# Why only 2 sig. fig.?

1. May 31, 2013

### e-zero

I have the following info for a question:

x0 = 380m
a = -9.80m/s^2
t = ?
v0 = 0

I use the formula x = x0 +v0t + (1/2)at^2

When I solve for 't' I get 8.80631

I figure I would round to 3 sig. figs. and have t = 8.81s, but my book is telling me t = 8.8 which is only 2 sig. figs.

Why are they reporting only 2 sig. figs. for this question?

2. May 31, 2013

### Staff: Mentor

I agree 8.81s looks better. Apparently authors of the book were not treating significant figures too religiously. Please remember they are only an approximated way of dealing with precision, and not a very good one.

3. May 31, 2013

### e-zero

Here is the full question in case I missed something:

Estimate how long it took King Kong to fall straight down from the top of the Empire State Building (380m high).

I assume a = -9.80m^2. The only other thing I see is that the height of the building can be seen as only 2 sig. figs. since the last digit is '0', but I would normally take that as 3 sig. figs. since it is not stated as 'approximate'.

Any input?

4. May 31, 2013

### Staff: Mentor

You haven't missed anything.

5. May 31, 2013

### MrAnchovy

In questions involving the acceleration due to gravity at the Earth's surface the key word "Estimate" usually means "assume a = 10 ms-2". So an estimate would be the square root of 76, or 8.7.

8.81 would be more appropriate if the question said "Calculate..."

6. May 31, 2013

### the_wolfman

There is an ambiguity when dealing with significant figures and different people/textbooks use different conventions.

Often 380m indicates two significant figures.
Including a decimal point after the last digit (380.m) would indicate three.

Using scientific notation removes the ambiguity
3.8 is 2 sig figs
3.80 is 3
3.8000 is 5

Many of the introductory physics books I've seen use this convention.

7. May 31, 2013