Setting aside what you were told to do, for the moment, what would be a more appropriate way to plot velocity against time? If you do that, do you see the same result as for D v. We know that your method of determining v at a given time is wrong, and it may be that this is somehow compensating for an error in the data, giving the appearance of greater accuracy. I only know the relative times, and perhaps even with the correction that we add in, it's just much more uncertain? In which case, squaring it will enhance the error on it? But I feel like i know the distances much better than I know the velocities. But that would make me think that plotting distance over time would give a better estimate, except my distance versus time^2 plot gives me a value that isn't as close to the known value of g than v versus t. Hmmm, well I know that the actual velocity at each point won't be the average over a segment. But i don't see this curve in the other plot, is this again, just because the t^2 is magnifying the errors? I also notice that in my D vs t^2 plot, there is a slight curve at the beginning, i'm guessing this is because the magnet when it's closer at the beginning of the drop. Is it accurate to say that the first method using v vs t is probably more reliable because, maybe errors would get propagated through the t^2 in the second method, but not for the first method since it's just t?
My question is, I'm supposed to think of which method seems better. The second method we plot Distance versus t^2 since D(t) =. The first method, we make a plot of velocity versus time, because V = V0 + gt, so the slope of the line will be g. And then we use the distances of the marks on the tape to calculate gravity.
#Acceleration due to gravity lab answers falling object free
For two of the methods, we have an object attached to a magnet, then the magnet is turned off so the object is in free fall and it makes a spark/mark every 1/60th of a second on a piece of tape. By so doing, we will be able to better focus on the conceptual nature of physics without too much of a sacrifice in numerical accuracy.So I did a lab where we are calculating the acceleration due to gravity using a variety of methods. We will occasionally use the approximated value of 10 m/s/s in The Physics Classroom Tutorial in order to reduce the complexity of the many mathematical tasks that we will perform with this number. In doing thislab, you will become more familiar with the effects of gravity on falling objects. It has the same value, irrespective of the mass of the object. There are slight variations in this numerical value (to the second decimal place) that are dependent primarily upon on altitude. These questions and others about gravity have yet to be answered. The acceleration due to gravity experienced by any object that is falling freely close to the Earths surface is a constant. The numerical value for the acceleration of gravity is most accurately known as 9.8 m/s/s. Acceleration was found by finding the slope of the velocity time graph. A matter of fact, this quantity known as the acceleration of gravity is such an important quantity that physicists have a special symbol to denote it - the symbol g.
It is known as the acceleration of gravity - the acceleration for any object moving under the sole influence of gravity. This numerical value for the acceleration of a free-falling object is such an important value that it is given a special name.
A free-falling object has an acceleration of 9.8 m/s/s, downward (on Earth). It was learned in the previous part of this lesson that a free-falling object is an object that is falling under the sole influence of gravity.