                                                m  mg  R
                                                 x
                R                                         R
                                                   g 
                                                 x
                                                          m
         m
          x     mg
                                    NOTE: terminal velocity occurs
                                              when   0
                                                    x
e.g. (i) A particle is projected vertically upwards with a velocity of u m/s
         in a resisting medium.
         Assuming that the retardation due to this resistance is equal to kv 2
         find expressions for the greatest height reached and the time
         taken to reach that height.

        m
         x

               mg
Case 3 (downwards motion)
                                                m  mg  R
                                                 x
                R                                         R
                                                   g 
                                                 x
                                                          m
         m
          x     mg
                                    NOTE: terminal velocity occurs
                                              when   0
                                                    x
e.g. (i) A particle is projected vertically upwards with a velocity of u m/s
         in a resisting medium.
         Assuming that the retardation due to this resistance is equal to kv 2
         find expressions for the greatest height reached and the time
         taken to reach that height.

        m
         x

               mg kv 2
Case 3 (downwards motion)
                                                m  mg  R
                                                 x
                R                                         R
                                                   g 
                                                 x
                                                          m
         m
          x     mg
                                    NOTE: terminal velocity occurs
                                              when   0
                                                    x
e.g. (i) A particle is projected vertically upwards with a velocity of u m/s
         in a resisting medium.
         Assuming that the retardation due to this resistance is equal to kv 2
         find expressions for the greatest height reached and the time
         taken to reach that height.

        m
         x

               mg kv 2

                    x  0, v  u
Case 3 (downwards motion)
                                                m  mg  R
                                                 x
                R                                         R
                                                   g 
                                                 x
                                                          m
         m
          x     mg
                                    NOTE: terminal velocity occurs
                                              when   0
                                                    x
e.g. (i) A particle is projected vertically upwards with a velocity of u m/s
         in a resisting medium.
         Assuming that the retardation due to this resistance is equal to kv 2
         find expressions for the greatest height reached and the time
         taken to reach that height.

        m
         x                                m   mg  kv 2
                                           x
                                                     k 2
               mg kv 2                        g  v
                                           x
                                                     m
                    x  0, v  u
dv  mg  kv 2
v 
 dx     m
dv  mg  kv 2
v 
 dx     m
 dx    mv
    
 dv  mg  kv 2
dv  mg  kv 2
v 
 dx        m
 dx      mv
    
 dv  mg  kv 2
         o
             vdv
  x  m 
         u
           mg  kv 2
         u
      m     2kvdv
      2k  mg  kv 2
    
         0
dv  mg  kv 2
v 
 dx        m
 dx      mv
    
 dv  mg  kv 2
         o
             vdv
  x  m 
         u
           mg  kv 2
         u
      m     2kvdv
      2k  mg  kv 2
    
         0


     logmg  kv 0
     m            2 u

     2k
dv  mg  kv 2
v 
 dx        m
 dx      mv
    
 dv  mg  kv 2
         o
             vdv
  x  m 
         u
           mg  kv 2
         u
      m     2kvdv
      2k  mg  kv 2
    
         0


     logmg  kv 0
      m             2 u

     2k
     logmg  ku 2   logmg 
     m
     2k
     m    mg  ku 2 
     log           
     2k   mg 
     m    ku 2 
     log1    
     2k   mg 
dv  mg  kv 2
v 
 dx        m
 dx      mv
    
 dv  mg  kv 2
         o
             vdv
  x  m 
         u
           mg  kv 2
         u
      m     2kvdv
      2k  mg  kv 2
    
         0


     logmg  kv 0
      m            2 u

     2k
     logmg  ku 2   logmg 
     m
     2k
     m    mg  ku 2               the greatest height
     log           
     2k      mg                       m      ku 2 
                                      is log1         metres
     m    ku 2                        2k     mg 
     log1     
     2k   mg 
k 2
   g  v
x
          m
k 2
    g  v
 x
           m
dv  mg  kv 2
    
dt         m
k 2
    g  v
 x
            m
dv  mg  kv 2
    
dt          m
          0
               dv
  t  m 
          u
            mg  kv 2
        u
    m      dv
    
    k 0 mg  v 2
         k
k 2
    g  v
 x
            m
dv  mg  kv 2
    
dt          m
          0
               dv
  t  m 
          u
            mg  kv 2
        u
    m      dv
    
