Recall that
[tex]{v_f}^2-{v_i}^2=2a\Delta x[/tex]
where [tex]v_i[/tex] is the luger's initial velocity, [tex]v_f[/tex] his final velocity, [tex]a[/tex] his acceleration, and [tex]\Delta x[/tex] the distance he traverses. Then
[tex]{v_f}^2-\left(25\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(9.5\,\dfrac{\mathrm m}{\mathrm s^2}\right)(10\,\mathrm m)[/tex]
[tex]\implies v_f\approx29\,\dfrac{\mathrm m}{\mathrm s}[/tex]
so the closest answer is C.