The acceleration due to gravity on the moon has not been specified. However if it is taken to be 1/6 th of the gravity on the Earth, it would be 1/6th of 9.8m/s^2.
Using the Second equation of motion S= ut +1/2 gt^2, where u is the initial velocity and S is the distance traveled in time t, we would have
S= ½ gt^2, because u=0
Substituting g= 1/6(9.8) and S =2, 2= ½ . 1/6(9.8) t^2
t^2=24/9.8
Or, t=1.56 seconds
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