![]() 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. 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. The velocity of the rock on its way down from y = 0 y = 0 is the same whether we have thrown it up or down to start with, as long as the speed with which it was initially thrown is the same.So I did a lab where we are calculating the acceleration due to gravity using a variety of methods. 10 m to be the same whether we have thrown it upwards at + 13. We would then expect its velocity at a position of y = − 5. That is, it has the same speed on its way down as on its way up. When its position is y = 0 y = 0 on its way back down, its velocity is − 13. Note that at the same distance below the point of release, the rock has the same velocity in both cases.Īnother way to look at it is this: In Example 2.14, the rock is thrown up with an initial velocity of 13. (b) A person throws a rock straight down from a cliff with the same initial speed as before, as in Example 2.15. The arrows are velocity vectors at 0, 1.00, 2.00, and 3.00 s. Astronauts training in the famous Vomit Comet, for example, experience free-fall while arcing up as well as down, as we will discuss in more detail later.įigure 2.42 (a) A person throws a rock straight up, as explored in Example 2.14. ![]() ![]() Both have the same acceleration-the acceleration due to gravity, which remains constant the entire time. Finally, note that free-fall applies to upward motion as well as downward. 67 m/s 2 size 12 are the positions (or displacements) of the rock, not the total distances traveled. Scott demonstrated on the Moon in 1971, where the acceleration due to gravity is only 1. This is a general characteristic of gravity not unique to Earth, as astronaut David R. Figure 2.38 A hammer and a feather will fall with the same constant acceleration if air resistance is considered negligible.
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