This is not a homework question, it is from a past paper which I am using to practice. The question is shown in the image below:
I really don't know much about mechanics, so I don't even know where to start with this one. Any help much appreciated.
Thanks
Well the particle will always be subject to gravity, so you've got the $-g$ term there, since we're 'pointing upwards'.
Then we're told the particle experiences a force of magnitude $mkv$ resistive to its upward motion. From Newton's second law with constant mass $F=m\frac{dv}{dt} \Rightarrow \frac{dv}{dt} = \frac{F}{m}$ so we get the second term $-kv$.
So we've done the first bit.
Now remember that $v = \frac{dx}{dt}$ which means that to get $x$ you're going to need to integrate $\frac{dv}{dt}$ twice with respect to $t$. We have the initial condition $v(0)=v_0$ too.
The differential equation you need to solve is therefore
$$ \frac{d^2 x}{dt}=-g-k\frac{dx}{dt} $$