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| <h2>Solutions of Savchenko Problems in Physics</h2> | | <h2>Solutions of Savchenko Problems in Physics</h2> |
| <p class="author"> | | <p class="author"> |
| Aliaksandr Melnichenka <br/> | | Aliaksandr Melnichenka <br/> |
| October 2023 | | October 2023 |
| </p> | | </p> |
| </header> | | </header> |
| | | |
| <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> |
| $1.3.8.$ A mortar is fired at objects located on the mountainside. At what distance from the mortar will the mines fall if their initial velocity is $v$, the angle of inclination of the mountain is $\alpha$, and the angle of fire relative to the horizon is $\beta$? | | $1.3.8.$ A mortar is fired at objects located on the mountainside. At what distance from the mortar will the mines fall if their initial velocity is $v$, the angle of inclination of the mountain is $\alpha$, and the angle of fire relative to the horizon is $\beta$? |
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| <img src="statement.png" | | <img src="statement.png" |
| loading="lazy" width="200" /> | | loading="lazy" width="200" /> |
| <figcaption> | | <figcaption> |
| For problem $1.3.8$ | | For problem $1.3.8$ |
| </figcaption> | | </figcaption> |
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| <p> | | <p> |
| </p> | | </p> |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| The equations of motion of the projectile can be written as follows: | | The equations of motion of the projectile can be written as follows: |
| $$ {x}={v}_{0}{t}\cos{\beta} $$ | | $$ {x}={v}_{0}{t}\cos{\beta} $$ |
| | | |
| $$ {y}={v}_{0}{t}\sin{\beta}-\frac{{g}{t}^{2}}{2} $$ | | $$ {y}={v}_{0}{t}\sin{\beta}-\frac{{g}{t}^{2}}{2} $$ |
| Substitute into the equations the coordinates of the target $x = L; \;y = L \tan \alpha$ | | Substitute into the equations the coordinates of the target $x = L; \;y = L \tan \alpha$ |
| $$ {L=v_{0}t\cos\beta} $$ | | $$ {L=v_{0}t\cos\beta} $$ |
| | | |
| $$ Ltg\alpha=v_{0}t\sin\beta-\frac{gt^{2}}{2} $$ | | $$ Ltg\alpha=v_{0}t\sin\beta-\frac{gt^{2}}{2} $$ |
| Let us express time from the first equation of the last system of equations and substitute its value into the second equation | | Let us express time from the first equation of the last system of equations and substitute its value into the second equation |
| $$ {t=\frac{L}{v_{0}\cos\beta}} $$ | | $$ {t=\frac{L}{v_{0}\cos\beta}} $$ |
| | | |
| $$ {Ltg\alpha=v_{0}\frac{L}{v_{0}\cos\beta}sin\beta-\frac{g}{2}\frac{L^{2}}{v_{0}^{2}\cos^{2}\beta} } $$ | | $$ {Ltg\alpha=v_{0}\frac{L}{v_{0}\cos\beta}sin\beta-\frac{g}{2}\frac{L^{2}}{v_{0}^{2}\cos^{2}\beta} } $$ |
| Where | | Where |
| $$ {v}_{0}=\sqrt{\frac{{gL}\cos\alpha}{2\cos\beta\sin(\beta-\alpha)}} $$ | | $$ {v}_{0}=\sqrt{\frac{{gL}\cos\alpha}{2\cos\beta\sin(\beta-\alpha)}} $$ |
| We express $L$, | | We express $L$, |
| $$ L = \frac{ 2\cos\beta\sin(\beta-\alpha)\cdot v^2_0}{g\cos\alpha} $$ | | $$ L = \frac{ 2\cos\beta\sin(\beta-\alpha)\cdot v^2_0}{g\cos\alpha} $$ |
| And we find the flight range along the wall: | | And we find the flight range along the wall: |
| $$ l = \frac{L}{\cos \alpha} $$ | | $$ l = \frac{L}{\cos \alpha} $$ |
| | | |
| $$ \fbox{$l = \frac{ 2v^2_0}{g} \frac{ \cos\beta\sin(\beta-\alpha)}{\cos^2\alpha}$} $$ | | $$ \fbox{$l = \frac{ 2v^2_0}{g} \frac{ \cos\beta\sin(\beta-\alpha)}{\cos^2\alpha}$} $$ |
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| </p> | | </p> |
| | | |
| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| $$L=\frac{2v^2}g\frac{\cos^2\beta}{\cos\alpha}(\text{tg}\beta-\text{tg}\alpha)$$ | | $$L=\frac{2v^2}g\frac{\cos^2\beta}{\cos\alpha}(\text{tg}\beta-\text{tg}\alpha)$$ |
| </p> | | </p> |
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