Physics - how to equate acceleration

I'm completely stuck on the following problem in my self-test beginners physics book -- I've got the question and the answer, but the book gives no real explanation for how it arrived at one from the other, so I'd LOVE for someone to please explain it to me.

The question:

An object falling from rest reaches a speed of 160 ft/s (or 49 m/s) after falling 5 seconds. (Ignore air resistance.) At this time it has fallen ____ ft (or _____ m).

It says the answers are 400 ft and 122.5 m respectively, but I can't figure out *how* they arrived at those (what the formula used was, etc), or what to do next time I come up against such a problem. Can someone please walk me through it?

Hi,
Use the formula S= ut + 1/2 a t^2

Here 'S' is the distance travelled by the body
,
'u' is its initial velocity but as the object is dropped from rest, u is 0,

'a' is the acceleration of the body, here as the body is accted upon by gravity, the acceleration would be acceleration due to gravity which is 9.8 mts per sec^2

and 't' is the time taken and here it is 5 secs.

Therefore distance travelled would be = [0*5] + [1/2 * 9.8 * 5^2]
which is 122.5 m

You should know about the equation:-

2gh=v^2-u^2

u will be 0 as it is falling from rest
g is 9.8m/s (acceleration due to gravity)
h is unknown
v is 49m/s

if u put the values in the equation, u'll get
2 * 9.8h = 49*49 - 0*0
19.6h = 2401
h = 2401/ 19.6

And then ull get the answer 122.5m