    k 0 mg  v 2
         k
                              u
    m k         1   k 
           tan    mg v 
                           
    k  mg                 0
k 2
    g  v
 x
            m
dv  mg  kv 2
    
dt          m
          0
               dv
  t  m 
          u
            mg  kv 2
        u
     m      dv
    
     k 0 mg  v 2
          k
                               u
     m k         1   k 
            tan    mg v 
                            
     k  mg                 0
       m  1  k               1 
         tan  mg u   tan 0
                         
       kg                        
     m     1   k 
       tan   mg u 
                     
     kg             
k 2
    g  v
 x
            m
dv  mg  kv 2
    
dt          m
          0
               dv
  t  m 
          u
            mg  kv 2
        u
     m      dv
    
     k 0 mg  v 2
          k
                               u
     m k         1   k 
            tan    mg v 
                            
     k  mg                 0
       m  1  k               1 
         tan  mg u   tan 0
                                                 m      1   k 
       kg                          it takes     tan    mg u  seconds
                                                                   
                                                  kg              
     m     1   k 
       tan   mg u 
                                       to reach the greatest height.
     kg             
(ii) A body of mass 5kg is dropped from a height at which the
     gravitational acceleration is g.
     Assuming that air resistance is proportional to speed v, the
     constant of proportion being 1 , find;
                                    8
(ii) A body of mass 5kg is dropped from a height at which the
      gravitational acceleration is g.
      Assuming that air resistance is proportional to speed v, the
      constant of proportion being 1 , find;
                                     8
a) the velocity after time t.
(ii) A body of mass 5kg is dropped from a height at which the
      gravitational acceleration is g.
      Assuming that air resistance is proportional to speed v, the
      constant of proportion being 1 , find;
                                     8
a) the velocity after time t.
(ii) A body of mass 5kg is dropped from a height at which the
      gravitational acceleration is g.
      Assuming that air resistance is proportional to speed v, the
      constant of proportion being 1 , find;
                                     8
a) the velocity after time t.




      5
       x
(ii) A body of mass 5kg is dropped from a height at which the
      gravitational acceleration is g.
      Assuming that air resistance is proportional to speed v, the
      constant of proportion being 1 , find;
                                     8
a) the velocity after time t.




      5
       x    5g
(ii) A body of mass 5kg is dropped from a height at which the
      gravitational acceleration is g.
      Assuming that air resistance is proportional to speed v, the
      constant of proportion being 1 , find;
                                     8
a) the velocity after time t.

             1
               v
             8
      5
       x    5g
(ii) A body of mass 5kg is dropped from a height at which the
      gravitational acceleration is g.
      Assuming that air resistance is proportional to speed v, the
      constant of proportion being 1 , find;
                                     8               1
a) the velocity after time t.            5  5 g  v
                                          x
                                                     8
               1                                    1
                 v                          g  v
                                          x
               8                                    40
      5
       x    5g
(ii) A body of mass 5kg is dropped from a height at which the
      gravitational acceleration is g.
      Assuming that air resistance is proportional to speed v, the
      constant of proportion being 1 , find;
                                     8               1
a) the velocity after time t.            5  5 g  v
                                          x
                                                     8
               1                                    1
                 v                          g  v
                                          x
               8                                    40
                                         dv 40 g  v
       5
         x    5g                             
                                         dt        40
(ii) A body of mass 5kg is dropped from a height at which the
      gravitational acceleration is g.
      Assuming that air resistance is proportional to speed v, the
      constant of proportion being 1 , find;
                                     8               1
a) the velocity after time t.            5  5 g  v
                                          x
                                                     8
               1                                     1
                 v                          g  v
                                          x
               8                                    40
                                         dv 40 g  v
       5
         x    5g                             
                                         dt        40
                                                   v
                                                       dv
                                            t  40 
                                                   0
                                                     40 g  v
(ii) A body of mass 5kg is dropped from a height at which the
      gravitational acceleration is g.
      Assuming that air resistance is proportional to speed v, the
      constant of proportion being 1 , find;
                                     8               1
a) the velocity after time t.            5  5 g  v
                                          x
                                                     8
               1                                     1
                 v                          g  v
                                          x
               8                                    40
                                         dv 40 g  v
       5
         x    5g                             
                                         dt        40
                                                   v
                                                       dv
                                            t  40 
                                                   0
                                                     40 g  v
                                             40log40 g  v 0
                                                                   v


