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Mathematics-Online course: Prepcourse Mathematics - Analysis - Functions

Flight Path

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The flight path of an object that is thrown under a certain angle $ \varphi$ with a speed $ v$ can be described by

$\displaystyle y=x\tan \varphi - \frac{g}{2v^2\cos^2 \varphi}x^2 $

with $ g$ being the acceleration of gravity.

\includegraphics[width=7.4cm]{schraeger_wurf_en}

The equation follows from splitting the uniform motion with initial speed $ v$ into $ x$- and $ y$-component as well as the interaction of the accelerated motion of the free fall:

$\displaystyle x(t)$ $\displaystyle =$ $\displaystyle vt\,\cos \varphi \quad \Rightarrow\quad t = \frac{x}{v \cos \varphi}$  
$\displaystyle y(t)$ $\displaystyle =$ $\displaystyle vt\,\sin \varphi - \frac{1}{2}\,gt^2$  
$\displaystyle \Rightarrow\,\,y(x)$ $\displaystyle =$ $\displaystyle \frac{vx\,\sin \varphi}{v\,\cos \varphi} - \frac{gx^2}{2v^2\,\cos^2 \varphi} \,=\, x\left(\tan \varphi -\frac{gx}{2v^2\,\cos^2 \varphi}\right).$  

The throwing range calculates as

$\displaystyle x=\frac{2v^2\cos^2\varphi}{g}\tan\varphi = \frac{v^2}{g} \sin (2\varphi)\,. $

It is maximal for $ \varphi = \pi/4 \,\widehat{=}\, 45^\circ$.
(Authors: Höllig/Hörner/Knesch/Abele)
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  automatically generated 9/18/2007