Answer
$v_{B} \approx2.13\frac{m}{s}$
Work Step by Step
İnitial kinetic energy of particle is:
$E_{k_{0}}=\dfrac {mv^{2}_{0}}{2}=\dfrac {0.9\times 7^{2}}{2}=22.05J$
Potential Energy difference between point A and B is
$\Delta U=U_{B}-U_{A}=35-15=20J$
$20 J < E_{k_{0}}$ so particle can reach $x=0$ and the kinetic energy at that time when it reaches $x=0$ will be:
$E_{k_{0}}+U_{A}=E_{k}+U_{B}\Rightarrow E_{k}=U_{A}+E_{k_{0}}-U_{B}=15+22.05-35=2.05J$
Then,
$\dfrac {mv^{2}_{B}}{2}=E_{k}\Rightarrow v_{B}=\sqrt {\dfrac {2E_{k}}{m}}=\sqrt {\dfrac {2\times 2.05}{0.9}}\approx 2.13\dfrac {m}{s}$
