The reelvant equation is
s = ut + 0.5 at*t
(disatnce = initial velocity multiplied by time falling, plus a half of acceleration times the square of the time falling - this is standard Newtoniain physics and applies to any accelerated body - not just objects falling under gravity.)
In this case u (the initial velocity) will be zero and the acceleration will be 9.81 m/s/s downwards. This simplifies the equation to...
h = 0.5 (9.81) *t*t where h = height of drop
now since you have specified h = 9.81 as well, this simplifies to
1 = 0.5 *t*t which rearranges to t*t = 2
so the time to fall will be the square root of 2, which is indeed 1.41
Sorry if the equations don't scan well, I don't know how to code superscripts (for squared) on this board.
[edit]
oops, just realised that you asked about hitting the 'groud' so the treatment above might not apply ;-)
s = ut + 0.5 at*t
(disatnce = initial velocity multiplied by time falling, plus a half of acceleration times the square of the time falling - this is standard Newtoniain physics and applies to any accelerated body - not just objects falling under gravity.)
In this case u (the initial velocity) will be zero and the acceleration will be 9.81 m/s/s downwards. This simplifies the equation to...
h = 0.5 (9.81) *t*t where h = height of drop
now since you have specified h = 9.81 as well, this simplifies to
1 = 0.5 *t*t which rearranges to t*t = 2
so the time to fall will be the square root of 2, which is indeed 1.41
Sorry if the equations don't scan well, I don't know how to code superscripts (for squared) on this board.
[edit]
oops, just realised that you asked about hitting the 'groud' so the treatment above might not apply ;-)