## Problem 317

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are requested to be thoughtful in not posting anything

that might explicitly give away how to solve a particular problem.

This forum is NOT meant to discuss solution methods for a problem.

In particular don't post any code fragments or results.

Don't start begging others to give partial answers to problems

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Don't start begging others to give partial answers to problems

Don't ask for hints how to solve a problem

Don't start a new topic for a problem if there already exists one

See also the topics:

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Comments, questions and clarifications about PE problems.

### Re: Problem 317

Could someone please confirm or deny that the answer lies in this range: <snipped by hk> ?

### Re: Problem 317

So sorry for posting the number, I saw others post incorrect results here so I though it would be ok to ask if I was near the answer at all. But since it was quickly removed by hk, I figured I was on the right track

### Re: Problem 317

Does it actually make sense to give g rounded to two decimal places but require the result with four decimal places? I learned in Physics that one should never have more precision in the result than in the given quantities.

EDIT: And the problem obviously does not mean a "velocity" of 20m/s, but a "speed" (velocity would be a vector, and the vectors of the speeds are definitely not the same).

EDIT: And the problem obviously does not mean a "velocity" of 20m/s, but a "speed" (velocity would be a vector, and the vectors of the speeds are definitely not the same).

### Re: Problem 317

Constants given in math problems are not measured. They are assumed to be exactly true, with perfect precision. Using short ones saves typing. If a constant is not exact, the problem will explicitly say so.oern wrote:Does it actually make sense to give g rounded to two decimal places but require the result with four decimal places? I learned in Physics that one should never have more precision in the result than in the given quantities.

### Re: Problem 317

What approximate value of $\pi$ should be used?

The friend key is 1004797_CkgowZqfFi0qQMGOqVjB0RP8htEhkWW2

### Re: Problem 317

You need 4 decimals in your solution. - So if you append the next digit of pi, the solution should not change at all.

Hint:

If you calculate using float / double: Use always the maximum possible in this format.

### Re: Problem 317

I submit many answers but they were not accepted.

So I did approximation of covered area as cylinder with height $h=Ymax=120m$ and radius $r=X_{max}=100m$

$Volume_{cylinder}=\pi*r^2*h$ Volume of cylinder. So the answer can not exceed $3,769,911.1843 m^3$

Could someone PM me to verify my approach to solve the problem?

Thanks,

Alex.

So I did approximation of covered area as cylinder with height $h=Ymax=120m$ and radius $r=X_{max}=100m$

$Volume_{cylinder}=\pi*r^2*h$ Volume of cylinder. So the answer can not exceed $3,769,911.1843 m^3$

Could someone PM me to verify my approach to solve the problem?

Thanks,

Alex.

The friend key is 1004797_CkgowZqfFi0qQMGOqVjB0RP8htEhkWW2

### Re: Problem 317

Your max. height is not correct (it is a little bit higher)

### Re: Problem 317

formula $h = Y_{max}=H+\frac{V_0^2}{2g}$

exact result is $120.3874m$

exact result is $120.3874m$

The friend key is 1004797_CkgowZqfFi0qQMGOqVjB0RP8htEhkWW2

### Re: Problem 317

This is my max. height too.

But (I think this is not to much hint) the problem did not describe some kind of cylinder - It's much less than a cylinder

But (I think this is not to much hint) the problem did not describe some kind of cylinder - It's much less than a cylinder

### Re: Problem 317

OK. Cylinder is very rough maximum approximation. The real volume is much less than the volume of the cylinder. I agree with you, v6ph1.

My results are still not accepted

What angle increment $1^o$ or $0.5^o$ should I use in my calculations?

This is the important question because an answer to it gives volume difference in $79.4676m^3$.

My results are still not accepted

What angle increment $1^o$ or $0.5^o$ should I use in my calculations?

This is the important question because an answer to it gives volume difference in $79.4676m^3$.

The friend key is 1004797_CkgowZqfFi0qQMGOqVjB0RP8htEhkWW2

### Re: Problem 317

I solved the problem.

v6ph1, Your suggestion to use

**5 decimals**in calculations gives

**an incorrect result**. The difference from the correct answer is $1.5681 m^3$. Rounding $\pi$ to 10 or more decimals and using $\pi$ without rounding gives the correct result.

