# Acceleration of a falling Ball

1. Nov 27, 2008

### ahrog

1. The problem statement, all variables and given/known data
A flash photograph of two tennis balls released simultaneously is taken. One is projected horizontally, and the other one is dropped from rest. The light flashes occur ever 0.2 seconds, and each box represents a distance of 0.4 s (they gave a graph...I can't get it on the computer, but I can get all the other details)

For the vertical ball:
Time (s) l vertical distance (m)
0 l 0
0.2 l 0.1
0.4 l 0.7
0.6 l 1.5
0.8 l 2.6
1.0 l 4.0

What is the vertical acceleration of the tennis ball?

2. Relevant equations
y= vyt + 1/2gt2
(<an equation my textbook gave me)
v=u+at (<this is an equation someone on this forum gave me)
possibly a=(vf-vi)/t

3. The attempt at a solution
I'm not quite sure how to go about this. Any tips would be appreciated! I was thinking that maybe I have to find the velocity using the first equation, then put it into the acceleration formula. If I did that, I would end up with -0.9m/s2.

Is this right?

2. Nov 28, 2008

### Mentallic

Use the formula $$s=ut+\frac{1}{2}at^2$$

thus $$a=\frac{2(s-ut)}{t^2}$$

where:
s = displacement (vertical position)
u = initial velocity
t = time
a = acceleration (due to gravity)

Using the data for the 1 second flash (longest time from drop) would most likely give you the closest real answer to the gravity. Testing for every interval of data will increase your reliability.

3. Nov 28, 2008

### ahrog

So, I would end up with the answer being 8 m/s, right? C: