# One-dimensional motion with drag

1. Nov 11, 2013

### 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.

http://puu.sh/5fOCO.png [Broken]

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.

Last edited by a moderator: May 6, 2017
2. Nov 11, 2013

### 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.

3. Nov 12, 2013

### ehild

Fd is proportional to v2, say Fd/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