помогите пожалуйста с задачами

konstantin.gundev

Please excuse me for writing in English! I can read Russian to some extent because I am from Bulgaria, but I do not dare to write or speak in Russian. My grammar is just too bad! I have been trying to solve the following problem, but without any success.
ЗАДАЧИ по ФИЗИКЕ
Издание третье, исправленное и дополненное Под редакцией
О. Я. Савченко
3.3.13∗. Снаряд, вылетев из орудия, попал в точку с координатами x по го- ризонтали и y по вертикали. Начальная скорость снаряда v. Найдите: б) границу области возможного попадания снаряда;
Указание. При решении воспользуйтесь тригонометрическим тождеством 1/ cos^2φ = tg^2φ + 1.
The answer is y=v^2/2g-gx^2/2v^2
I have had difficulties understanding the meaning of the question. I think that I should find the relation between the maximum possible x and the maximum possible y at a random initial angle.
y=v^2(1-cos2a)/4g
x=v^2sin2a/g,so
(xg/v^2)^2+(1-4gy/v^2)^2=1 which is apparently different from the given answer by Savchenko.
Could someone give me hint about the meaning of the question?Or may be I understand the question, but my solution is wrong?
Пожалуйста!

student_kiev

Developer писал(а):Source of the post I have had difficulties understanding the meaning of the question. I think that I should find the relation between the maximum possible x and the maximum possible y at a random initial angle.

Well, the maximum possible $$x$$

and the maximum possible $$y$$

won't give you the boundary of the region you're looking for. Just think about: the maximum possible $$y$$

is always the vertex of the parabola for any given $\alpha$ and there's only one $\alpha$ (equal to $\pi/2$ ) for which the maximum possible $$y$$

is actualy on the boundary of the region.
You should find the locus of points (lying on parabolas) that are furtherst from the origin for any given $\alpha$ . To do this, first express $$y$$

as a function of $$x$$

and $\alpha$ : $y(x,\alpha)$ (don't forget the hint given in the problem to express everything in terms of $\tan \alpha$ ). Then define a function $R(x,\alpha) \equiv x^2 + y^2(x,\alpha)$ that represents the distance from the origin to any point on any given parabola specified by $\alpha$ . Now maximize $R(x,\alpha)$ as a function of $$x$$

treating $\alpha$ as a parameter. Finally, find $\tan \alpha$ in terms of $$x$$

and substitute it back into $y(x,\alpha)$ . You'll get $$y(x)$$

which represents the boundary of the region you're looking for.

grigoriy · Сообщение **grigoriy** » 01 ноя 2015, 10:56

По поводу...
Если не изменяет память, то "парабола безопасности" для начальной скорости $$V_0$$

совпадает
c траекторией тела, брошенного горизонтально с той же начальной скоростью с высоты $H=H_{max}(V_0,\,\alpha=\frac{\pi}{2})$

balans · Сообщение **balans** » 03 ноя 2015, 03:40

konstantin.gundev писал(а):Source of the post zadach.split_.zip

"Архив поврежден или имеет неизвестный формат."

konstantin.gundev

student_kiev, grigoriy, спасибо!
I was finally able to understand the Russian of the problem and realized what I am supposed to find.
I am sorry I am not able to attach The photo of the whole solution.
in short: write the equation of y(x,α、v、).Next substitute y with rcosφ and x with rsinφ.find the maximum of of r by differentiating to α.Substitue the the α in the r(φ) equation...

[img]/modules/file/icons/application-pdf.png[/img] convert-jpg-to-pdf.net_2015-11-04_01-33-13.pdf

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