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?
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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

radius of Earth = 6 378.1 kilometers

therefore 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.

[color=blue][b]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 .:)[/b][/color]

understand?