Q15 (c) (i)

Criteria	Marks
Provides correct solution	3
Integrates to obtain an expression for $t$ in terms of $v$, or equivalent merit	2
Provides an expression for the resultant force, or equivalent merit	1

Sample answer:

$\text{Force}=-mg-kv^2$
$ma=-mg-kv^2$
$a=-g-\frac{kv^2}{m}=-g-kv^2\quad\text{given }m=1$
$\frac{dv}{dt}=-g-kv^2$
$\frac{dt}{dv}=-\frac1{g+kv^2}$
$t=-\frac1{\sqrt{gk}}\tan^{-1}\left(v\sqrt{\frac kg}\right)+c$

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

$c=\frac1{\sqrt{gk}}\tan^{-1}\left(u\sqrt{\frac kg}\right)$
$t=\frac1{\sqrt{gk}}\tan^{-1}\left(u\sqrt{\frac kg}\right)-\frac1{\sqrt{gk}}\tan^{-1}\left(v\sqrt{\frac kg}\right)$

At maximum height, $v=0$:

$t=\frac1{\sqrt{gk}}\tan^{-1}\left(u\sqrt{\frac kg}\right)\quad\text{as required.}$

(Either notation, arctan or $\tan^{-1}$, is acceptable.)