I'm trying to understand this. At 9.8m/sec/sec, does this mean that a falling object will travel 147 meters in 5 seconds? (in a vacuum, of course).

My results are as follows:
At 1 second it has fallen 9.8 meters.
At 2, it has gone another 19.6m, a total of 29.4m.
At 3, another 29.4, totalling 58.8
At 4, another 39.2, totalling 98.0
At 5, another 49.0, totalling 147 meters.

Is this correct?

No. The distance formula derived from calculus is distance =1/2(g)(t)². Thus, it will have fallen 1/2(9.8)25= 122.5 meters.

An acceleration of 9.8 m/sec/sec implies that the object's speed changes by 9.8 meters/sec each second. So after 5 seconds it's travelling 49m/sec. This gives it an average speed during descent of 24.5 m/sec, confirming the the distance over 5 seconds to be 122.5 meters.

Thanks gerry. But where's the flaw in my thinking? I get an average speed of 29.4 m/sec.

at 0 s, the object is travelling 0 m/s

at 1 s, the object is traveling 9.8 m/s.
--> this means that over the course of this second, it has had an average speed of 4.9 m/s, so it has travelled 4.9 m.

at 2 s, the object is travelling 19.6 m/s
--> this means that over the course of this second, it has had an average speed of 14.7 m/s, adding another 14.7 m for a total of 19.6 m.

using the same technique, you get
3 s -> 29.4 m/s, 44.1 m
4 s -> 39.2 m/s, 78.4 m
5 s -> 49 m/s, 122.5 m

basically, you took the speed at the end of the second instead of the average speed during the second. The simpler way to do it is not to figure each second individually (there's nothing special about an interval of a second instead of a tenth of a second or 5 seconds) and just take the initial speed and final speed and averaging them, like gerry did.

Totally clear now methos. Thanks. I was assuming, (rather ignorantly, I admit), that at the one second mark, the object travelled 9.8 meters. One has to account for acceleration the entire way. (Not just every second).

So let me understand; If the speed at each 1 second interval instantly increased per the formula, then we'd have the 147 meters covered, eh?

If the speed were constant at 9.8 m/s during the first second, then constant at 19.6 m/s between the first and second second, then 29.4 m/s between the second and third second, etcetera, then yes, the 147 meter distance would be correct. However, this isn't how it works, it would not be possible for an object to increase it's speed by a finite amount instantaneously, as this would require infinite force and infinite acceleration.

Yes, gerry, I know. I was being silly. Thanks for the reply.