One-dimensional motion with drag

[cupparamen]

Hi, this is my first time posting on this forum. This is a question about how to approach questions about one-dimensional motion with drag.

A link to the question:
http://puu.sh/5fOCO.png

While attempting this question, I thought the two equations they were referring to were:
dy/dt = v
dv/dt = -g

However, this would be far too basic for 4 marks (this is a past exam).

It would be much appreciated if you could help me out with this question.

Thanks.

 

[cupparamen]

After some thinking, I got:

ma = -sign(v)*F_d - mg

So, a = -sign(v)*F_d/m - g

And of course I can sub in F_d which is given in the question. Not sure about how to get dy/dt though.

 

[ehild]

F_d is proportional to v², say F_d/m=b.

You need to determine the functions v(t) and y(t), by solving the differential equation

dv/dt = -sign(v)*b v² - g,

and then integrating v(t) = dy/dt to get y(t).


ehild