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William Tell must split the apple atop his son's head from a distance of $\SI{27}{m}$. When William aims directly at the apple, the arrow is horizontal. At what angle must he aim it to hit the apple if the arrow travels at a speed of $\SI{35}{\meter\per\second}$?
The question we have to answer is: At what angle is the maximum width $\SI{27}{m}$, concerning $\SI{35}{\meter\per\second}$? As in every ballisticexercise, \emph{the vertical motion determines the time of the motion}. The arrow will travel for \begin{align} t &= t_\uparrow + t_\downarrow = 2 t_\uparrow\\ &= 2\frac{v_y}{g} = 2 \frac{v_0\sin\alpha}{g} \end{align} Knowing that, the width the arrow reaches at any angle $\alpha$ is: \begin{align} s_x &= v_x \cdot t\\ &= v_0\cos\alpha \cdot 2\frac{v_0\sin\alpha}{g} = \frac{v_0^2}{g}\sin(2\alpha) \end{align} (Here we used $\sin\alpha\cos\alpha = \frac12 \sin 2\alpha$.) We can solve this last equation for the angle $\alpha$: \begin{align} \alpha &= \frac{1}{2} \cdot \arcsin\left( \frac{gs_x}{v_0^2} \right)\\ &= \ang{6.24} \end{align}
09:11, 31. Aug. 2018  emph  Urs Zellweger (urs)  Current Version 
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