                                             40log40 g  v   log40 g 
                                                     40 g 
                                             40 log          
                                                     40 g  v 
t       40 g 
    log          
40       40 g  v 
t       40 g 
         log          
     40       40 g  v 
             t
 40 g
          e 40
40 g  v
t       40 g 
         log          
     40       40 g  v 
             t
 40 g
          e 40
40 g  v
40 g  v
               t
             
          e 40
 40 g
t       40 g 
         log          
     40       40 g  v 
             t
 40 g
          e 40
40 g  v
40 g  v
               t
             
          e 40
 40 g              t
                 
40 g  v  40 ge 40
                               t
                          
       v  40 g  40 ge       40


                      
                        t
       v  40 g 1  e 40 
                         
                         
t       40 g 
           log          
       40       40 g  v 
              t
  40 g
           e 40
 40 g  v
 40 g  v
                t
              
           e 40
  40 g              t
                  
 40 g  v  40 ge 40
                                t
                           
        v  40 g  40 ge       40


                       
                         t
        v  40 g 1  e 40 
                          
                          
b) the terminal velocity
t       40 g 
           log          
       40       40 g  v 
              t
  40 g
           e 40
 40 g  v
 40 g  v
                t
              
           e 40
  40 g              t
                  
 40 g  v  40 ge 40
                                t
                           
        v  40 g  40 ge       40


                       
                         t
        v  40 g 1  e 40 
                          
                          
b) the terminal velocity
   terminal velocity occurs when   0
                                 x
t       40 g 
           log          
       40       40 g  v 
               t
   40 g
            e 40
  40 g  v
  40 g  v
                 t
               
            e 40
   40 g              t
                   
  40 g  v  40 ge 40
                                 t
                            
         v  40 g  40 ge       40


                        
                          t
         v  40 g 1  e 40 
                           
                           
b) the terminal velocity
   terminal velocity1 occurs when   0
                                  x
         i.e.0  g  v
                     40
             v  40 g
t       40 g 
           log          
       40       40 g  v 
               t
   40 g
            e 40
  40 g  v
  40 g  v
                 t
               
            e 40
   40 g              t
                   
  40 g  v  40 ge 40
                                 t
                            
         v  40 g  40 ge       40


                        
                          t
         v  40 g 1  e 40 
                                         OR
                           
b) the terminal velocity
   terminal velocity1 occurs when   0
                                  x
         i.e.0  g  v
                     40
             v  40 g
t       40 g 
           log          
       40       40 g  v 
               t
   40 g
            e 40
  40 g  v
  40 g  v
                 t
               
            e 40
   40 g              t
                   
  40 g  v  40 ge 40
                                 t
                            
         v  40 g  40 ge       40


                        
                          t
         v  40 g 1  e 40 
                                                  OR
                           
                                                                  
                                                                    t
b) the terminal velocity                   lim v  lim 40 g 1  e 40 
                                                                     
                                           t     t 
   terminal velocity1 occurs when   0
                                  x                                  
         i.e.0  g  v                            40 g
                     40
             v  40 g
t       40 g 
           log          
       40       40 g  v 
              t
  40 g
           e 40
 40 g  v
 40 g  v
                t
              
           e 40
  40 g              t
                  
 40 g  v  40 ge 40
                                t
                           
        v  40 g  40 ge       40


                       
                         t
        v  40 g 1  e 40 
                                                     OR
                          
