= Solution
The required constant deceleration follows from $v^2=u^2+2as$:
$$
a=-\frac{4.8^2}{2(2)}=-5.76\,\mathrm{m\,s^{-2}}.
$$
Both gravity and sliding <friction> oppose ascent, so $g(\sin\theta+0.1\cos\theta)=5.76$. Write
$$
\sin\theta+0.1\cos\theta=\sqrt{1.01}\sin(\theta+\arctan0.1).
$$
The physical incline solution is \b[$\theta=30.0^\circ$] more precisely $30.0389^\circ$. The combined <mass> cancels from the equation. The top of the track is consequently $H=2\sin\theta=1.00117\,\mathrm m$ above its base.
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