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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