Answer
The car will coast 260 meters up the hill before starting to roll back down.
Work Step by Step
The acceleration directed down the incline is $g~sin(\theta)$. Therefore;
$d = \frac{v^2-v_0^2}{2a}$
$d = \frac{0-(30~m/s)^2}{(2)(-9.80~m/s^2)~sin(10^{\circ})}$
$d = 260~m$
The car will coast 260 meters up the hill before starting to roll back down.
