Tricky Projectiles M2 Question Watch

SamKeene
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#1
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Here it is, in it's full glory:



I'm able to derive the equation, but am unsure going about showing the next steps. I've considered differentiating the equation, but in terms of x or theta?
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ghostwalker
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(Original post by SamKeene)
...
Maximum value of x as theta varies, imples differentiate wrt theta, and set equal to 0.

Then
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You'll get x in terms of theta, and sub back into your original equation to get x in terms of a,b.
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TeeEm
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(Original post by SamKeene)
Here it is, in it's full glory:



I'm able to derive the equation, but am unsure going about showing the next steps. I've considered differentiating the equation, but in terms of x or theta?
beautiful question

follow ghostwalker's suggestion.

I just did a neat solution for my own resources which you are welcome to it should you need to check your answer or are still stuck
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SamKeene
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(Original post by ghostwalker)
Maximum value of x as theta varies, imples differentiate wrt theta, and set equal to 0.

Then
Spoiler:
Show

You'll get x in terms of theta, and sub back into your original equation to get x in terms of a,b.
(Original post by TeeEm)
beautiful question

follow ghostwalker's suggestion.

I just did a neat solution for my own resources which you are welcome to it should you need to check your answer or are still stuck
Hi guys, thanks for both of your replies. And I'm happy to introduce to you Tee to the question, it is a great one. Kinda reminds me of STEP!

Alright so I differentiated the equation with respect to x, and in the end I ended up with:

x=\frac{a}{\tan{\theta}}

Subbing this into the original equation I ended up with:

a^2 + 2ab = x^2

a^2 + 2ab = \frac{a^2}{\tan^2{\theta}}

Which gave me both results!

Now before I try to tackle the next bit.

Why didn't I use implicit differentiation? Is it because x is dependent on theta and thus can be treated as a constant in the differentiation?
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ghostwalker
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(Original post by SamKeene)
Why didn't I use implicit differentiation?
I don't know, I used implicit differentiation, and chain rule, and product rule.

Is it because x is dependent on theta and thus can be treated as a constant in the differentiation?
If x is dependent on theta, as it is, you can't treat it as a constant.

Unless I've misunderstood something, you were just lucky.
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SamKeene
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(Original post by ghostwalker)
I don't know, I used implicit differentiation, and chain rule, and product rule.



If x is dependent on theta, as it is, you can't treat it as a constant.

Unless I've misunderstood something, you were just lucky.
Alright, I did it implicitly and got the same result.

I'm having trouble setting up the inequality for my next bit. I was thinking.. well if the guy hears the shell before it hits him at Q, then the time it takes the sound at speed\sqrt{cg} to reach him along PQ, must be less than the time it takes the shell to at initial speed \sqrt{ag} to reach Q.

calculating PQ as a+b the closest I got was:

\frac{a+b}{\sqrt{cg}}<\frac{???}  {\sqrt{ag}}

trying ??? as \sqrt{a(a+2b)} didn't work..

any pointers?
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WishingChaff
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Calculate the time taken for the particle to reach your set value of x. As a hint try using a SUVAT equation.

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 \sqrt{a(a+2b)} = \sqrt{ag} \sqrt{\frac{a+2b}{2(a+b)}} t


(Original post by SamKeene)
Alright, I did it implicitly and got the same result.

I'm having trouble setting up the inequality for my next bit. I was thinking.. well if the guy hears the shell before it hits him at Q, then the time it takes the sound at speed\sqrt{cg} to reach him along PQ, must be less than the time it takes the shell to at initial speed \sqrt{ag} to reach Q.

calculating PQ as a+b the closest I got was:

\frac{a+b}{\sqrt{cg}}<\frac{???}  {\sqrt{ag}}

trying ??? as \sqrt{a(a+2b)} didn't work..

any pointers?
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SamKeene
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(Original post by WishingChaff)
Calculate the time taken for the particle to reach your set value of x. As a hint try using a SUVAT equation.

Spoiler:
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 \sqrt{a(a+2b)} = \sqrt{ag} \sqrt{\frac{a+2b}{2(a+b)}} t
did you use x=u\cos{\theta}t

Since wouldn't that require me expressing cos in terms of tan?
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WishingChaff
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Yes, I used that. However, you know the value of tan from the previous part. So, constructing a right angle triangle you should be able to find an expression for cos in terms of a and b.

(Original post by SamKeene)
did you use x=u\cos{\theta}t

Since wouldn't that require me expressing cos in terms of tan?
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SamKeene
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(Original post by WishingChaff)
Yes, I used that. However, you know the value of tan from the previous part. So, constructing a right angle triangle you should be able to find an expression for cos in terms of a and b.
Wow I got it.

Looks like this was one of those questions where one little tick of the mind was preventing me from doing it, in this case it was drawing right angle triangle to find the other trig rations.. god damn that's C2 stuff.

At least I got the left part of my original inequality correct :P

Thanks for the help!
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WishingChaff
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No problem, happy to help.

(Original post by SamKeene)
Wow I got it.

Looks like this was one of those questions where one little tick of the mind was preventing me from doing it, in this case it was drawing right angle triangle to find the other trig rations.. god damn that's C2 stuff.

At least I got the left part of my original inequality correct :P

Thanks for the help!
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