Dig It Out

One should know that a body executes simple harmonic motion, if it is dropped from the surface. The angular frequency $ω$ is defined as -

$ω = \sqrt{\dfrac{g}{R}}$ where $R$ is the radius of the earth.

I can derive the formula if someone asks for it.

Maximum velocity in an oscillation is $v_{max} = Aω$ where A is the amplitude of the oscillation.

Note that since the body is dropped from the surface, its amplitude of oscillation is $R$

Therefore,

$v_{max} = \sqrt{\dfrac{g}{R}}R$

$v_{max} = \sqrt{gR}$

Escape velocity is $v_{escape} = \sqrt{2gR}$

Thus,

$v_{max} = \dfrac{v_{escape}}{\sqrt{2}}$