Translated 1.3.3-1.3.15; Fixed GUI image generator

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+ <meta name="description" content="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?">
+ <meta name="author" content="Aliaksandr Melnichenka">
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+ <h2>Solutions of Savchenko Problems in Physics</h2>
+ <p class="author">
+ Aliaksandr Melnichenka <br/>
+ October 2023
+ </p>
+ </header>
+
+ <h3 id="back-link"><a href="../../#1.3">$\leftarrow$Back</a></h3>
+
+ <h3> Statement </h3>
+ <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?
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+ The horizontal component of velocity remains unchanged:
+$$ v_{x}(\varphi) = v_{x}(\alpha) = v \cdot \cos{\varphi} $$
+And the horizontal component decreases, depending on time, according to the law:
+$$ v_{y}(t) = vt \sin{\varphi} - gt $$
+From where the angle that the velocity makes with the horizon is determined as:
+$$ \tan{\alpha} = \frac{v_{y}(t)}{v_x} $$
+Or,
+$$ \tan{\alpha} \cdot v \cdot \cos{\varphi} = v \cdot \sin{\varphi} - gt $$
+From where we obtain the required moment of time:
+$$ \fbox{$t= \frac{v}{g}(\sin{\varphi} - \cos{\varphi}\tan{\alpha})$} $$
+
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ $$t= \frac{v}{g}(\sin{\varphi} - \cos{\varphi}\tan{\alpha})$$
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