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| <span><img src="../../img/book.png"></span><span>Savchenko Solutions</span> | | <span><img src="../../img/book.png"></span><span>Savchenko Solutions</span> |
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| <p class="author"> | | <p class="author"> |
| Solutions of Savchenko Problems in Physics <br> | | Solutions of Savchenko Problems in Physics <br> |
| <i><b>knowledge must be free</b></i> | | <i><b>knowledge must be free</b></i> |
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| <h3 id="back-link"><a href="../../#1.3">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#1.3">$\leftarrow$Back</a></h3> |
| | | |
| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <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? | | $1.3.10.$ 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? |
| </p> | | </p> |
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| <figure> | | <figure> |
| <img src="https://savchenkosolutions.com/1/1.3.10/statement.png" | | <img src="https://savchenkosolutions.com/1/1.3.10/statement.png" |
| loading="lazy" width="250" /> | | loading="lazy" width="250" /> |
| <figcaption> | | <figcaption> |
| For problem $1.3.10$ | | For problem $1.3.10$ |
| </figcaption> | | </figcaption> |
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| <p> | | <p> |
| </p> | | </p> |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
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| loading="lazy" width="300" /> | | loading="lazy" width="300" /> |
| <figcaption> | | <figcaption> |
| Direction of duck's flying | | Direction of duck's flying |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| Let's consider this equation of motion: | | Let's consider this equation of motion: |
| $$ y = h = \nu \sin \alpha - \frac{gt^{2}}{2} $$ | | $$ y = h = \nu \sin \alpha - \frac{gt^{2}}{2} $$ |
| It is known that: | | It is known that: |
| $$ \left\{\begin{matrix} | | $$ \left\{\begin{matrix} |
| S_{yt}=ut \\ | | S_{yt}=ut \\ |
| S_{o}=\nu t \cos \alpha | | S_{o}=\nu t \cos \alpha |
| \end{matrix}\right. $$ | | \end{matrix}\right. $$ |
| For angle $\alpha$: | | For angle $\alpha$: |
| $$ \tan \alpha = \frac{h}{S_{o} - S_{yt}} = \frac{h}{t(\nu \cos \alpha - u)} $$ | | $$ \tan \alpha = \frac{h}{S_{o} - S_{yt}} = \frac{h}{t(\nu \cos \alpha - u)} $$ |
| We substitute the value of the angle $\alpha$ into the equation for the height: | | We substitute the value of the angle $\alpha$ into the equation for the height: |
| $$ h = t (\nu \cos \alpha - u) \tan \alpha = \nu t \sin \alpha - \frac{gt^{2}}{2} $$ | | $$ h = t (\nu \cos \alpha - u) \tan \alpha = \nu t \sin \alpha - \frac{gt^{2}}{2} $$ |
| From this equation we can express time $t$: | | From this equation we can express time $t$: |
| $$ t = \frac{2u \tan \alpha}{g} $$ | | $$ t = \frac{2u \tan \alpha}{g} $$ |
| We substitute the time value $t$ back into the equation for height: | | We substitute the time value $t$ back into the equation for height: |
| $$ \fbox{$h = \frac{2u \tan^{2} \alpha}{g} (\nu \cos \alpha - u)$} $$ | | $$ \fbox{$h = \frac{2u \tan^{2} \alpha}{g} (\nu \cos \alpha - u)$} $$ |
