Here’s an interesting challenge you can give to a friend. Hold a $1 (or larger!) bill by an upper corner. Have a friend prepare to pinch a lower corne

arenceabigns 2021-06-10 Answered
Here’s an interesting challenge you can give to a friend. Hold a $1 (or larger!) bill by an upper corner. Have a friend prepare to pinch a lower corner, putting her fingers near but not touching the bill. Tell her to try to catch the bill when you drop it by simply closing her fingers. This seems like it should be easy, but it’s not. After she sees that you have released the bill, it will take her about 0.25 s to react and close her fingers-which is not fast enough to catch the bill. How much time does it take for the bill to fall beyond her grasp? The length of a bill is 16 cm.

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Expert Answer

ottcomn
Answered 2021-06-11 Author has 22523 answers
Step 1
In this case the bill is doing a free fall.
Let, the distance traveled be denoted by 's', the acceleration due to gravity be 'g' , time taken to fall be 't' , and the initial velocity be denoted by 'u'
Now, the given quantities:
\(s=16\) cm
\(u=0\) (as initially at rest)
\(g=9.8\ m/s^{2}\)
Apply the equation of motion to get,
\(s=ut+\left(\frac{1}{2}\right)gt^{2}\)
Step 2
put the values, to get
\(16\times10^{-2}=\frac{1}{2}\times9.8\times t^{2}\)
\(\Rightarrow t^{2}=\frac{2\times16\times10^{-2}}{9.8}\)
\(\Rightarrow t=0.18s\)
So, the time taken for the bill to fall beyond her grasp is 0.18s.
And as the average reaction time is 0.25s, so the friend would never catch it.
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content_user
Answered 2021-10-05 Author has 11086 answers

The time taken by the bill to travel throught the fingers is,

\(s=\frac{1}{2}gt^2\)

Here, g is acceleration due to gravity, t is time and s is length of the bill

\(t=\sqrt{\frac{2s}{g}}\)

\(=\sqrt{\frac{2(0.16 m)}{9.8 m/s^2}}\)

=0.18s

Which is less than the reaction time (0.25 s)  of the person. So ,she cannot catch the bill

Therefore , the time taken by the bill to fall beyond her grasp is 0.18 s

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