In a small vacuum tube, electrons and protons start from rest and fall through 1.8 kV…

Step 1. The energy. $K = eV = 1800\ \text{eV} = 1800 \times 1.6 \times 10^{-19}\ \text{J} = 2.88 \times 10^{-16}\ \text{J}$.

Step 2. The electron. From $\tfrac{1}{2}mv^2 = K$ we want $v$, so $v = \sqrt{\dfrac{2K}{m}} = \sqrt{\dfrac{2 \times 2.88 \times 10^{-16}\ \text{J}}{9.1 \times 10^{-31}\ \text{kg}}} = 2.52 \times 10^{7}\ \text{m/s}$.

Step 3. The proton. $v = \sqrt{\dfrac{2 \times 2.88 \times 10^{-16}\ \text{J}}{1.67 \times 10^{-27}\ \text{kg}}} = 5.87 \times 10^{5}\ \text{m/s}$.