http://www.phy.duke.edu/~rgb/Class/review53/node10.html
In addition to the question in the link, please colculate:
same calculation but assuming moon is sphere radius 1738 km
gravitational constant 4.90287e3. Thanks

a) Since the initial horizontal velocity zero, vertical motion is 
equivalen to a free fall on the moon. 
H=1/2 * a_m * t_f^2
t_f = squareroot (2*H/a_m) = 29.97 sec

b) R= v_o * t_f = 3596.41 meters

c) total velocity v_f = squareroot(v_o^2 +v_vertical^2) 
              and v_vertical = a_m * t_f = 50.05 m/sec
                  v_f = 130.02 m/sec

___________________________________________________

As to the answer to your last question, I assume the unit of the
gravitational constant you gave is meter/sec^2, and very very high
value.

mountain top is only 750 meter high, only 0.75km/1783 =0.00043 less
than one thousandth of the radius of the moon which is negligible, so
we can disregard the curvature of the moon, and can accept the
gravitational constant unchanged during the motion.
so 
instead od a_m=1.67 m/sec^2  we will assume a_m=4.90287e3 = 4902.87 m/sec^2
a)t_f= 0.55 sec
b)R= 66.37 meters
c)v_f= 2699.25 m/sec