### Re: Problem 317

Posted:

**Wed Jul 24, 2013 1:51 am**Okay thanks

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Posted: **Wed Jul 24, 2013 1:51 am**

Okay thanks

Posted: **Thu Dec 12, 2013 12:21 am**

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

Posted: **Thu Dec 12, 2013 3:05 pm**

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

Posted: **Mon Dec 16, 2013 3:16 pm**

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).

Posted: **Thu Jan 02, 2014 6:54 am**

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.

Posted: **Fri Jul 14, 2017 9:46 pm**

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

Posted: **Fri Jul 14, 2017 11:54 pm**

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.

Posted: **Fri Jul 28, 2017 4:55 pm**

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.

Posted: **Fri Jul 28, 2017 10:54 pm**

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

Posted: **Sat Jul 29, 2017 12:32 am**

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

exact result is $120.3874m$

exact result is $120.3874m$

Posted: **Sat Jul 29, 2017 10:31 pm**

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

Posted: **Sun Jul 30, 2017 1:23 pm**

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 $79.4676m^3$.

Posted: **Tue Aug 01, 2017 9:48 pm**

I solved the problem.

v6ph1, Your suggestion to use

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.

Posted: **Wed Aug 02, 2017 2:20 am**

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.

Posted: **Wed Aug 02, 2017 12:14 pm**

Hi japp,

My**initial **question in the forum "Clarifications on Project Euler Problems" was
**initial ** answer was

jaap, here's your comment

*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.**

My

The answer must be a number, for example $3.14$. v6ph1'sWhat approximate value of $\pi$ should be used?

This 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.

Thank you for understanding.

Regards,

Alex.

Posted: **Wed Aug 02, 2017 8:01 pm**

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.

Posted: **Thu Aug 03, 2017 11:43 am**

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.

Posted: **Thu Aug 03, 2017 11:57 am**

Posted: **Sat Aug 17, 2019 11:14 pm**

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

?

Posted: **Mon Mar 08, 2021 10:35 pm**

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.