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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> |
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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> |
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| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $1.3.7.$ A ball is launched along a smooth inclined plane at a speed of $v$. How much horizontal distance will it travel before it rolls off the plane? The plane is inclined to the horizon at an angle of $45^\circ$ . The initial velocity of the ball forms an angle of $45^\circ$ with the horizontal edge of the plane. | | $1.3.7.$ A ball is launched along a smooth inclined plane at a speed of $v$. How much horizontal distance will it travel before it rolls off the plane? The plane is inclined to the horizon at an angle of $45^\circ$ . The initial velocity of the ball forms an angle of $45^\circ$ with the horizontal edge of the plane. |
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| <img src="statement.png" | | <img src="statement.png" |
| loading="lazy" width="230" /> | | loading="lazy" width="230" /> |
| <figcaption> | | <figcaption> |
| For problem $1.3.7$ | | For problem $1.3.7$ |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
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| <p> | | <p> |
| </p> | | </p> |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
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| On an inclined plane, the body is acted upon by the component of acceleration due to gravity: | | On an inclined plane, the body is acted upon by the component of acceleration due to gravity: |
| $$ a = g^* = g \cdot \cos{45^{\circ}} $$ | | $$ a = g^* = g \cdot \cos{45^{\circ}} $$ |
| From the formula for flight range obtained in <a href="../1.3.6" target="_blank">1.3.6</a> it follows: | | From the formula for flight range obtained in <a href="../1.3.6" target="_blank">1.3.6</a> it follows: |
| $$ L=\frac{v^2}{g^*}\operatorname{sin}90^{\circ} $$ | | $$ L=\frac{v^2}{g^*}\operatorname{sin}90^{\circ} $$ |
| We substitute $g^*$: | | We substitute $g^*$: |
| $$ \fbox{$L=\frac{v^2 \sqrt{2}}{g}$} $$ | | $$ \fbox{$L=\frac{v^2 \sqrt{2}}{g}$} $$ |
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| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| $$L= \sqrt {2} v^{2} / g$$ | | $$L= \sqrt {2} v^{2} / g$$ |
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