Hi there!
We can use the work-energy theorem to solve.
Recall:
[tex]\large\boxed{E_i = E_f}}[/tex]
Initial energy = final energy
The initial energy is purely potential (she starts from rest), and, if we assign the bottom of the slope as the 0 line, her energy at the bottom is purely kinetic.
PE = mgh
KE = 1/2mv²
We can begin by setting the two equal:
mgh = 1/2mv²
Cancel out the mass and rearrange to solve for velocity:
2gh = v²
v = √2gh
Plug in given values and use g ≈ 10 m/s²:
v = √2(10)(20) = 20 m/s
