Wizard
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Wizard
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April 11th, 2020 at 9:11:58 PM permalink
I suggest we have a separate thread for the 538.com math puzzles. Here is the one for this week:

Quote: 538.com

The rows of seats in the audience are all on the same level (i.e., there is no incline), they are numbered 1, 2, 3, etc., and the T-shirts are being launched from directly in front of Row 1. Assume also that there is no air resistance (yes, I know, thatís a big assumption). You also happen to know quite a bit about the particular model of T-shirt cannon being used ó with no air resistance, it can launch T-shirts to the very back of Row 100 in the audience, but no farther.

The crew member aiming in your direction is still figuring out the angle for the launch, which you figure will be a random angle between zero degrees (straight at the unfortunate person seated in Row 1) and 90 degrees (straight up). Which row should you be sitting in to maximize your chances of nabbing the T-shirt?



Link: Can You Catch The Free T-Shirt?

Here is my answer. If some of the other math geniuses on the forum agree, I'll explain my solution.

Row 100. I figure the shirt will land there if the angle is between about 41.4 and 48.6. This represents about a 7.2% chance. It would take some integration to get the exact probability, but I'm pretty sure row 100 is right.
It's not whether you win or lose; it's whether or not you had a good bet.
ssho88
ssho88
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April 11th, 2020 at 11:42:55 PM permalink
Quote: Wizard

I suggest we have a separate thread for the 538.com math puzzles. Here is the one for this week:



Link: Can You Catch The Free T-Shirt?

Here is my answer. If some of the other math geniuses on the forum agree, I'll explain my solution.

Row 100. I figure the shirt will land there if the angle is between about 41.4 and 48.6. This represents about a 7.2% chance. It would take some integration to get the exact probability, but I'm pretty sure row 100 is right.




It all depends on the initial speed of the T-shirt and the shooting angle, and then the trajectory, Y = Xtanθ - gX^2/(2v^2)/(cosθ)^2 (it can be defined easily) . . .

When Y =0, X = 2 * v^2 * cosθ * sinθ/g = v^2 * sin2θ/g

Xmax = v^2 /g, when θ = 45 degree

If the initial speed of T-shirt is very high, definitely it will reach row 100.....

If no air resistance, then 1/2mv^2 = mgh....

AM I MISSING SOMETHING ? Please explain further. Thnaks
Last edited by: ssho88 on Apr 12, 2020
Wizard
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Wizard
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April 12th, 2020 at 9:06:01 AM permalink
Quote: ssho88

It all depends on the initial speed of the T-shirt and the shooting angle, and then the trajectory, Y = Xtanθ - gX^2/(2v^2)/(cosθ)^2 (it can be defined easily) . . .



We're told we can ignore air resistance and the maximum row the t-shirt can reach is 100. That should be enough to fill in the unknowns.
It's not whether you win or lose; it's whether or not you had a good bet.
ssho88
ssho88
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April 12th, 2020 at 10:45:42 AM permalink

Assuming each row is 1m apart, so Xmax = 100, then X = 100sin2θ, if we plot the graph, the max distance(row 100) area will cover "WIDEEST" range of angle near θ=45 degree(peak of the graph).

1) scenario 1, audience sit at row 100 will get the T-shirt if 99.5 <X<100, so 42.13 degree < θ < 45 degree, the θ range = 2.87 degree.

2) scenario 2, audience sit at row 99 will get the T-shirt if 98.5 <X<99, so 40.03 degree < θ < 40.95 degree, the θ range = 0.92 degree.

Therefore, audience sit at row 100 have more than 3 times higher chance to catch the T-shirt compare to audience sit at row 99.

So row 100 should have highest θ range compare to any other row.


EDITED.

After some thoughts, I think it can be proved this way, X = 100sin2θ, dX/dθ = 2cos2θ, for a fix value of dX, the higher the dθ, the lower the dX/dθ, but we know that the hgiher the dθ, the higher the probability to catch the T-shirt. Therefore, probability become maximum when dX/dθ is minimum, from the graph, the minimum value of dX/dθ = 0, 2cos2θ=0, θ=45 degree, X = 100 * sin90 = 100. So audience sit at row 100 have highest probability.

Last edited by: ssho88 on Apr 12, 2020
ThatDonGuy
ThatDonGuy
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April 12th, 2020 at 11:18:34 AM permalink

Assume the rows are equally spaced apart, and the cannon is at "row 0".

Let V be the velocity of the cannon, and X be the angle.
The time that the T-shirt is in the air is the solution to 0 = (V sin X) t - 1/2 g t^2
t = 2V/g sin X
The horizontal distance = t cos X = 2V/g sin X cos X = V/g sin 2X
The maximum horizontal distance is where X = 45 degrees -> sin X = cos X = sqrt(2)/2

Express distances in terms of rows:
100 = V/g sin 2X = V/g -> V = 100g
The horizontal distance = 100 sin 2X
It will land in row N if the horizontal distance is between N - 1/2 and N + 1/2
N - 1/2 < 100 sin 2X < N + 1/2
(N - 1/2) / 100 < sin 2x < (N + 1/2) / 100
arcsin ((N - 1/2) / 100) < 2X < arcsin((N + 1/2) / 100)
arcsin ((N - 1/2) / 100) / 2 < X < arcsin((N + 1/2) / 100) / 2

For what N in {1, 2, ..., 99} is (arcsin((N + 1/2) / 100) - arcsin((N + 1/2) / 100)) / 2 a maximum, and are any of them less than (arcsin 1 - arcsin 0.995)?

I needed a little help with Excel for this one, but the maximum is at N = 100, so sit in row 100.

ssho88
ssho88
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April 13th, 2020 at 1:55:51 AM permalink

Another interesting finding, audience at Row 100 have 20 times higher chance to catch the T-shirt compare to Row 1 !
Doc
Doc
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April 13th, 2020 at 9:39:08 PM permalink
No secret insights here to hide behind spoilers, just a practical consideration. If (a) the canon has a launch speed capable of reaching row #100, (b) I have a seat in row #1, and (c) the shooter aims at very nearly zero loft, then I suspect that the in-flight time until the projectile passes my row will be so short that I would have no hope of having a reaction speed capable of snagging the shirt. My only hope of being a winner would be for the shooter to aim almost vertical so I get to watch the straight-up-and-down trajectory and wait for it to land in my lap.

I think some aspect of this reaction time factor may influence the optimal seat selection and not just calculations of how far a shirt will fly.
ssho88
ssho88
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April 14th, 2020 at 4:46:51 AM permalink
We are not talking about reaction of snagging but landing infront of the audience, and equal chance for any shooting angle from 0 to 90 degrees.
charliepatrick
charliepatrick
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April 14th, 2020 at 6:02:59 AM permalink
I've assumed the shooter is on the 0 foot line and the furthest the cannon can fire is to land on the 100 foot line (I accept that chairs being 1ft apart is unlikely, but then a stadium having 100 rows is!).

As has been said the angle that shoots the furthest is 45o and the distance is K sin(2*45).

Thus if the value of sin(x) is between 0 and 0.01, seat #1 gets the t-shirt. As the derivative of sin is cos, then when the angle is small x increases fairly fast for a change in angle. Thus earlier seats have a smaller angle range that covers their seat compared to later seats. Thus earlier seats have less chance that seats "behind" them.


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Wizard
Administrator
Wizard
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April 14th, 2020 at 6:07:34 AM permalink
You have to choose your seat before the cannon is fired and must stay in that seat. In other words, no running after the cannon is fired to another seat. Reaction time should not play into it.
It's not whether you win or lose; it's whether or not you had a good bet.

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