Answer
4.89 m/s.
Work Step by Step
The sled’s mechanical energy is conserved. Take y = 0 at the bottom. The initial KE + initial PE equals the final KE + final PE. The sled comes to a momentary stop at the end.
$$\frac{1}{2}mv_i^2 + mgy_i=\frac{1}{2}mv_f^2 + mgy_f $$
$$\frac{1}{2}mv_i^2 + 0 = 0 + mgy_f $$
$$v_i=\sqrt{2gy_f}=4.89 m/s$$
