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