Answer
$v_f=65.3\frac{Km}{s}$
Work Step by Step
$dK.E=-dU$
$\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2=-q(V_2-V_1)$
This simplifies to:
$v_f=\sqrt{v_i^2-(\frac{2q}{m})(V_2-V_1)}$
We plug in the known values to obtain:
$v_f=\sqrt{(90\times 10^3)^2-(\frac{2\times 1.6\times 10^{-19}}{1.67\times 10^{-27}}) (-50-(-70))}=65.3\frac{Km}{s}$
