Q15 (c) (ii)

Criteria	Marks
Provides correct solution	3
Integrates to obtain an expression for $x$ in terms of $v$, or equivalent merit	2
Provides an integral expression for $x$ in terms of $v$, or equivalent merit	1

Sample answer:

$v\frac{dv}{dx}=-g-kv^2$
$\frac{dx}{dv}=-\frac{v}{g+kv^2}$
$x=-\frac1{2k}\ln(g+kv^2)+c$

When $x=0$, $v=u$:

$c=\frac1{2k}\ln(g+ku^2)$
$x=\frac1{2k}\left(\ln(g+ku^2)-\ln(g+kv^2)\right)=\frac1{2k}\ln\left(\frac{g+ku^2}{g+kv^2}\right)$

At maximum height, substitute $v=0$:

$\therefore\text{maximum height }x=\frac1{2k}\ln\left(\frac{g+ku^2}{g}\right)$