Projectile Motion Question (Envelope of trajectories) query

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PeroK said:
Yes.
hmm, can't see the algebra for some reason.
 
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erobz said:
I think we are talking about the solutions to different problems. I plotted them, these aren't equivalent expressions.

$$ \sqrt{2gh+\frac{gl^2}{2h}} \neq \sqrt{gh + g \sqrt{h^2 + l^2} } $$
Yes,there are two different problems in this thread.
 
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kuruman said:
Below I am addressing the second question, 3.6.
But 3.6 explicitly asks for a trajectory that “just clears the wall at the top of its parabolic motion”. This to me suggests that they are not asking for the smallest possible speed, but the lowest speed that clears the wall with the apex just above it. This is also consistent with
kuruman said:
the optimization condition yields v0=gh+gl2+h2 which verifies what you got.
not being one of the multiple choice options.
 
I think problem (b) and problem 3.6 from post #1 are equivalent questions. The projectile apex ##h## of each problem is ##\ell## distance away from the thrower, and it will land ##\ell## distance past the wall in both cases.
 
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erobz said:
I think problem (b) and problem 3.6 from post #1 are equivalent questions. The projectile apex ##h## of each problem is ##\ell## distance away from the thrower, and it will land ##\ell## distance past the wall in both cases.
Quite so. But the OP applied a method (zero determinant) used by @laser's prof in a problem that asked for minimum velocity. This is what led to the right hand expression in your post #33.
That the method works for that problem can be seen thus:
Minimising y wrt x in ##ax^2+xy+c=0## produces ##2ax=-y##, whence ##y^2=4ac##.
 
haruspex said:
That the method works for that problem can be seen thus:
Minimising y wrt x in ##ax^2+xy+c=0## produces ##2ax=-y##, whence ##y^2=4ac##.
I don't follow, what is meant by the method works? They yield different expressions for ##v##.

laser said:
Interesting, I just rechecked the example the professor did, and indeed, that question did ask for the "minimum speed"! Here was the exact wording:

1703016628381.png


The method that the OP and @kuruman use solves this ^^ problem, which is equivalent to neither problem in the opening post.
 
erobz said:
I don't follow, what is meant by the method works? They yield different expressions for ##v##.
Avoiding the pronouns, that the determinant method works for the minimum velocity problem. It gives the wrong answer for both questions in post #1.
 
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Orodruin said:
But 3.6 explicitly asks for a trajectory that “just clears the wall at the top of its parabolic motion”. This to me suggests that they are not asking for the smallest possible speed, but the lowest speed that clears the wall with the apex just above it. This is also consistent with

not being one of the multiple choice options.
Yes, they are equivalent. The funny thing is that I initially confirmed one of the choices, but then I got turned around by the phantom problem of optimizing ##v_0##.
erobz said:
@kuruman says they are addressing ##\dots##
Thank you for being considerate. For the record, my preferred pronouns are "he, him, his."
 
I agree with @erobz that there are three problems mentioned here:
1. Problem (b) in post #1.
2. Problem 3.6 in post#1.
3. The professor's problem in post #13.

Small wonder I got confused. To avoid further confusion, can we refer to them respectively as (b), 3.6 and prof? I think we all agree that the first two are equivalent but the third is not for reasons already explained.
 
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kuruman said:
I agree with @erobz that there are three problems mentioned here:
1. Problem (b) in post #1.
2. Problem 3.6 in post#1.
3. The professor's problem in post #13.
Yes.

kuruman said:
Small wonder I got confused. To avoid further confusion, can we refer to them respectively as (b), 3.6 and prof? I think we all agree that the first two are equivalent but the third is not for reasons already explained.
Indeed.