Here's the proof. The solution range is $3,769,911.1843m^3$ $10^7$ (see this post) plus 4 decimal digits. Therefore, we have $10^{11}$ range of significant digits. $\pi$ range is $10^1$. So a result will be not be affected if we round to $\log_{10}{(10^{11}/10^1)} = \log_{10}{10^{10}} = 10$ digits.

The friend key is 1004797_CkgowZqfFi0qQMGOqVjB0RP8htEhkWW2

### Re: Problem 317

I think you misunderstood v6ph1, though the post was somewhat ambiguous. I think the intended suggestion was to use as many decimals in the value of pi as you need such that adding an extra decimal does not change the four decimals in the calculated answer.

_{Jaap's Puzzle Page}

### Re: Problem 317

Hi japp,

My

jaap, here's your comment

Thank you for understanding.

Regards,

Alex.

My

**initial**question in the forum "Clarifications on Project Euler Problems" wasThe answer must be a number, for example $3.14$. v6ph1'sWhat approximate value of $\pi$ should be used?

**initial**answer wasThis is a general answer that can be applied to any constant, for example $e$.You need 4 decimals in your solution. - So if you append the next digit of pi, the solution should not change at all.

jaap, here's your comment

The 317 problem can be solved by using scientific calculator. It may not have enough decimals of $\pi$. It may lead to an incorrect result.I think the intended suggestion was to use as many decimals in the value of pi as you need such that adding an extra decimal does not change the four decimals in the calculated answer.

*Therefore, the value of $\pi$ must be stated in the problem description the same way it stated for $g$.*Thank you for understanding.

Regards,

Alex.

**P.S. I suggest to moderators that problem 317 wording has to be updated by adding value of $\pi$ constant.**The friend key is 1004797_CkgowZqfFi0qQMGOqVjB0RP8htEhkWW2

### Re: Problem 317

My intention was to calculate with as many decimals as possible/needed.

And this could be checked if the next digit of pi would not change the result.

Suggesting a specific precise approximation will gave the others too much hints.

The wording should not change as constants like pi and e have an exactly defined value.

And this could be checked if the next digit of pi would not change the result.

Suggesting a specific precise approximation will gave the others too much hints.

The wording should not change as constants like pi and e have an exactly defined value.

### Re: Problem 317

The problem is asking for the value you'd get by calculating the volume exactly, then rounding it to four decimal places. It doesn't mention $\pi$ at all. Any method of finding that rounded value is acceptable. You might use the exact value of $\pi$, some approximation, or not use $\pi$ at all.

Whichever method you use, you'll need to satisfy yourself that it gives the same result as rounding the exact volume. If you decide to use an approximation to $\pi$, finding a suitable approximation is up to you.

$g$ is a different matter. In the real world, it's a physical value that must be measured. The value you get will depend on the time and place of measurement, the equipment used, and the care and skill of the person taking the measurement. To ensure that everyone solving this problem gets the same answer, we all need to be using the same value. So that has to be given in the problem.

### Re: Problem 317

v6ph1,

**I understand your answer now,**not to give the others to much hints.

The friend key is 1004797_CkgowZqfFi0qQMGOqVjB0RP8htEhkWW2

- RishadanPort
**Posts:**79**Joined:**Mon Jun 10, 2013 7:31 am

### Re: Problem 317

Can someone confirm to me, that the Max X distance, is ~<

*removed by moderator**>, at t = ~xxx,*

with Theta = ~xxx radians

?with Theta = ~xxx radians

?

*Rishada is the gateway to free trade—but the key will cost you.*

### Re: Problem 317

I'm sure I've done this correctly, especially after reading old comments on here. I think I have a precision problem, could someone help me troubleshoot by checking that the 6 decimals after the 4 decimals asked for in the problem statement are 170668? In other words, the solution looks something like X.abcd170668.

Edit: nevermind, I had a mistake copying a number towards the end from my calculator to paper. Solved.

Edit: nevermind, I had a mistake copying a number towards the end from my calculator to paper. Solved.

1691991_rIEOKCNEDBtm7EzRUeWtIZDvhFNxQVp1