What is the time taken for an iron ball to reach the surface of the moon when it is dropped from height of 10m

The 'g' in moon is 1/6 of 'g' in earth.


{S = ut + 陆at虏 } General equation


h = ut + 陆(g/6)t虏


10 = 0 x t + 陆 (g/6)t虏


10 = (gt虏) / 12


t虏 = (10 x 12)/g


t虏 = 120 / 9.8 = 12.24


t = 鈭?2.24 = 3.5 sec


=====================What is the time taken for an iron ball to reach the surface of the moon when it is dropped from height of 10m
t = square root of [ 2 s / g]


t = square root of [ 2*10 / 1.63]


t = 3.5 s


------------------------What is the time taken for an iron ball to reach the surface of the moon when it is dropped from height of 10m
You'll need to know that the acceleration due to gravity on the moon is roughly 1.6 m/s/s. It's about 9.8 m/s/s on Earth. Now it should just be a matter of plugging your numbers into a kinematics equation.
Exactly the same as the time for an aluminum ball. Or a Teflon cube. Or a wooden cylinder.





The equivalence principle of general relativity says that the acceleration of a test mass is independent of the composition of that mass.





Now that is an important thing to know!





How long it takes for anything to fall on the moon, on the other hand, is a pretty boring question.
We dont know we havn't been there.JK