                                                                     
                                                                       t
b) the terminal velocity                      lim v  lim 40 g 1  e 40 
                                                                        
                                              t     t 
   terminal velocity1 occurs when   0
                                  x                                     
         i.e.0  g  v                               40 g
                     40
             v  40 g                     terminal velocity is 40 g m/s
c) The distance it has fallen after time t
c) The distance it has fallen after time t

        dx              
                          t
            40 g 1  e 40 
                           
        dt                 
c) The distance it has fallen after time t

        dx              
                          t
            40 g 1  e 40 
                           
        dt                 
                   t
                           
                             t
          x  40 g  1  e 40 dt
                              
                   0          
c) The distance it has fallen after time t

        dx              
                          t
            40 g 1  e 40 
                           
        dt                 
                   t
                           
                             t
          x  40 g  1  e 40 dt
                              
                   0          
                                      t
                             
                                 t
                                     
          x  40 g t  40e     40
                                     
                                    0
c) The distance it has fallen after time t

        dx              
                          t
            40 g 1  e 40 
                           
        dt                 
                   t
                           
                             t
          x  40 g  1  e 40 dt
                              
                   0          
                                       t
                           
                               t
                                
          x  40 g t  40e  40

                               0
                          
                              t
                                    
          x  40 g t  40e  0  40
                             40

                                   
                                        t
                                   
          x  40 gt  1600 ge          40
                                             1600 g
c) The distance it has fallen after time t

        dx              
                          t
            40 g 1  e 40 
                           
        dt                 
                   t
                           
                             t
          x  40 g  1  e 40 dt
                              
                   0          
                                       t
                           
                               t
                                
          x  40 g t  40e  40                            Exercise 8C;
                               0                     2, 4, 5, 6, 8, 10, 13, 16
                          
                              t
                                    
          x  40 g t  40e  0  40
                             40

                                   
                                        t
                                   
          x  40 gt  1600 ge          40
                                             1600 g

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X2 t06 02 resisted motion (2012)

