Need Physics help with velocity, acceleration, etc..?

Question

A student wants to determine gravity, so he drops a ball and records the time of fall.
Data:
Distance(m): 0.10 0.50 1.0 1.7 2.0
Time(s): 0.14 0.32 0.46 0.59 0.63
If only distance and time are used, what quantities should the student graph in order to produce a linear relationship between the two quantities? Also, how can I find the acceleration using the data I already have? please help!
Any information would be helpful. If you give a website, please be specific as to where I can find help for this problem.

thanks everyone for explaining the answer, they were all helpful for me to understand the question and the answer

Answer 1

Look at the equation s = (1/2)gt^2
The(1/2)g should be constant, so you should have a linear relationship between s and t^2.
Make a chart of your data. Put your distance values in the first column. Put the time values in the second column. Make a third column for t^2. Square all your time values and put the results in the t^2 column.
Now you're going to make a graph of s versus t^2. Put s on the vertical axis and t^2 on the horizontal axis. Choose your scale carefully to make the graph as big as possible without going off the paper. Plot all the points (s,t^2) from your data table. Include (0,0) as one of your points because you know t^2 should be zero when s is zero. Use a straight-edge and draw a line that passes as close as possible to all the points.
Calculate the slope of that line, m.
m = (1/2)g
Therefore g = 2m
The value of g found in that manner is your approximate value of the acceleration of gravity based on your experimental data.

Answer 2

We have the values for distance versus time.

Distance (m): 0.10, 0.50, 1.0, 1.7, and 2.0

Time(s): 0.14, 0.32, 0.46, 0.59 and 0.63

Now find the square of the time and note them as follows.

Distance (m): 0.10, 0.50, 1.0, 1.7, and 2.0

[Time]^2 (t^2): 0.0196, 0.1024, 0.2116, 0.3481and 0.3969

Now draw a graph between t^2 versus distance.

THE GRAPH NOW WILL BE A STRIGHT LINE.

Find the slope of this straight line drawn.

The acceleration is twice this slope.
----------------------------------------------------------------------

S = .5 at^2

If S is plotted along y axis and t^2 is plotted along x axis

We have X = .5a Y

This is a straight line and its slope is 0.5 a where a is the acceleration.
------------------------------------------------------------------------------------

This is the same method we do for finding acceleration due to gravity using simple pendulum. There we draw a L -T^2 graph and find the g acceleration due to gravity.

Here also we use distance L and t^2 to find the acceleration.

------------------------------------------------------------

Answer 3

Use the equation X=.5at^2 and plug in each distance and the time it corresponds to. Then solve for a which will give you acceleration. You will get a bunch of different numbers for each pair. Then take the average

Answer 4

no longer plenty to describe. a majority of those ought to be formulae you have learnt at college or have on your textbook. a) t = d / v = one hundred / 25 = 4seconds b) a = (v - u) / t = (50 - 25) / 5 = 25 / 5 = 5ms^-2 c) s = ut + a million/2 at^2 = 25*5 + a million/2 * 5 * 5^2 = 187.5m d) d = vt = 50*5 = 250m