The document discusses resisted motion and Newton's third law. It presents three cases of resisted motion: (1) horizontal motion, (2) upward motion, and (3) downward motion. For each case, it derives equations showing that resistance is always in the opposite direction of motion. It also provides an example of finding the greatest height and time to reach it for a particle projected vertically upwards in a resisting medium.

1. Resisted Motion

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 x  g 

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 x  g 

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
xg

m
m
x mg

25. Case 3 (downwards motion)
m  mg  R
x
R R
xg

m
m
x mg
NOTE: terminal velocity occurs
when   0
x

26. Case 3 (downwards motion)
m  mg  R
x
R R
xg

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
xg

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
xg

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
xg

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
xg

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
xg

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
xg

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

m
2k
logmg  kv 2 u0

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

m
2k
logmg  kv 2 u0
 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

m
2k
logmg  kv 2 u0
 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  mg  v 2

0
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  mg  v 2

0
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  mg  v 2

0
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  mg  v 2

0
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