Chapter 8 - Potential Energy and Conservation of Energy - Problems and Conceptual Exercises - Page 249: 78

$4.4m$

We can find the required distance as follows: $x=v_{bottom}\sqrt{2y_{bottom}/g}$ $\implies x=\sqrt{(v_A^2+2gh)(2y_{bottom}/g)}$ We plug in the known values to obtain: $x=\sqrt{(0.54)^2+2(9.81)(3.2)(2(1.50)/9.81)}$ $x=4.4m$