The duck was flying in a horizontal straight line with a constant speed u. An inexperienced "hunter" threw a stone at it, and the throw was made without pre-emption, i.e. at the time of the throw, the speed of the stone v was directed just at the duck at an angle \alpha to the horizon. At what height did the duck fly, if the rock still hit it?

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Solutions of Savchenko Problems in Physics
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    <h3> Statement </h3>
    <p>
        $1.3.10.$ The duck was flying in a horizontal straight line with a constant speed $u$. An inexperienced "hunter" threw a stone at it, and the throw was made without pre-emption, i.e. at the time of the throw, the speed of the stone $v$ was directed just at the duck at an angle $\alpha$ to the horizon. At what height did the duck fly, if the rock still hit it?

    <h3>Solution</h3>
    <p>

Let's consider this equation of motion: $$y = h = \nu \sin \alpha - \frac{gt^{2}}{2}$$ It is known that: $$\left\{\begin{matrix} S_{yt}=ut \\ S_{o}=\nu t \cos \alpha \end{matrix}\right.$$ For angle $\alpha$: $$\tan \alpha = \frac{h}{S_{o} - S_{yt}} = \frac{h}{t(\nu \cos \alpha - u)}$$ We substitute the value of the angle $\alpha$ into the equation for the height: $$h = t (\nu \cos \alpha - u) \tan \alpha = \nu t \sin \alpha - \frac{gt^{2}}{2}$$ From this equation we can express time $t$: $$t = \frac{2u \tan \alpha}{g}$$ We substitute the time value $t$ back into the equation for height: $$\fbox{$h = \frac{2u \tan^{2} \alpha}{g} (\nu \cos \alpha - u)$}$$

    <h4>Answer</h4>
    <p>
        $$h=\frac{2utg^{2}\alpha}{g}(\nu \cos\alpha -u)$$
    </p>

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