Make a diagram to visualize the problem.
As you can observe, at point A the roller-coater has both mechanical energies, while at point B, it has just potential energy because it stops once reaches the 24-meters hill.
Using the law of conservation of energy, we have'
[tex]\begin{gathered} E_A=E_B \\ \text{mgh}_A+\frac{1}{2}m(v_A)^2=\text{mgh}_B \end{gathered}[/tex]We can cancel out because they are the same, and solve for v_A.
[tex]\begin{gathered} gh_A+\frac{1}{2}v^2_A=gh_B \\ \frac{1}{2}v^2_A=gh_B-gh_A \\ v^2_A=2(gh_B-gh_A) \\ v_A=\sqrt[]{2(gh_B-gh_A)} \end{gathered}[/tex]Where g = 9.8 m/s^2, h_B = 24 m, and h_A = 11 m.
[tex]\begin{gathered} v_A=\sqrt[]{2(9.8\cdot24-9.8\cdot11)}(\frac{m}{s}) \\ v_A=\sqrt[]{2(235.2-107.8)}(\frac{m}{s}) \\ v_A\approx16(\frac{m}{s}) \end{gathered}[/tex]Therefore, the minimum velocity the roller-coaster would need when going over the 11-meters hill is 16 m/s.
