1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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 :)

    Matt
     
  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

    [tex]\Delta{s}=\frac{1}{2}at^{2}[/tex]

    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:

    [tex]g=2m[/tex]

    This is due to the [itex]\frac{1}{2}[/itex] in the equation of motion.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



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

  2. Freefall acceleration (Replies: 1)

Loading...