A body dropped from a height h onto the floor makes elastic collision with the floor. The frequency of oscillation of its periodic motion is 

Let the time taken to reach the ground be t,

$\mathrm{h}=\frac{1}{2}{\mathrm{gt}}^2$

$\mathrm{t}=\sqrt{\frac{2\mathrm{h}}{\mathrm{g}}}$

The ball will move between the ground and height h because of this. There will be a total time duration of 2t.

$\mathrm{T}=2\mathrm{t}=2\sqrt{\frac{2\mathrm{h}}{\mathrm{g}}}$

$\mathrm{Frequency}=\frac{1}{\mathrm{time} \mathrm{period}}=\frac{1}{\mathrm{T}}=\frac{1}{2}\sqrt{\frac{\mathrm{g}}{2\mathrm{h}}}$

Hence the correct answer is $\frac{1}{2}\sqrt{\frac{\mathrm{g}}{2\mathrm{h}}}.$