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An athlete executing a long jump leaves the ground at a \ang{28.0} angle and travels $\SI{7.80}{m}$. \begin{abcliste} \abc What was the takeoff speed? \abc If this speed were increased by just 5.0\%, how much longer would the jump be? \end{abcliste}
\begin{abcliste} \abc If we solve the formula for the maximum width, i.e. \begin{align} s_x &= v_x \cdot t = v_x \cdot 2\frac{v_y}{g}\\ &= v_0\cos\alpha \cdot 2\frac{v_0\sin\alpha}{g} = \frac{v_0^2}{g} 2\cdot \sin\alpha\cos\alpha\\ &= \frac{v_0^2}{g} \cdot \sin 2\alpha \end{align} for the speed $v_0$, we end up with \begin{align} v_0 &= \sqrt{\frac{gs_x}{\sin2\alpha}}\\ &= \SI{9.61}{\meter\per\second} \end{align} \abc Increasing the speed by 5\% means a takeoff by $\SI{10.09}{\mps}$. This would result in a width of \begin{align} s_x &= \frac{(\SI{10.09}{\meter\per\second})^2}{\SI{9.81}{\meter\per\second\squared}} \cdot \sin(2\cdot \ang{28})\\ &= \SI{8.6}{m}, \end{align} i.e. the jump would be $\SI{80}{cm}$ longer. \end{abcliste}
09:17, 31. Aug. 2018  lsg verbessert  Urs Zellweger (urs)  Current Version 
23:09, 13. June 2018  lsg, si  Urs Zellweger (urs)  Compare with Current 