I wonder if both ball 2 experiment is conducted at the same time, however one is at ground level, the other is at 10km above ground level, will they arrive at the same time?

Based on the concept of Newton’s law of universal gravitation,

http://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation g = G m1.M2 / r^2

where:

* F is the magnitude of the gravitational force between the two point masses,

* G is the gravitational constant, G is approximately equal to 6.67 × 10^?11 N m^2 kg^-2

* m1 is the mass of the first point mass,

* m2 is the mass of the second point mass,

* r is the distance between the two point masses.

Let's assume m1 = mass of ball say = 1 kg for easy substitution and calculation of g.

mass of Earth = 5.9742 × 10^24 kilograms = M_2

http://www.google.com.sg/search?q=mass+of+earth&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-aradius of Earth = 6 378.1 kilometers

http://www.google.com.sg/search?hl=en&client=firefox-a&rls=org.mozilla:en-US:official&hs=vAF&pwst=1&sa=X&oi=spell&resnum=0&ct=result&cd=1&q=radius+of+earth&spell=1therefore using, g = G m1.M2 / r^2

g = (6.67 × 10^?11)(1)(5.9742× 10^24) / (6378 x 10^3)^2 = 9.80 m/s^2

at a point where R' = R + 10 km =

g' = (6.67 × 10^?11)(1)(5.9742× 10^24) / ([6378+10] x 10^3)^2 = 9.77 m/s^2

so assuming the ball is on a slope tilt of angle teta,

a = g.sin (teta) = 9.80.sin (teta)

a' g'.sin(teta) = 9.77. sin (teta) where teta is say = 30 degrees

assuming motion is under constant acceleration,

equation of motion says, s = u.t + 1/2.a.t^2 and s' =u'.t + 1/2.a'.t'^2

subs in s = s' = say 1 m of simple substitution , u = u' =0

simplified........

solving which gives t = Math.sqrt[(1)(2)/9.81.sin(30^o)] = 0.639 s & t' = 0.640 s approximately.

**in conclusion to answer your question, the answer should be roughly the same time unless you can conduct the experiment at a height above Earth where the g' is very different of the sea-level g .**understand?