ssho88 Joined: Oct 16, 2011
• Posts: 422
April 14th, 2020 at 6:46:16 AM permalink
Shooting angle to reach seat 100 is 45 degree while shooting angle to reach seat 99, X = 100sin2θ, sin2θ = 99/100, θ =40.945.

dθ = 45 - 40.945 = 4.055, prob of seat 100 = 4.055/45 *100 = 9.01%.
ThatDonGuy Joined: Jun 22, 2011
• Posts: 4020
April 15th, 2020 at 7:53:28 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 = 2V2/g sin X cos X = V2/g sin 2X
Let y = 2X; the first derivative of V2/g sin y = V2/g cos y, which, for 0 < y < 180 (since 0 < X < 90), is 0 at y = 90 ->X = 45
Note that this is a maximum point, so the maximum horizontal distance = 100 = V2/g sin (2 * 45) = V2/g

For angle X, the horizontal distance = V2/g sin 2X = 100 sin 2X
Assume the T-shirt lands in row N if the distance is between N - 1/2 and N + 1/2 for N in {1, 2, 3, ..., 99}, and the distance is between 99.5 and 100 for N = 100
For N <= 99:
N - 1/2 < 100 sin 2X < N + 1/2
(N - 1/2) / 100 < sin 2X < (N + 1/2)
Since arcsin > 0 for all 2X in the possible values, taking the arcsin of all three values does not change the inequality signs:
arcsin ((N - 1/2) / 100) / 2 < X < arcsin((N + 1/2) / 100) / 2
The range of X for a given N is (arcsin((N + 1/2) / 100) - arcsin ((N - 1/2) / 100)) / 2

The first derivative of arcsin x is 1 / sqrt(1 - x2) dx
Let f(N) = (arcsin((N + 1/2) / 100) - arcsin ((N - 1/2) / 100)) / 2
f'(N) = (1 / 100 * 1 / sqrt(1 - (N + 1/2)2 / 10,000) - 1 / 100 * 1 / sqrt(1 - (N - 1/2)2 / 10,000)) / 2
= (100 / sqrt(10,000 - (N + 1/2)2) - 100 / sqrt(10,000 - (N - 1/2)2)) / 200
= (sqrt(10,000 - (N + 1/2)2) - sqrt(10,000 - (N - 1/2)2)) / 2
This is strictly increasing as N increases, so for rows 1-99, the largest range is at N = 99

The question becomes, which is larger - the angle range for N = 99, or for N = 100?
For N = 99, it is (arcsin 199/200 - arcsin 197/200) / 2
For N = 100, it is 2 (arcsin 1 - arcsin 199/200) / 2 = arcsin 1 - arcsin 199/200
(it is multiplied by 2 to cover the ranges < 45 degrees and > 45 degrees)
I can't find a way to determine this off the top of my head other than calculating the values directly
N = 100 is 90 - 84.268 = 5.732 degrees
N = 99 is (84.268 - 80.064) / 2 = 2.102 degrees
Thus, N = 100 is the most likely result

TomG Joined: Sep 26, 2010
• Posts: 2018
April 15th, 2020 at 8:55:27 AM permalink
Shouldn't there be three variables?: speed of the shirt when it leaves the cannon, angle of the cannon, and height of the cannon. A 5' gymnast holding it at her hip is going to shoot the t-shirt a different distance than 7' center holding it over his head, even with the same launch angle. We can assume it's from a fixed height, but wouldn't you need to know what that height is? Or is my intersection of physics and calculus all messed up?
ssho88 Joined: Oct 16, 2011
• Posts: 422
April 15th, 2020 at 8:58:56 AM permalink
Quote: TomG

Shouldn't there be three variables?: speed of the shirt when it leaves the cannon, angle of the cannon, and height of the cannon. A 5' gymnast holding it at her hip is going to shoot the t-shirt a different distance than 7' center holding it over his head, even with the same launch angle. We can assume it's from a fixed height, but wouldn't you need to know what that height is? Or is my intersection of physics and calculus all messed up?

Height of cannon = height of seats.
rsactuary Joined: Sep 6, 2014
• Posts: 1582
May 22nd, 2020 at 5:41:57 PM permalink
Official write up of the T-Shirt problem.. (new problem down below): You were seated in an audience, when T-shirts were being launched via cannon in your direction. The rows of seats in the audience were all on the same level, they were numbered 1, 2, 3, etc., and the T-shirts were being launched from directly in front of Row 1.

Additionally, there was no air resistance, and the particular model of T-shirt cannon being used was able to launch T-shirts to the very back of Row 100 in the audience, but no farther.

If the T-shirt was launched at a random angle between zero degrees (straight at the unfortunate person seated in Row 1) and 90 degrees (straight up), which row should you have been sitting in to maximize your chances of nabbing the T-shirt?

