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| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $1.3.5.$ A stone is thrown at a velocity $v$ at an angle $\varphi$ to the horizon. After what time will the velocity be at the angle $\alpha$ with the horizon? | | $1.3.5.$ A stone is thrown at a velocity $v$ at an angle $\varphi$ to the horizon. After what time will the velocity be at the angle $\alpha$ with the horizon? |
| </p> | | </p> |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| The horizontal component of velocity remains unchanged: | | The horizontal component of velocity remains unchanged: |
| $$ v_{x}(\varphi) = v_{x}(\alpha) = v \cdot \cos{\varphi} $$ | | $$ v_{x}(\varphi) = v_{x}(\alpha) = v \cdot \cos{\varphi} $$ |
| And the horizontal component decreases, depending on time, according to the law: | | And the horizontal component decreases, depending on time, according to the law: |
| $$ v_{y}(t) = vt \sin{\varphi} - gt $$ | | $$ v_{y}(t) = vt \sin{\varphi} - gt $$ |
| From where the angle that the velocity makes with the horizon is determined as: | | From where the angle that the velocity makes with the horizon is determined as: |
| $$ \tan{\alpha} = \frac{v_{y}(t)}{v_x} $$ | | $$ \tan{\alpha} = \frac{v_{y}(t)}{v_x} $$ |
| Or, | | Or, |
| $$ \tan{\alpha} \cdot v \cdot \cos{\varphi} = v \cdot \sin{\varphi} - gt $$ | | $$ \tan{\alpha} \cdot v \cdot \cos{\varphi} = v \cdot \sin{\varphi} - gt $$ |
| From where we obtain the required moment of time: | | From where we obtain the required moment of time: |
| $$ \fbox{$t= \frac{v}{g}(\sin{\varphi} - \cos{\varphi}\tan{\alpha})$} $$ | | $$ \fbox{$t= \frac{v}{g}(\sin{\varphi} - \cos{\varphi}\tan{\alpha})$} $$ |
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| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| $$t= \frac{v}{g}(\sin{\varphi} - \cos{\varphi}\tan{\alpha})$$ | | $$t= \frac{v}{g}(\sin{\varphi} - \cos{\varphi}\tan{\alpha})$$ |
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