  • 2. Resisted Motion Resistance is ALWAYS in the OPPOSITE direction to the motion.
  • 3. Resisted Motion Resistance is ALWAYS in the OPPOSITE direction to the motion. (Newton’s 3rd Law)
  • 4. Resisted Motion Resistance is ALWAYS in the OPPOSITE direction to the motion. (Newton’s 3rd Law) Case 1 (horizontal line)
  • 5. Resisted Motion Resistance is ALWAYS in the OPPOSITE direction to the motion. (Newton’s 3rd Law) Case 1 (horizontal line)
  • 6. Resisted Motion Resistance is ALWAYS in the OPPOSITE direction to the motion. (Newton’s 3rd Law) Case 1 (horizontal line) m x
  • 7. Resisted Motion Resistance is ALWAYS in the OPPOSITE direction to the motion. (Newton’s 3rd Law) Case 1 (horizontal line) m x R
  • 8. Resisted Motion Resistance is ALWAYS in the OPPOSITE direction to the motion. (Newton’s 3rd Law) Case 1 (horizontal line) mx m   R x R
  • 9. Resisted Motion Resistance is ALWAYS in the OPPOSITE direction to the motion. (Newton’s 3rd Law) Case 1 (horizontal line) mx m   R x R    x R m
  • 10. Resisted Motion Resistance is ALWAYS in the OPPOSITE direction to the motion. (Newton’s 3rd Law) Case 1 (horizontal line) mx m   R x R    x R m Case 2 (upwards motion)
  • 11. Resisted Motion Resistance is ALWAYS in the OPPOSITE direction to the motion. (Newton’s 3rd Law) Case 1 (horizontal line) mx m   R x R    x R m Case 2 (upwards motion)
  • 12. Resisted Motion Resistance is ALWAYS in the OPPOSITE direction to the motion. (Newton’s 3rd Law) Case 1 (horizontal line) mx m   R x R    x R m Case 2 (upwards motion) m x
  • 13. Resisted Motion Resistance is ALWAYS in the OPPOSITE direction to the motion. (Newton’s 3rd Law) Case 1 (horizontal line) mx m   R x R    x R m Case 2 (upwards motion) m x mg
  • 14. Resisted Motion Resistance is ALWAYS in the OPPOSITE direction to the motion. (Newton’s 3rd Law) Case 1 (horizontal line) mx m   R x R    x R m Case 2 (upwards motion) m x mg R
  • 15. Resisted Motion Resistance is ALWAYS in the OPPOSITE direction to the motion. (Newton’s 3rd Law) Case 1 (horizontal line) mx m   R x R    x R m Case 2 (upwards motion) m   mg  R x m x mg R
  • 16. Resisted Motion Resistance is ALWAYS in the OPPOSITE direction to the motion. (Newton’s 3rd Law) Case 1 (horizontal line) mx m   R x R    x R m Case 2 (upwards motion) m   mg  R x m R x    g  x m mg R
  • 17. Resisted Motion Resistance is ALWAYS in the OPPOSITE direction to the motion. (Newton’s 3rd Law) Case 1 (horizontal line) mx m   R x R    x R m Case 2 (upwards motion) m   mg  R x m R x    g  x m mg R NOTE: greatest height still occurs when v = 0
  • 18. Case 3 (downwards motion)
  • 19. Case 3 (downwards motion)
  • 20. Case 3 (downwards motion) m x
  • 21. Case 3 (downwards motion) m x mg
  • 22. Case 3 (downwards motion) R m x mg
  • 23. Case 3 (downwards motion) m  mg  R x R m x mg
  • 24. Case 3 (downwards motion) m  mg  R x R R   g  x m m x mg
  • 25. Case 3 (downwards motion) m  mg  R x R R   g  x m m x mg NOTE: terminal velocity occurs when   0 x
  • 26. Case 3 (downwards motion) m  mg  R x R R   g  x m m x mg NOTE: terminal velocity occurs when   0 x e.g. (i) A particle is projected vertically upwards with a velocity of u m/s in a resisting medium. Assuming that the retardation due to this resistance is equal to kv 2 find expressions for the greatest height reached and the time taken to reach that height.
  • 27. Case 3 (downwards motion) m  mg  R x R R   g  x m m x mg NOTE: terminal velocity occurs when   0 x e.g. (i) A particle is projected vertically upwards with a velocity of u m/s in a resisting medium. Assuming that the retardation due to this resistance is equal to kv 2 find expressions for the greatest height reached and the time taken to reach that height.
  • 28. Case 3 (downwards motion) m  mg  R x R R   g  x m m x mg NOTE: terminal velocity occurs when   0 x e.g. (i) A particle is projected vertically upwards with a velocity of u m/s in a resisting medium. Assuming that the retardation due to this resistance is equal to kv 2 find expressions for the greatest height reached and the time taken to reach that height. m x
