How can I accurately measure g using freefall and a simple formula?

[Matt.D]

I’m trying to measure g using freefall. I’ve conducted the experiment and now I’m struggling with the formula. The formula we’ve been told to use is:

S = ut ½ at2

Substituting a for g (because that’s what I’m trying to find).

S = ut ½ gt2

The initial Speed is 0 therefore I can remove ut (right?) to leave:

S = ½ gt2

(right so far?)

Using my results I’ve drawn a graph that has the height of freefall (S) on the y-axis and t2 on the x axis. Working out the gradient as ΔS ∕ Δt2 = 0.40 ∕ 0.105 = 3.81ms-2

My lecturer showed me what to do next but my notes look hazy.

I have written down y = 0 + m x which I think is the same as s = ut + gradient x t2

Then underneath I have m = ½ g which I have equating to g = 2 gradient

I did try and use the directions that I’ve written down earlier, but I calculated g to be about 7, which I know to be wrong from textbooks and another experiment I conducted using a simple pendulum.

I hope the information I’ve given is enough. Any help as usual is always appreciated :)

Matt

 

[christinono]

What exactly is "u" in the formula?

 

[Sirus]

Initial velocity. He means \Delta{s}=v_{o}t+\frac{1}{2}at^{2}.

Matt, since initial velocity is zero, you are right in taking out the first term in the equation to leave

\Delta{s}=\frac{1}{2}at^{2}

You are plotting displacement as a function of the square of the elapsed time, which should be a linear function y=mx +b (in this case b=0 and x is time squared) which means the slope of the graph should be \frac{1}{2}g, due to the formula above (do you see this?). Therefore you can calculate g when you find the slope of your graph as follows:

g=2m

This is due to the \frac{1}{2} in the equation of motion.