Answer
$\theta = 12^{\circ}$
Work Step by Step
We can use kinematics to find the acceleration:
$x = v_0t + \frac{1}{2}at^2$
$a = \frac{2(x - v_0t)}{t^2} = \frac{2(18 ~m - (2.0 ~m/s)(3.3 ~s))}{(3.3 ~s)^2} = 2.1~m/s^2$
We can use the acceleration to find the angle:
$F = ma$
$mg ~sin(\theta) = ma$
$sin(\theta) = \frac{a}{g}$
$\theta = sin^{-1}(\frac{a}{g}) = sin^{-1}(\frac{2.1 ~m/s^2}{9.80 ~m/s^2}) = 12^{\circ}$
The incline has an angle of $12^{\circ}$ above the horizontal.
