Solution

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