# Calculating Gravitational Acceleration

1. Sep 27, 2010

### nobodyuknow

I'm not 100% if I had worded the title correctly, but I was wondering how to calculate the acceleration of a ball dropping at height.

I have it's weight, time and height.
0.45kg
67 milliseconds
3m height

2. Sep 28, 2010

### cragar

$$d=(.5)gt^2$$

d=distance
t=time
g= acceleration due to gravity
and i'm sure you know that mass doesn't matter

3. Sep 28, 2010

### HallsofIvy

The original equation was about calculating the acceleration. For an object, like a ball, dropped at the earth's surface, the acceleration is that constant, g, approximately 9.81 m/s2, in cragar's post.

4. Sep 28, 2010

### nobodyuknow

That's what I was thinking. But it dropped from a height of 3m in 67 milliseconds. I did the above equation:

d = (0.5)gt^2
g = 9.8
t = 0.67

Therefore,

0.5 x 9.8 x 0.67 x 0.67 should equate to 3 - or am I mistaken and/or made some errors?

5. Sep 29, 2010

67ms = 0.067s not 0.67
And how did you measure the time? You should repeat the experiment several times and see how big is the error.

6. Sep 29, 2010

### nobodyuknow

It was recorded on a Digital Camera and I viewed it through Windows Movie Maker. I watched it again several times it's definitely within the 60 - 70 millisecond range.

7. Sep 30, 2010

### cjl

You mean 670-700 millisecond range.

1 millisecond = 1/1000 of a second. There's no way that it fell 3 meters in 67 milliseconds, at least not on earth.