Answer
$v = 56~m/s$
Work Step by Step
The initial kinetic energy is equal to the gravitational potential energy at the highest point on the slope.
We can find the initial speed:
$K_1 = U_2$
$\frac{1}{2}mv^2 = mgh$
$\frac{1}{2}v^2 = gh$
$v^2 = 2gh$
$v = \sqrt{2gh}$
$v = \sqrt{2gL~sin~\theta}$
$v = \sqrt{(2)(9.8~m/s^2)(920~m)~(sin~10^{\circ})}$
$v = 56~m/s$
