It takes 2.40 s for a small ball with a mass of 0.120 kg released from rest from a tall building to reach the ground. Calculate the height from which the ball is released.
If that ball had been released from the same height, but this time above the surface of the moon, how long would it have taken for the ball to hit the ground?

Use:

s = ut + \frac12 at^2

u, the initial velocity is zero since the ball was released,
t is the time that you have,
a is the acceleration due to gravity,
s is the height of the building.

On the moon, the acceleration due to gravity is about 6 times less than that of the Earth (if i remember well). Use the height you obtained in the first part and the new acceleration to find the time.

Post what you get! :)

sorry the first part I solved it.. I had the problem with thew moon thing..
the s = 28.224 m

but I tried to do it with the moon , it did not work !
I did
28.223 = 0.5/1.67*t
and I did not get the correct answer
1.67 moon a

Got it ! Sorry

I did not dp t^2 haha

:)

I give you the right to think I'm fool (in physics ), it's the only thing I'm bad at ! I'm really good in math and chemistry ! Hahaha

Lol, I'm worse in physics than in Maths and Chemistry :p

With SHM, Electricity, electromagnetism and photoelectricity, in brief, most topics not concerning kinematics and dynamics, I'm not very good at. I don't think you are a fool in physics. It's in fact good that you can spot your own mistakes :)