Tricky Projectiles M2 Question

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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?

Maximum value of x as theta varies, imples differentiate wrt theta, and set equal to 0.

Then

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

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?

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?

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

did you use $x=u\cos{\theta}t$

Since wouldn't that require me expressing cos in terms of tan?

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!