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)
mx 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)
mx 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)
mx 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)
mx 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)
mx 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)
mx 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)
mx 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)
mx 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)
mx 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)
mx 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
25. Case 3 (downwards motion)
m mg R
x
R R
xg
m
m
x mg
NOTE: terminal velocity occurs
when 0
x
26. Case 3 (downwards motion)
m mg R
x
R R
xg
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
xg
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
xg
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
xg
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
xg
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
xg
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
xg
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
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
40log40 g v 0
v
40log40 g v log40 g
40 g
40 log
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
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