Updated greek laters @ latex compiling

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<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
−$$ v_x = v \cdot \cos{\varphi}\;(1) $$
+$$ v_x = v \cdot \cos{\varphi}\quad(1) $$
And the vertical component depends as a function of time
−$$ v_y(t) = v \cdot \sin{\varphi} - gt\;(2) $$
+$$ v_y(t) = v \cdot \sin{\varphi} - gt\quad(2) $$
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:
−$$ y(t) = vt \cdot \sin{\varphi} - \frac{gt^2}{2}\;(3) $$
+$$ y(t) = vt \cdot \sin{\varphi} - \frac{gt^2}{2}\quad(3) $$
−$$ x(t) = vt \cdot \cos{\varphi}\;(4) $$
+$$ x(t) = vt \cdot \cos{\varphi}\quad(4) $$
c) From $(4)$ we express $t$:
−$$ t = \frac{x}{v \cdot \cos{\varphi}} \;(5) $$
+$$ t = \frac{x}{v \cdot \cos{\varphi}} \quad(5) $$
And we substitute in $(3)$
−$$ y(x) = x \tan \varphi-\frac{gx^2}{2v^2\cos^2\varphi}\;(6) $$
+$$ y(x) = x \tan \varphi-\frac{gx^2}{2v^2\cos^2 \varphi}\quad(6) $$
d) The maximum height is achieved when the velocity is directed horizontally $(v_y = 0)$:
−$$ t_0 = \frac{v\sin\varphi}{g}\;(7) $$
+$$ t_0 = \frac{v\sin\varphi}{g}\quad(7) $$
Substituting $(7)$ into $(3)$:
−$$ H = \frac{v^2}{2g}\operatorname{sin}^2\varphi\;(8) $$
+$$ H = \frac{v^2}{2g}\operatorname{sin}^2\varphi\quad(8) $$
From the symmetry of the parabola, the time after which the body will again be on the ground$(y = 0)$, we find as:
−$$ T = 2t_0 = \frac{2v\sin\varphi}{g}\;(9) $$
+$$ T = 2t_0 = \frac{2v\sin\varphi}{g}\quad(9) $$
Substituting $(9)$ into $(4)$:
−$$ L=\frac{v^2}{g}\operatorname{sin}2\varphi\;(10) $$
+$$ L=\frac{v^2}{g}\operatorname{sin}2\varphi\quad(10) $$
</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>
+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>
</p>
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