# Dollar bill Free fall problem

1. Feb 6, 2014

### kenji1992

1. The problem statement, all variables and given/known data
Your friend makes a wager with you. He holds a $50 bill between your thumb and finger, and says you can keep the$100 bill if you can catch it when it drops. The bill is 16cm long. Using your reaction time do you catch the bill?

2. Relevant equations

3. The attempt at a solution
Rxn time: 0.20 s
I assume acceleration and velocity are relevant, but I don't know how to solve it. We solved a similar example in class, but I can't remember.

2. Feb 6, 2014

### grzz

The approximate distance fallen and the acceleration due to gravity are both known. Hence one can calculate an approximate value for the reaction time and then a decision can be made if this is possible.

3. Feb 6, 2014

### kenji1992

d=16cm
a=9.8 m/s^2 (would this be negative?)
t=?

4. Feb 6, 2014

### kenji1992

Would I use this formula: y(t) = vi*t - 1/2gt^2

0.16m = 0m/s*t - 1/2*9.8 m/s^2*t^2
0.16 m=4.9 m/s^2 *t^2
0.16 m/4.9 m/s^2 = t^2
then square it?

5. Feb 6, 2014

### grzz

Note that units of distance have to be consistent.

6. Feb 6, 2014

### grzz

Yes that is the correct formula except that I would have put (1/2) instead of 1/2.

7. Feb 6, 2014

### grzz

If you square your answer you get a value for t$^{4}$ and not for the value of t.

8. Feb 6, 2014

### kenji1992

Oops, I typed the wrong formula. In my text, it shows that it is y(t)=yi +vi*t -(1/2)gt^2

What do i do about the yi?

9. Feb 6, 2014

### grzz

It is OK. Just take y$_{i}$ to be 0.

10. Feb 6, 2014

### grzz

That is the distance fallen is going to be measured from the top of the dollar bill.

11. Feb 6, 2014

### kenji1992

y(t) = vi*t - 1/2gt^2

0.16m = 0m/s*t - 1/2*9.8 m/s^2*t^2
0.16 m=4.9 m/s^2 *t^2
0.16 m/4.9 m/s^2 = t^2
t=0.18 s

so then theoretically i could not catch the dollar bill? right?

12. Feb 6, 2014

### grzz

The calculation is ok.

The final decision is for you to decide!

13. Feb 6, 2014

### tycoon515

Kinematics

To clarify, I think there are some additional conditions: your thumb and index finger are aligned with the bottom of the bill, and you cannot move your hand.
When your friend drops the bill, you must have pinched your thumb and index finger together at or before the time the top of the bill falls to their height. In other words, your reaction time must be less than or equal to the time it takes the bill to fall 16 cm.
We know that objects at or near Earth's surface experience an acceleration of ~9.8 meters per second every second toward the Earth's center of mass. Since the bill remains close to Earth's surface throughout its fall, we may assume acceleration is a constant -9.8 m/s2 as long as we define the upward direction to be positive. Therefore, we may define a function that relates the acceleration of this bill to the elapsed time, a(t) = -9.8. From this, we can antidifferentiate a(t) to find a family of functions that relate the velocity of the bill to time, v(t) = -9.8t + vi, where vi is the initial velocity of the bill. Since it is at rest before being dropped, vi = 0, and v(t) = -9.8t. Antidifferentiate v(t) to find a family of functions that relate the position of the top of the bill to time, p(t) = -4.9t2 + hi, where hi is the initial height of the top of the bill. If you let your hand be the reference point, then the hi = .16 m, and p(t) = -4.9t2 + .16. We want to know the elapsed time when the top of the bill has reached your hand, so we set p(t) = 0.
-4.9t2 + .16 = 0
t = $\sqrt{}(.16/4.9)$ ≈ .18 seconds, so with a reaction time of .2 seconds, the bill will fall through your fingers before you can react, and your friend will win the bet.

14. Feb 6, 2014

### grzz

One final difficulty to be solved.

Reading the first post one wonders how the original bill of $50 changes into one of$100 when it is falling!

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