  • 29. Case 3 (downwards motion) m  mg  R x R R   g  x m m x mg NOTE: terminal velocity occurs when   0 x e.g. (i) A particle is projected vertically upwards with a velocity of u m/s in a resisting medium. Assuming that the retardation due to this resistance is equal to kv 2 find expressions for the greatest height reached and the time taken to reach that height. m x mg
  • 30. Case 3 (downwards motion) m  mg  R x R R   g  x m m x mg NOTE: terminal velocity occurs when   0 x e.g. (i) A particle is projected vertically upwards with a velocity of u m/s in a resisting medium. Assuming that the retardation due to this resistance is equal to kv 2 find expressions for the greatest height reached and the time taken to reach that height. m x mg kv 2
  • 31. Case 3 (downwards motion) m  mg  R x R R   g  x m m x mg NOTE: terminal velocity occurs when   0 x e.g. (i) A particle is projected vertically upwards with a velocity of u m/s in a resisting medium. Assuming that the retardation due to this resistance is equal to kv 2 find expressions for the greatest height reached and the time taken to reach that height. m x mg kv 2 x  0, v  u
  • 32. Case 3 (downwards motion) m  mg  R x R R   g  x m m x mg NOTE: terminal velocity occurs when   0 x e.g. (i) A particle is projected vertically upwards with a velocity of u m/s in a resisting medium. Assuming that the retardation due to this resistance is equal to kv 2 find expressions for the greatest height reached and the time taken to reach that height. m x m   mg  kv 2 x k 2 mg kv 2    g  v x m x  0, v  u
  • 33. dv  mg  kv 2 v  dx m
  • 34. dv  mg  kv 2 v  dx m dx mv  dv  mg  kv 2
  • 35. dv  mg  kv 2 v  dx m dx mv  dv  mg  kv 2 o vdv x  m  u mg  kv 2 u m 2kvdv 2k  mg  kv 2  0
  • 36. dv  mg  kv 2 v  dx m dx mv  dv  mg  kv 2 o vdv x  m  u mg  kv 2 u m 2kvdv 2k  mg  kv 2  0  logmg  kv 0 m 2 u 2k
  • 37. dv  mg  kv 2 v  dx m dx mv  dv  mg  kv 2 o vdv x  m  u mg  kv 2 u m 2kvdv 2k  mg  kv 2  0  logmg  kv 0 m 2 u 2k  logmg  ku 2   logmg  m 2k m  mg  ku 2   log  2k  mg  m  ku 2   log1   2k  mg 
  • 38. dv  mg  kv 2 v  dx m dx mv  dv  mg  kv 2 o vdv x  m  u mg  kv 2 u m 2kvdv 2k  mg  kv 2  0  logmg  kv 0 m 2 u 2k  logmg  ku 2   logmg  m 2k m  mg  ku 2   the greatest height  log  2k  mg  m  ku 2  is log1   metres m  ku 2  2k  mg   log1   2k  mg 
  • 39. k 2    g  v x m
  • 40. k 2    g  v x m dv  mg  kv 2  dt m
  • 41. k 2    g  v x m dv  mg  kv 2  dt m 0 dv t  m  u mg  kv 2 u m dv   k 0 mg  v 2 k
  • 42. k 2    g  v x m dv  mg  kv 2  dt m 0 dv t  m  u mg  kv 2 u m dv   k 0 mg  v 2 k u m k 1  k    tan   mg v   k  mg   0
  • 43. k 2    g  v x m dv  mg  kv 2  dt m 0 dv t  m  u mg  kv 2 u m dv   k 0 mg  v 2 k u m k 1  k    tan   mg v   k  mg   0 m  1  k  1   tan  mg u   tan 0  kg     m 1  k   tan   mg u   kg  
  • 44. k 2    g  v x m dv  mg  kv 2  dt m 0 dv t  m  u mg  kv 2 u m dv   k 0 mg  v 2 k u m k 1  k    tan   mg v   k  mg   0 m  1  k  1   tan  mg u   tan 0  m 1  k  kg      it takes tan   mg u  seconds  kg   m 1  k   tan   mg u   to reach the greatest height. kg  
  • 45. (ii) A body of mass 5kg is dropped from a height at which the gravitational acceleration is g. Assuming that air resistance is proportional to speed v, the constant of proportion being 1 , find; 8
  • 46. (ii) A body of mass 5kg is dropped from a height at which the gravitational acceleration is g. Assuming that air resistance is proportional to speed v, the constant of proportion being 1 , find; 8 a) the velocity after time t.
  • 47. (ii) A body of mass 5kg is dropped from a height at which the gravitational acceleration is g. Assuming that air resistance is proportional to speed v, the constant of proportion being 1 , find; 8 a) the velocity after time t.
  • 48. (ii) A body of mass 5kg is dropped from a height at which the gravitational acceleration is g. Assuming that air resistance is proportional to speed v, the constant of proportion being 1 , find; 8 a) the velocity after time t. 5 x
  • 49. (ii) A body of mass 5kg is dropped from a height at which the gravitational acceleration is g. Assuming that air resistance is proportional to speed v, the constant of proportion being 1 , find; 8 a) the velocity after time t. 5 x 5g
