Answer:
[tex]D.v= \± \sqrt{\frac{2E}{m}}[/tex]
Step-by-step explanation:
[tex]We\ are\ given\ that,\\Kinetic\ Energy\ possessed\ by\ an\ object=\frac{1}{2}mv^2\\Hence,\\E_k=\frac{1}{2}mv^2\\2E=2*\frac{1}{2}mv^2 [Multiplying\ both\ the\ sides\ with\ 2]\\2E=mv^2\\\frac{2E}{m}=\frac{mv^2}{m}[Dividing\ both\ the\ sides\ by\ m]\\\frac{2E}{m}=v^2\\Now,\\As\ v\ could\ be\ positive\ or\ negative\ but\ its\ square(v^2)\ would\ always\ be\ positive.\\[/tex]
[tex](\± v)^2= v^2\\Hence,\\\frac{2E}{m}= (\±v)^2 \\\sqrt{\frac{2E}{m}} = \±v\\Or,\\ v= \± \sqrt{\frac{2E}{m}}[/tex]
