- #1
Matt.D
- 25
- 0
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
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