Consider the kinematic equation

Δy = v_{0}t + 1/2at^{2}

Assuming the ball was dropped (v_{0} = 0), we can solve for time t such that

t = √(2Δy)/(a)

From this, you can see that the time on Earth, t_{E}, and the time on Mars, t_{M}, are related such that for the same drop distance Δy on each planet,

t_{E }= √0.38 * t_{M}

Thus, if t_{E} is know, computing t_{M} becomes trivial.