Hi there
I have been going through the examples in different books and I found same question being repeated in few books but they had different values for the acceleration of the particle and I am not sure which one I should follow. I will type up the question from every book
Cambridge
A particle is projected vertically upwards in a medium where the resistance is proportional to the square of the velocity. The velocity of projection is V, mass is m and initial resistance is mkV2(why is m present in the resistance??
F = ma = -mg -mkv2 and so a = -g -kv2
Terry Lee
A particle is projected vertically upwards from the ground with the initial velocity U. It experiences an air resistance which is proportional to the square of its speed
ma = -mg – kv2 (no m present with the kv2.. why?) and so a = -g -kv2/m
Fitzpatrick
A particle is projected vertically upwards with a velocity of u m/s in a revisiting medium. Assuming that the retardation sue to this resistance is equal to kv2 (why is this different form the Cambridge assumption???), find greatest height and time taken to reach it.
ma = -mg –mkv2 and so a = -g –kv2
Excel book has ma= -mg – R (where R is the resistance)
So which one should I follow ?? could it be that Terry lee’s one is just a typo.
Anyway are you supposed to work the acceleration first and then think about ma? Because I always draw a diagram and from that what I would have got what terry lee has as I do not understand why the extra m is present with the kv2
I would be glad for any help if you could make sense of all that
I have been going through the examples in different books and I found same question being repeated in few books but they had different values for the acceleration of the particle and I am not sure which one I should follow. I will type up the question from every book
Cambridge
A particle is projected vertically upwards in a medium where the resistance is proportional to the square of the velocity. The velocity of projection is V, mass is m and initial resistance is mkV2(why is m present in the resistance??
F = ma = -mg -mkv2 and so a = -g -kv2
Terry Lee
A particle is projected vertically upwards from the ground with the initial velocity U. It experiences an air resistance which is proportional to the square of its speed
ma = -mg – kv2 (no m present with the kv2.. why?) and so a = -g -kv2/m
Fitzpatrick
A particle is projected vertically upwards with a velocity of u m/s in a revisiting medium. Assuming that the retardation sue to this resistance is equal to kv2 (why is this different form the Cambridge assumption???), find greatest height and time taken to reach it.
ma = -mg –mkv2 and so a = -g –kv2
Excel book has ma= -mg – R (where R is the resistance)
So which one should I follow ?? could it be that Terry lee’s one is just a typo.
Anyway are you supposed to work the acceleration first and then think about ma? Because I always draw a diagram and from that what I would have got what terry lee has as I do not understand why the extra m is present with the kv2
I would be glad for any help if you could make sense of all that