If altitude is just few meters then the answer is T = √2h/g.
But what if an object is released on very high altitude, h >> R_earth?
‘It depends, because once you hit terminal velocity, you will stop accelerating’

Folks, since you start at h>>Re like
1,000,000 km above the ground, by
the time you hit thin 10km atmosphere
you will be almost done.
Oh, dear… I did not mean exact solution here,
just acurate to O(h/R), or O((h/R)²).
That’s usualy what they mean when they say h>>R. My fault.

If someone still want to find exact soulution,
please go here:

http://answers.yahoo.com/question/index;_ylt=AjBahyytsLKFsO_786LqkyXty6IX?qid=20070312114334AAjk9Kj

T = [ (-15 *R0) / (2 * sqrt (MG)) ] ^ 3/5

It does not even look like dimensions match here.

If altitude is just few meters then the answer is T = √2h/g.
But what if an object is released on very high altitude, h >> Rearth?

A boy launches an arrow at an initial velocity of 31.2 m/s at an angle of 22.0o above the horizontal.
How high above the projection point is the arrow after 1.18 s?

On a hot summer day in the state of Washington while kayaking, I saw several swimmers jump from a railroad bridge into the Snohomish River below. The swimmers stepped off the bridge, and I estimated that they hit the water 1.00 later.

1. How high was the bridge?
2.How fast were the swimmers moving when they hit the water?
3.What would the swimmer’s drop time be if the bridge were twice as high?