Translated 1.3.3-1.3.15; Fixed GUI image generator

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+ <title>The gun is fired at an angle \varphi to the horizon. Initial velocity of the projectile v. The ground surface is horizontal. Find: <br>a) the horizontal and vertical projections of velocity as a function of time; <br>b) the dependence of the x and y coordinates on time; <br>c) the trajectory equation, i.e. the dependence of y on x; <br>d) the flight time, the highest altitude and range of the projectile.</title>
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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>
+ $1.3.6.$ The gun is fired at an angle $\varphi$ to the horizon. Initial velocity of the projectile $v.$ The ground surface is horizontal. Find: <br>a) the horizontal and vertical projections of velocity as a function of time; <br>b) the dependence of the $x$ and $y$ coordinates on time; <br>c) the trajectory equation, i.e. the dependence of $y$ on $x$; <br>d) the flight time, the highest altitude and range of the projectile.
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+ a) Since the external force (gravitational force) acts only in the vertical direction, the horizontal component of the velocity $v_x$ remains unchanged
+$$ v_x = v \cdot \cos{\varphi}\;(1) $$
+And the vertical component depends as a function of time
+$$ v_y(t) = v \cdot \sin{\varphi} - gt\;(2) $$
+b) Based on the obtained expressions for the speed $(1)$ and $(2)$, we obtain the dependence of the horizontal and vertical coordinates on time, respectively:
+$$ y(t) = vt \cdot \sin{\varphi} - \frac{gt^2}{2}\;(3) $$
+
+$$ x(t) = vt \cdot \cos{\varphi}\;(4) $$
+c) From $(4)$ we express $t$:
+$$ t = \frac{x}{v \cdot \cos{\varphi}} \;(5) $$
+And we substitute in $(3)$
+$$ y(x) = x \tan \varphi-\frac{gx^2}{2v^2\cos^2\varphi}\;(6) $$
+d) The maximum height is achieved when the velocity is directed horizontally $(v_y = 0)$:
+$$ t_0 = \frac{v\sin\varphi}{g}\;(7) $$
+Substituting $(7)$ into $(3)$:
+$$ H = \frac{v^2}{2g}\operatorname{sin}^2\varphi\;(8) $$
+From the symmetry of the parabola, the time after which the body will again be on the ground$(y = 0)$, we find as:
+$$ T = 2t_0 = \frac{2v\sin\varphi}{g}\;(9) $$
+Substituting $(9)$ into $(4)$:
+$$ L=\frac{v^2}{g}\operatorname{sin}2\varphi\;(10) $$
+
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ a) $v_x= v\cos\varphi, v_y = v\sin\varphi-gt$<br>
+b) $x = (v\cos\varphi)t, y = (v\sin{\varphi})t - gt^2/2$<br>
+c) $y = x \tan \varphi-\frac{gx^2}{2v^2\cos^2\varphi} =x\operatorname{tg}\varphi-\frac{gx^2}{2v^2}(\operatorname{tg}^2\varphi+1)$<br>
+d) $T=\frac{2v}{g}\operatorname{sin}\varphi, H=\frac{v^2}{2g}\operatorname{sin}^2\varphi, L=\frac{v^2}{g}\operatorname{sin}2\varphi$<br>
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