Guerinsylvie

Range of a projectile

--Guerinsylvie (talk) 10:46, 8 August 2010 (UTC) : - Good morning, I don't know the rules here. So I discuss the article in this page.

First of all, I say my thought : the redaction, for the present time, is not good. I am going to try to improve it. BUT, I apologize, this subject is very well-known since 1642! Koyré has extensively studied it. My intervention is perhaps very ridiculous ; but we must improve the article : it is my only goal.

• from the beginning, I take the "good" units : without restriction of generality, I take v=g=1.

• First, I am going to explicit surprisingly, in the paragraph about the maximum of the range. It is not surprising !! The algebra was taken in "brute force", and it is not the best manner ! I propose 2 another ways( of course, I have found these because I know the results and the geometry)

• Second, I suggest another manner to present the fall of "graves".

• First : The text says :

$u+{\frac {u{\sqrt {1-u}}}{\sqrt {1-u+C}}}-(1-u)-({\sqrt {1-u}}){\sqrt {1-u+C}}=0$

Which reduces to the surprisingly simple expression:

$u={\frac {C+1}{C+2}}$

I contest the word : surprisingly.It is just because the algebra is not good-driven : the question was :

Prove that :

$d(\theta )={\frac {v\cos \theta }{g}}\left(v\sin \theta +{\sqrt {(v\sin \theta )^{2}+2gy_{0}}}\right)$ has maximum equal to $d_{m}={\sqrt {1+2y}}$ when $1/sin^{2}\,\theta =2+2y~$ (v=g=1).

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• first demonstratio :

LET BE $t(\theta )$ , the fonction defined by the positive root of : $-y=-t^{2}/2+t\cdot \sin(\theta )$

then, $0={\frac {dt}{d\theta }}\cdot [\sin \,\theta -t]+t\cdot [cos\,\theta ]$.

And for the case of d_max, because $x=t\cdot \cos \,\theta$,

• $0={\frac {dt}{d\theta }}\cdot [\cos \,\theta ]+t\cdot [-\sin \,\theta ]$

This manner of writing shows that :

$[sin\theta -t][-sin\theta ]-[cos\theta ][cos\theta ]=0$, therefore,

$t\cdot \sin \,\theta =1~$

therefore x, therefore y , with the results, cqfd.

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• second demonstratio :

I don't want use the derivative because I am too young-pupil ( in France, if you are less than fifteen old, you have not learn the calculus ).

So I write better : I know if I change the function by another function monotone, the value of $\theta _{max}$ is not changing . Idem, if I take a new variable strictly-monotone. So take a glance to these "manipulations" ( of course, it is by "tour de force", don't blame me ! ):

$LETBE\,\,{\sqrt {1+{\frac {2y}{\sin ^{2}\,\theta }}}}=1+X$ , then ${\frac {2y}{\sin ^{2}\,\theta }}=X(2+X)$

and re-write at d with some attention :

$d=[cos\theta ]/[sin\theta ]\cdot (\sin ^{2}\,\theta )[2+X]=[cos\theta ]/[sin\theta ]\cdot ({\frac {2y}{X(2+X)}}[2+X]=[cos\theta ]/[sin\theta ]\cdot {\frac {2y}{X}}$

and "the trick" is played : simplification is made :

$d^{2}=[{\frac {1}{\sin ^{2}\,\theta }}-1]^{2}\cdot {\frac {4y^{2}}{X^{2}}}={\frac {1}{X^{2}}}((2X+X^{2})2y-4y^{2})=2y+1-P$ with $P=-4y/X+4y^{2}/X^{2}+1~$ is positive or null.

the maximum is

$d_{max}^{2}=2y+1$

for :

$2y=X~$

cqfd.

Note : the distance OI = OF+FI = 1/2+(1/2+y) = 1+y , and it is the only one thing to memorize : it contains THE demonstration AND gives THE geometric_figure.

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The victory is bitter : none of these are beautiful : what a lot of algebra !

So, I have the trick of X ; but I known this one because the geometry ! So, why don't take the historical demonstratio of Torricelli, 350 years old !

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Presentation "à la Torricelli"

--Guerinsylvie (talk) 11:18, 8 August 2010 (UTC):- Now, I try to make another presentation of the article : my essential critical thought is : too direct from the sratch, so I make operations and I find. Yes! I have proved that I known how to do.But I have no satisfaction, because I have not taken TIME_TIME_time for discuss the physical things. Of course, the whole part I read is EXACT and firmly logic. BUT...

• So, new try. But before, read the oldest versions : Well, I have read all. It is OK : as the Trajectory of projectile, the level of math is low. So I maintain my project : do the algebra for sixteen people.

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Derivation

• First v=g=1
• Then the figure : Name the points on the figure : O departure, I ground_Impact, S summit of the parabola, H ={0,1/2} ( the vertical maximal point.
• If the projectile is launched vertically, it arrive at H {0, h= 1/2 *Vo^2/g = 1/2}.

The line z= h plays an important role because all the trajectories of same Vo are parabolas of the same directrices : this line !

• Note also on the figure the point F, focus of the parabola : OF = OH = h , et angle {Oz, OF} = 2 (Pi/2 -\theta).
• Let be T, point of Torricelli of the trajectory the intersection of the line OF with the parabola : O,F and T are aligned.

In fact, I want to demonstrate that the point of impact I_max is the point T.

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Then $a$ Name the durée of climbing the arc OS $t_{u}$ , and $t_{f}$ the durée of fall along the arc SI.

We have : $t_{u}=\sin \,\theta$ , and because the durée are like the sqrt of height , $t_{f}=t_{u}\cdot {\sqrt {1+2y/\sin ^{2}\,\theta }}$

So the range is :

$x=\cos \,\theta \cdot (t_{u}+t_{f})=\cos \,\theta \cdot \sin \,\theta \cdot [1+{\sqrt {1+2y/\sin ^{2}\,\theta }}]$

There was no mystery in "my" precedent factorization ! But I don't want to use any algebra : think about this : the trajectory is a parabola, with the focus F, and the directrix z= h . So for all the points M of the trajectory, FM = h- z(M)=1/2-z.

Now, if I want the maximum, no need of algebra, just think : if I want l'abscissa of I max, then I need the focus F on the line OI : that is the KEY-point : stop here ; and look at this argument : because OF = 1/2 and IF = 1/2+y , OI will be max iff OI = 1+y . End !!!

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• After that, you see easyly : the rapport ${\frac {t_{f}}{t_{u}}}=x_{f}/x_{u}=FI/OF=(1/2+y)/(1/2)=1+2y$, by definition of the focus ! therefore look at the X defined before , then X = 2y . End.

¤¤¤ It seems to me that this demonstration by Torricelli has less algebra. (PS: of course, you need the picture).