  • 50. (ii) A body of mass 5kg is dropped from a height at which the gravitational acceleration is g. Assuming that air resistance is proportional to speed v, the constant of proportion being 1 , find; 8 a) the velocity after time t. 1 v 8 5 x 5g
  • 51. (ii) A body of mass 5kg is dropped from a height at which the gravitational acceleration is g. Assuming that air resistance is proportional to speed v, the constant of proportion being 1 , find; 8 1 a) the velocity after time t. 5  5 g  v x 8 1 1 v   g  v x 8 40 5 x 5g
  • 52. (ii) A body of mass 5kg is dropped from a height at which the gravitational acceleration is g. Assuming that air resistance is proportional to speed v, the constant of proportion being 1 , find; 8 1 a) the velocity after time t. 5  5 g  v x 8 1 1 v   g  v x 8 40 dv 40 g  v 5 x 5g  dt 40
  • 53. (ii) A body of mass 5kg is dropped from a height at which the gravitational acceleration is g. Assuming that air resistance is proportional to speed v, the constant of proportion being 1 , find; 8 1 a) the velocity after time t. 5  5 g  v x 8 1 1 v   g  v x 8 40 dv 40 g  v 5 x 5g  dt 40 v dv t  40  0 40 g  v
  • 54. (ii) A body of mass 5kg is dropped from a height at which the gravitational acceleration is g. Assuming that air resistance is proportional to speed v, the constant of proportion being 1 , find; 8 1 a) the velocity after time t. 5  5 g  v x 8 1 1 v   g  v x 8 40 dv 40 g  v 5 x 5g  dt 40 v dv t  40  0 40 g  v  40log40 g  v 0 v  40log40 g  v   log40 g   40 g   40 log   40 g  v 
  • 55. t  40 g   log  40  40 g  v 
  • 56. t  40 g   log  40  40 g  v  t 40 g  e 40 40 g  v
  • 57. t  40 g   log  40  40 g  v  t 40 g  e 40 40 g  v 40 g  v t   e 40 40 g
  • 58. t  40 g   log  40  40 g  v  t 40 g  e 40 40 g  v 40 g  v t   e 40 40 g t  40 g  v  40 ge 40 t  v  40 g  40 ge 40    t v  40 g 1  e 40     
  • 59. t  40 g   log  40  40 g  v  t 40 g  e 40 40 g  v 40 g  v t   e 40 40 g t  40 g  v  40 ge 40 t  v  40 g  40 ge 40    t v  40 g 1  e 40      b) the terminal velocity
  • 60. t  40 g   log  40  40 g  v  t 40 g  e 40 40 g  v 40 g  v t   e 40 40 g t  40 g  v  40 ge 40 t  v  40 g  40 ge 40    t v  40 g 1  e 40      b) the terminal velocity terminal velocity occurs when   0 x
  • 61. t  40 g   log  40  40 g  v  t 40 g  e 40 40 g  v 40 g  v t   e 40 40 g t  40 g  v  40 ge 40 t  v  40 g  40 ge 40    t v  40 g 1  e 40      b) the terminal velocity terminal velocity1 occurs when   0 x i.e.0  g  v 40 v  40 g
  • 62. t  40 g   log  40  40 g  v  t 40 g  e 40 40 g  v 40 g  v t   e 40 40 g t  40 g  v  40 ge 40 t  v  40 g  40 ge 40    t v  40 g 1  e 40    OR   b) the terminal velocity terminal velocity1 occurs when   0 x i.e.0  g  v 40 v  40 g
  • 63. t  40 g   log  40  40 g  v  t 40 g  e 40 40 g  v 40 g  v t   e 40 40 g t  40 g  v  40 ge 40 t  v  40 g  40 ge 40    t v  40 g 1  e 40    OR      t b) the terminal velocity lim v  lim 40 g 1  e 40    t  t  terminal velocity1 occurs when   0 x   i.e.0  g  v  40 g 40 v  40 g
  • 64. t  40 g   log  40  40 g  v  t 40 g  e 40 40 g  v 40 g  v t   e 40 40 g t  40 g  v  40 ge 40 t  v  40 g  40 ge 40    t v  40 g 1  e 40    OR      t b) the terminal velocity lim v  lim 40 g 1  e 40    t  t  terminal velocity1 occurs when   0 x   i.e.0  g  v  40 g 40 v  40 g  terminal velocity is 40 g m/s
  • 65. c) The distance it has fallen after time t
  • 66. c) The distance it has fallen after time t dx    t  40 g 1  e 40    dt  
  • 67. c) The distance it has fallen after time t dx    t  40 g 1  e 40    dt   t    t x  40 g  1  e 40 dt   0 
  • 68. c) The distance it has fallen after time t dx    t  40 g 1  e 40    dt   t    t x  40 g  1  e 40 dt   0  t   t  x  40 g t  40e 40   0
  • 69. c) The distance it has fallen after time t dx    t  40 g 1  e 40    dt   t    t x  40 g  1  e 40 dt   0  t   t  x  40 g t  40e  40  0   t  x  40 g t  40e  0  40 40   t  x  40 gt  1600 ge 40  1600 g
  • 70. c) The distance it has fallen after time t dx    t  40 g 1  e 40    dt   t    t x  40 g  1  e 40 dt   0  t   t  x  40 g t  40e  40 Exercise 8C;  0 2, 4, 5, 6, 8, 10, 13, 16   t  x  40 g t  40e  0  40 40   t  x  40 gt  1600 ge 40  1600 g