Answer
$10\;\rm m/s$
Work Step by Step
First, we need to find the minimum speed that allows the skier to fly the horizontal 5 meters without falling into the water pool.
We know that the initial speed of this trip is horizontal. Thus, $v_{iy}=0$ m/s. And we know that her initial height is 2 m above the water pool.
So we can find the time of this trip by calculating the time she takes during her vertical trip to the ground when falling from rest.
$$y=y_i+v_{iy}t+\frac{1}{2}a_yt^2$$
thus,
$$(0)=y_i+(0)t+\frac{-1}{2}gt^2$$
Thus,
$$t=\sqrt{\dfrac{2y_i}{g}}$$
Plugging the known;
$$t=\sqrt{\dfrac{2\times 2}{9.8}}=\bf 0.639\;\rm s$$
We assume that the horizontal velocity component is constant since air resistance is assumed to be negligible.
$$v_{1}=\dfrac{\Delta x}{t}=\dfrac{5}{0.639}$$
$$v_{1}=\bf 7.83\;\rm m/s$$
This is the speed of the skier at the top of the ramp, now we need to find her speed at the bottom of the ramp while we know its height and that it is frictionless which means that the energy is conserved.
$$E_0=E_1$$
$$K_0+U_{0g}=K_1+U_{g1}$$
where $y_0=0$ at the bottom of the ramp, thus
$$\frac{1}{2}mv_{0}^2+mg(0)=\frac{1}{2}mv_{1}^2+mgy_1$$
Solving for $v_0$;
$$ \color{red}{\bf\not} mv_{0}^2 = \color{red}{\bf\not} mv_{1}^2+2 \color{red}{\bf\not} mgy_1$$
$$ v_{0} = \sqrt{v_{1}^2+2gy_1}$$
Plugging the known;
$$ v_{0} = \sqrt{ 7.83^2+2(9.8)(2)}$$
$$ v_{0} = \color{red}{\bf 10}\;\rm m/s$$