While some solvers took an analytical approach, there was no shortage of simulations. For example, if you were to launch a T-shirt at whole number values of degrees between zero and 90, and you recorded which rows they landed in, here�s what you�d find: (sorry, graph didn't copy over)

Each blue dot represents a T-shirt fired at a different angle, and the red bars show how many T-shirts landed in each row. (Some rows didn�t get any T-shirts because this simulation only used whole number values of the angles.)

Already, it appears that the last few rows received more T-shirts than the others. This result was confirmed by Eli Luberoff, Jason Shaw, Ravi Chandrasekaran and Angelos Tzelepis, the last of whom sampled the angles from zero to 90 degrees at 0.01-degree steps. Every single one of them found that Row 100 had the greatest chance of receiving a T-shirt.

So what�s going on here? As solver J. D. Roaden explained, it helped If we looked at a graph of launch angle θ versus row number R, which physics tells us are related by the equation R = 100·sin(2θ), shown by the blue curve in the graph below:
range of angles that will land in Row 100

Meanwhile, the vertical red bar shows the range in angles that will launch a T-shirt to Row 100. Because the blue curve is flattest at the top, that means larger variations in angle have relatively little effect on row number. And so sure enough, Row 100 corresponds to the widest swath of angles. Solver Jonah Majumder specifically found that the chances of nabbing the T-shirt in Row 100 were just over 9 percent. Not bad!

Finally, as an aside, this phenomenon (whereby the maximum possible value is more likely because that�s where it�s changing the least), also explains why rainbows exist. 🌈Seriously! 🌈

Here is this week's puzzle:

To share a cylindrical muffin equally with his two toddlers, Robert makes three vertical cuts in a �Y� pattern, producing three equal pieces.

The next morning, his wife wants in on the fun. But before he can cut the muffin into quarters with an �X� pattern, one of his children suggests using an �A� pattern. If Robert were to produce equal fourths in this manner, what would be the ratio of length of the A�s middle bar to the radius of the muffin?

I noodled around and got pi/2, but I could easily be wrong.
Last edited by: rsactuary on May 22, 2020
ssho88 Joined: Oct 16, 2011
• Posts: 422
May 22nd, 2020 at 7:17:43 PM permalink
Here is this week's puzzle:

To share a cylindrical muffin equally with his two toddlers, Robert makes three vertical cuts in a �Y� pattern, producing three equal pieces.

The next morning, his wife wants in on the fun. But before he can cut the muffin into quarters with an �X� pattern, one of his children suggests using an �A� pattern. If Robert were to produce equal fourths in this manner, what would be the ratio of length of the A�s middle bar to the radius of the muffin?

I noodled around and got pi/2, but I could easily be wrong.

sin(2θ) = (1/2 - θ/90) *pi, where θ is in degree -------Eq1

ratio of middle bar length to the radius = (pi * tanθ)^0.5 ------Eq2

solve Eq1 to find θ, then replace it into Eq2, my estimate answer is 1.178
ChesterDog Joined: Jul 26, 2010
• Posts: 834
May 22nd, 2020 at 7:21:51 PM permalink
Quote: rsactuary

...Here is this week's puzzle:

To share a cylindrical muffin equally with his two toddlers, Robert makes three vertical cuts in a �Y� pattern, producing three equal pieces.

The next morning, his wife wants in on the fun. But before he can cut the muffin into quarters with an �X� pattern, one of his children suggests using an �A� pattern. If Robert were to produce equal fourths in this manner, what would be the ratio of length of the A�s middle bar to the radius of the muffin?

I noodled around and got pi/2, but I could easily be wrong.

I get approximately pi / 2.667197. This agrees with ssho88.
ssho88 Joined: Oct 16, 2011
• Posts: 422
May 22nd, 2020 at 7:34:49 PM permalink
Quote: ChesterDog

I get approximately pi / 2.667197. This agrees with ssho88.

Could you(or anyone) please teach me how to solve the below equation ?

sin(2θ) = (1/2 - θ/90) *pi, where θ is in degrees
ChesterDog Joined: Jul 26, 2010
• Posts: 834
May 22nd, 2020 at 8:21:10 PM permalink
Quote: ssho88

...Could you(or anyone) please teach me how to solve the below equation ?...

The page says, "...most equations where the variable appears both as an argument to a transcendental function and elsewhere in the equation are not solvable in closed form..."

When I put your equation into WolframAlpha, it gives a numerical solution, so probably there is no closed-form solution.

By the way, how did you make the symbol for "theta" in your post?
ssho88 Joined: Oct 16, 2011
• Posts: 422
May 22nd, 2020 at 9:25:15 PM permalink
Quote: ChesterDog

The page says, "...most equations where the variable appears both as an argument to a transcendental function and elsewhere in the equation are not solvable in closed form..."

When I put your equation into WolframAlpha, it gives a numerical solution, so probably there is no closed-form solution.

By the way, how did you make the symbol for "theta" in your post?

Thanks

symbol for "theta" :-