Drawing Blanks

Premature Optimization is a Prerequisite for Success

45 degrees proof

with 2 comments

Everybody knows that you should launch things at a 45 degree angle in order to maximize the distance. For some reason all proofs of this fact that I see on the web require either the knowledge of trigonometric identities or differentiation, or even both. In fact, all that is needed for the proof is Pythagoras theorem and some kindergarten algebra.

First, we do the usual physics. We say that the velocity is the vector sum of the horizontal and vertical components: V = Vx + Vy. We say that Vx is constant because no forces act in the horizontal direction. So the distance traveled equals Vx times the time traveled. And Vy at first decreases at the constant rate g, and then increases at the same rate. So the time until the fall is 2*Vy/g. Hence the distance is 2*Vx*Vy/g.

We want to maximize Vx*Vy keeping in mind that V is fixed. If you draw a rectangle with the sides Vx and Vy, you readily see that V is its diagonal and Vx*Vy is its area. So out of all rectangles having the given diagonal, we want to find one with the largest area. Obviously, it is the square. But we want to prove this. There is a number of ways to do that without using cosines and derivatives. There are probably some nice geometric demonstrations, but I’ll resort to algebra.

45

Let a (=Vx) and b (=Vy) are the sides of our rectangle and d (=|V|) is the diagonal. Let e=a-b. Then a=b+e. So the area S=ab=b^2+be. The diagonal squared d^2 = a^2+b^2 = (b+e)^2 + b^2 = 2(b^2+be) + e^2. But b^2+be=S. So, S=(d^2-e^2)/2. S is maximal when e=0, or a=b.

Brought to you by Fret of Ivy (an anagram of Forty Five, via Xworder)

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Written by bbzippo

01/14/2010 at 6:36 am

Posted in math

2 Responses

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  1. […] a comment » In a previous post https://bbzippo.wordpress.com/2010/01/14/45-degrees-proof/ we saw that the 45 degree launch angle results in the maximum distance, because out of all […]


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