Answer
45.4 m/s.
Work Step by Step
The skier’s mechanical energy is conserved. Take y = 0 at the bottom of the hill. The initial KE + initial PE equals the final KE + final PE.
$$\frac{1}{2}mv_i^2 + mgy_i=\frac{1}{2}mv_f^2 + mgy_f $$
$$0 + mg(105 m)=\frac{1}{2}mv_f^2 + 0$$
The mass cancels out. Solve for the final speed.
$$v_f=\sqrt{2gy_i}=45.4 m/s$$
