When dropped from the same height,a body reaches the ground quicker at the poles than at the equator. Why?

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(N/A) The acceleration due to gravity $(g)$ is greater at the poles than at the equator because the Earth is flattened at the poles and bulges at the equator,making the distance from the center of the Earth to the surface smaller at the poles.
According to the equation of motion $s = ut + (1/2)gt^2$,if a body is dropped from rest $(u = 0)$,then $s = (1/2)gt^2$,which implies $t = \sqrt{2s/g}$.
Since the distance $(s)$ is the same and the acceleration due to gravity $(g)$ is higher at the poles,the time $(t)$ taken to reach the ground is smaller at the poles compared to the equator.

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