
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!

Measuring g in freefall

  1. Jan 24, 2005 #1
    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 text books 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 :)

  2. jcsd
  3. Jan 24, 2005 #2
    What exactly is "u" in the formula?
  4. Jan 24, 2005 #3
    Initial velocity. He means [tex]\Delta{s}=v_{o}t+\frac{1}{2}at^{2}[/tex].

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


    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 [itex]\frac{1}{2}g[/itex], due to the formula above (do you see this?). Therefore you can calculate g when you find the slope of your graph as follows:


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

Similar Discussions: Measuring g in freefall
  1. Freefall Problems. (Replies: 5)

  2. Freefall acceleration (Replies: 1)