Measuring Reaction Time with a Simple Device

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The discussion centers on a device for measuring reaction time using a cardboard strip and two dots. Participants debate the correct application of the formula h = 1/2gt^2 to determine the distance for a 50 ms reaction time. Initial calculations mistakenly suggest distances that are too large, prompting corrections and clarifications. The correct distance for a 50 ms reaction time is ultimately established as 1.225 cm. Accurate arithmetic and understanding of the formula are emphasized as crucial for correct results.
lgen0290
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Homework Statement


The figure below shows a simple device for measuring your reaction time. It consists of a cardboard strip marked with a scale and two large dots. A friend holds the strip vertically, with thumb and forefinger at the dot on the right in the figure below. You then position your thumb and forefinger at the other dot (on the left in the figure), being careful not to touch the strip. Your friend releases the strip, and you try to pinch it as soon as possible after you see it begin to fall. The mark at the place where you pinch the strip gives your reaction time.
(a) How far from the lower dot should you place the 50.0 ms mark?


Homework Equations





The Attempt at a Solution



I tried to use the equation h=1/2gt.
I got 24.5 cm=(.5)9.8m/s(5s)

Looking at it further, it seems that is 24.5 m, not cm, so I'm not sure. 24.5 meters is not right for sure, and neither is 24.5 cm.
 
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lgen0290 said:
[b
I tried to use the equation h=1/2gt.

It should be (1/2)gt^2.
 
But wouldn't that make the number even bigger? Is the reaction time after 50 ms really 122.5 cm? It just doesn't sound right.
 
lgen0290 said:
But wouldn't that make the number even bigger? Is the reaction time after 50 ms really 122.5 cm? It just doesn't sound right.

You are right -- it doesn't. But the value comes out to be 1.225 cm, which does. Please at least do your arithmetic correctly.
 
Thanks. You could have simply pointed out that my conversion was incorrect on the seconds.
 
Sorry. I didn't mean to be rude, but so many people take our help for granted . I'm sure you are not one of them, and your mistake was genuine. We are :cool:, I hope.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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